Vector Algebra

Given the vectors $\mathbf{f} = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 3 \hat{\mathbf{k}}$ and $\mathbf{g} = 2 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + a \hat{\mathbf{k}}$, we know that the sine of the angle $\theta$ between them is $\sin \theta = \sqrt{\frac{24}{28}}$.

Since $\sin^2 \theta + \cos^2 \theta = 1$, we can find $\cos \theta$ as follows:

$ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{\frac{1}{7}} $

The cosine of the angle $\theta$ is also given by:

$ \cos \theta = \frac{\hat{\mathbf{f}} \cdot \hat{\mathbf{g}}}{|\hat{\mathbf{f}}| \cdot |\hat{\mathbf{g}}|} $

Calculating the dot product:

$ \hat{\mathbf{f}} \cdot \hat{\mathbf{g}} = (1 \cdot 2) + (2 \cdot -3) + (-3 \cdot a) = 2 - 6 - 3a = -4 - 3a $

Calculating the magnitudes:

$ |\hat{\mathbf{f}}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14} $

$ |\hat{\mathbf{g}}| = \sqrt{2^2 + (-3)^2 + a^2} = \sqrt{13 + a^2} $

Thus, we have:

$ \frac{1}{\sqrt{7}} = \frac{-4 - 3a}{\sqrt{14} \cdot \sqrt{13 + a^2}} $

Simplifying, we find:

$ 2(13 + a^2) = (4 + 3a)^2 $

Expanding and rearranging gives:

$ 26 + 2a^2 = 16 + 9a^2 + 24a $

This simplifies to the quadratic equation:

$ 7a^2 + 24a - 10 = 0 $

Solving for $a$ using the quadratic formula:

$ a = \frac{-24 \pm \sqrt{24^2 - 4 \times 7 \times (-10)}}{2 \times 7} $

$ a = \frac{-24 \pm \sqrt{576 + 280}}{14} $

$ a = \frac{-24 \pm \sqrt{856}}{14} $

Finally, calculating $7a^2 + 24a$:

Substitute back into the quadratic result:

$ 7a^2 + 24a = 10 $

To determine the relationship between the points $A$, $B$, and $C$ given their position vectors:

$ \mathbf{r}_1 = \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} $ for point $A$

$ \mathbf{r}_2 = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - \hat{\mathbf{k}} $ for point $B$

$ \mathbf{r}_3 = -3 \hat{\mathbf{i}} - \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $ for point $C$

We can calculate the distances between these points:

Distance between $A$ and $B$:

$ \begin{aligned} |\mathbf{A B}| &= \left|\mathbf{r}_2 - \mathbf{r}_1\right| \\ &= |(2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - \hat{\mathbf{k}}) - (\hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}})| \\ &= |\hat{\mathbf{i}} + 5 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}}| \\ &= \sqrt{1^2 + 5^2 + 4^2} \\ &= \sqrt{1 + 25 + 16} \\ &= \sqrt{42} \end{aligned} $

Distance between $B$ and $C$:

$ \begin{aligned} |\mathbf{B C}| &= \left|\mathbf{r}_3 - \mathbf{r}_2\right| \\ &= |(-3 \hat{\mathbf{i}} - \hat{\mathbf{j}} - 2 \hat{\mathbf{k}}) - (2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - \hat{\mathbf{k}})| \\ &= |-5 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - \hat{\mathbf{k}}| \\ &= \sqrt{5^2 + 4^2 + 1^2} \\ &= \sqrt{25 + 16 + 1} \\ &= \sqrt{42} \end{aligned} $

Distance between $C$ and $A$:

Since

$ \mathbf{C A} = \mathbf{r}_1 - \mathbf{r}_3 = (\hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) - (-3 \hat{\mathbf{i}} - \hat{\mathbf{j}} - 2 \hat{\mathbf{k}}) $

you find

$ |\mathbf{C A}| = \sqrt{42}. $

All three distances are equal,

$ |\mathbf{AB}| = |\mathbf{BC}| = |\mathbf{CA}| = \sqrt{42} $

Thus, points $A$, $B$, and $C$ form an equilateral triangle because all three sides are of equal length.

Given, the equation,

$ \begin{aligned} & 2 a+3 b+5 c-10 d=0 \Rightarrow 5 e=10 d-2 a-3 b \\ & c=2 d-\frac{2}{5} a-\frac{3 b}{5} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, consider the points $\mathbf{a}$ and $\mathbf{b}$ with the line joining them divided by point $\mathbf{c}$ and $\mathbf{d}$ the position vector of any point dividing $\mathbf{a}$ and $\mathbf{b}$ internally in the ratio $\mathrm{m}: \mathrm{n}$ given by

$ P=\frac{m \mathbf{b}+n \mathbf{a}}{m+n} $

Similarly, let the point $\mathbf{c}$ dividing the line joining $\mathbf{a}$ and $\mathbf{b}$ in the ratio $K: 1$.

$ \mathbf{c}=\frac{K \mathbf{b}+\mathrm{a}}{K+1} \Rightarrow \frac{K \mathbf{b}+\mathrm{a}}{K+1}=2 \mathrm{~d}-\frac{2}{5} \mathrm{a}-\frac{3}{5} \mathrm{~b} $

$ \Rightarrow K \mathbf{b}+\mathbf{a}=(K+1)\left(2 \mathbf{d}-\frac{2}{5} \mathbf{a}-\frac{3}{5} \mathbf{b}\right) $

$ \begin{aligned} & \Rightarrow K \mathbf{b}+\mathbf{a}=2(K+1) \mathbf{d}-\frac{2(K+1)}{5} \mathbf{a}-\frac{3(K+1)}{5} \mathbf{b} \\ & \Rightarrow\left(K+\frac{3(K+1)}{5}\right) \mathbf{b}+\left(1+\frac{2(K+1)}{5}\right) \mathbf{a}=\frac{3(K+1)}{5} \mathbf{d} \\ & \Rightarrow\left(\frac{8 K+3}{5}\right) \mathbf{b}+\left(\frac{7+2 K}{5}\right) \mathbf{a}=3(K+1) \mathbf{d} \end{aligned} $

On comparing the coefficients and solve for $K$, we get

$ K+1=\frac{5}{2} $

Thus, $\frac{K}{1}$ given the ratio in which the point C divides a and b . So, $K: 1=3: 2$

Given three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ with magnitudes $ |\mathbf{a}| = 5 $, $ |\mathbf{b}| = 8 $, and $ |\mathbf{c}| = 11 $, and given that they satisfy the equation $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$, we need to determine the angle between vectors $\mathbf{a}$ and $\mathbf{b}$.

First, from the equation $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$, we can derive:

$ \mathbf{a} + \mathbf{b} = -\mathbf{c} $

By squaring both sides, we have:

$ (\mathbf{a} + \mathbf{b})^2 = (-\mathbf{c})^2 $

This can be expanded to:

$ \mathbf{a}^2 + \mathbf{b}^2 + 2 \mathbf{a} \cdot \mathbf{b} = \mathbf{c}^2 $

Substituting the magnitudes of the vectors, we get:

$ |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2 \mathbf{a} \cdot \mathbf{b} = |\mathbf{c}|^2 $

$ 5^2 + 8^2 + 2 \mathbf{a} \cdot \mathbf{b} = 11^2 $

$ 25 + 64 + 2 \mathbf{a} \cdot \mathbf{b} = 121 $

Solving for $2 \mathbf{a} \cdot \mathbf{b}$, we find:

$ 2 \mathbf{a} \cdot \mathbf{b} = 121 - 89 = 32 $

$ \mathbf{a} \cdot \mathbf{b} = \frac{32}{2} = 16 $

The cosine of the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is given by:

$ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{16}{5 \times 8} = \frac{2}{5} $

Thus, the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is:

$ \theta = \cos^{-1}\left(\frac{2}{5}\right) $

$ \begin{aligned} & \mathbf{a}=\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \mathbf{c}=3 \mathbf{i}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}},|\mathbf{a}|=\sqrt{14} \end{aligned} $

Since, $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are linearly dependent the determinant of the matrix formed by these vector must be zero.

$ \begin{aligned} & {[a b c] }=0 \\ &\left|\begin{array}{ccc} \alpha & \beta & 3 \\ 0 & 1 & 2 \\ 3 & 2 & 1 \end{array}\right|=0 \\ & \alpha(1-4)-\beta(0-6)+3(0-3)=0 \\ &-3 \alpha+6 \beta-9=0 \\ &-\alpha+2 \beta=3 \Rightarrow \alpha=2 \beta-3 \\ &|\alpha|=\sqrt{14} \\ & \alpha^2+\beta^2+9=14 \\ & \alpha^2+\beta^2=14-9 \\ & \alpha^2+\beta^2=5 \\ &(2 \beta-3)^2+\beta^2=5 \\ & 4 \beta^2+9-12 \beta+\beta^2=5 \\ & 5 \beta^2-12 \beta+4=0 \\ & 5 \beta^2-(10+2) \beta+4=0 \\ & 5 \beta^2-10 \beta-2 \beta+4=0 \\ & 5 \beta(\beta-2)-2(\beta-2)=0 \\ &(5 \beta-2)(\beta-2)=0 \\ & \beta=22 / 5 \end{aligned} $

On substitute into $-\alpha+2 \beta=3$

$ \begin{aligned} & \beta=2-\alpha+4=3 \\ & \alpha=1 \\ & \alpha+\beta=1+2=3 \\ & \beta=\frac{2}{5} \Rightarrow-\alpha+\frac{4}{5}=3 \end{aligned} $

On substitute $-\alpha=3-\frac{4}{5}=\frac{15-4}{5}$

Neglecting $\alpha=\frac{-11}{5}$

Given, $\mathbf{a}=4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and

$ \mathbf{b}=12 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}},|\mathbf{c}|=3 \sqrt{13} $

The required vectors $\mathbf{c}$ is given by

$ \begin{aligned} c & =\lambda(\mathbf{a}+\mathbf{b}) \\ & =\lambda\left(\frac{\mathbf{a}}{|\mathbf{a}|}+\frac{\mathbf{b}}{|\mathbf{b}|}\right) \\ & =\lambda\left(\frac{4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}}{\sqrt{16+49+16}}+\frac{12 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}}{\sqrt{144+9+16}}\right) \\ & =\lambda\left[\frac{4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}}{9}+\frac{12 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}}{13}\right] \\ \mathbf{c} & =\lambda\left[\frac{52 \hat{\mathbf{i}}+91 \hat{\mathbf{j}}-52 \hat{\mathbf{k}}+108 \hat{\mathbf{i}}-27 \hat{\mathbf{j}}+36 \hat{\mathbf{k}}}{9 \times 13}\right] \\ \mathbf{c} & =\lambda\left[\frac{160 \hat{\mathbf{i}}+64 \hat{\mathbf{j}}-16 \hat{\mathbf{k}}}{9 \times 13}\right] \\ \mathbf{c} & =\frac{16 \lambda}{9 \times 13}[10 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}] \\ |\mathbf{c}| & =\frac{16 \lambda}{13 \times 9} \sqrt{100+16+1}=\frac{16 \lambda}{13 \times 9} \sqrt{117} \\ \mathbf{c} & =3 \sqrt{13} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(given)\\ 3 \sqrt{13} & =\frac{16 \lambda}{13 \times 9} \sqrt{117} \Rightarrow 3 \sqrt{13}=\frac{16 \lambda}{13 \times 9} \sqrt{13 \times 9} \\ 3 \sqrt{13} & =\frac{16 \lambda}{13 \times 9} 3 \sqrt{13} \\ \lambda & =\frac{13 \times 9}{16} \end{aligned} $

On putting in Eq.(i), we get

Hence, $c=10 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}$.

The vector $\mathbf{c}$ is a unit vector that is also perpendicular to $\mathbf{b}$, as expressed by:

$ \mathbf{b} \cdot \mathbf{c} = 0 $

Additionally, the magnitude of $\mathbf{c}$ is 1:

$ |\mathbf{c}| = 1 $

To find $\mathbf{c}$, which lies in the plane formed by the vectors $\mathbf{a}$ and $\mathbf{b}$, we can use the cross product of these vectors:

$ \mathbf{c} = \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 2 & 1 & 1 \end{array}\right| $

Calculating the determinant gives us:

$ = \hat{\mathbf{i}}(-1-1) - \hat{\mathbf{j}}(1-2) + \hat{\mathbf{k}}(1+2) $

This simplifies to:

$ = -2 \hat{\mathbf{i}} + \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} $

Finally, we compute the dot product of $\mathbf{c}$ with the vector $\hat{\mathbf{i}} + \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}$:

$ c \cdot (\hat{\mathbf{i}} + \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) = (-2 \hat{\mathbf{i}} + \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) \cdot (\hat{\mathbf{i}} + \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) $

This yields:

$ = 5 $

Thus, $\mathbf{c}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) = 5$.

To solve for $(\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{d})$, we will begin by calculating the cross products $\mathbf{a} \times \mathbf{c}$ and $\mathbf{b} \times \mathbf{d}$ separately.

We have the vectors defined as:

$\mathbf{a} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}$

$\mathbf{b} = \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2\hat{\mathbf{k}}$

$\mathbf{c} = 2\hat{\mathbf{i}} - 3\hat{\mathbf{j}} - \hat{\mathbf{k}}$

$\mathbf{d} = 2\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$

Step 1: Calculate $\mathbf{a} \times \mathbf{c}$

The cross product $\mathbf{a} \times \mathbf{c}$ is determined using the determinant:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 2 & -3 & -1 \end{vmatrix} $

Calculating the determinant, we have:

$ \hat{\mathbf{i}}[( -1)(-1) - (1)(-3)] - \hat{\mathbf{j}}[(1)(-1) - (1)(2)] + \hat{\mathbf{k}}[(1)(-3) - (-1)(2)] $

This simplifies to:

$ \hat{\mathbf{i}}(1 + 3) - \hat{\mathbf{j}}(-1 - 2) + \hat{\mathbf{k}}(-3 + 2) $

Resulting in:

$ 4\hat{\mathbf{i}} + 3\hat{\mathbf{j}} - \hat{\mathbf{k}} $

Step 2: Calculate $\mathbf{b} \times \mathbf{d}$

Similarly, compute the cross product $\mathbf{b} \times \mathbf{d}$ using the determinant:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & -2 \\ 2 & 1 & 1 \end{vmatrix} $

Calculating this determinant, we find:

$ \hat{\mathbf{i}}[(1)(1) - (-2)(1)] - \hat{\mathbf{j}}[(1)(1) - (-2)(2)] + \hat{\mathbf{k}}[(1)(1) - (1)(2)] $

This gives:

$ \hat{\mathbf{i}}(1 + 2) - \hat{\mathbf{j}}(1 + 4) + \hat{\mathbf{k}}(1 - 2) $

Resulting in:

$ 3\hat{\mathbf{i}} - 5\hat{\mathbf{j}} - \hat{\mathbf{k}} $

Step 3: Calculate $(\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{d})$

Now use the result from both cross products to find their cross product:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & -1 \\ 3 & -5 & -1 \end{vmatrix} $

This determinant is computed as follows:

$ \hat{\mathbf{i}}[(3)(-1) - (3)(-1)] - \hat{\mathbf{j}}[(4)(-1) - (-1)(-1)] + \hat{\mathbf{k}}[(4)(-5) - (3)(3)] $

Which results in:

$ \hat{\mathbf{i}}(-3 + 5) - \hat{\mathbf{j}}(-4 - 1) + \hat{\mathbf{k}}(-20 - 9) $

Simplifying, we find:

$ -8\hat{\mathbf{i}} + \hat{\mathbf{j}} - 29\hat{\mathbf{k}} $

Thus, the final result of $(\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{d})$ is:

$ -8\hat{\mathbf{i}} + \hat{\mathbf{j}} - 29\hat{\mathbf{k}} $

To find the angle between the diagonals of a parallelogram, we start with the given vectors representing the adjacent sides of the parallelogram:

Let:

$\mathbf{a} = 2 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}$

$\mathbf{b} = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}$

The diagonal vectors of the parallelogram can be calculated as follows:

First diagonal, $\mathbf{c}$:

$ \mathbf{c} = \mathbf{a} + \mathbf{b} = (2 + 1) \hat{\mathbf{i}} + (4 + 2) \hat{\mathbf{j}} + (-5 + 3) \hat{\mathbf{k}} = 3 \hat{\mathbf{i}} + 6 \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $

Second diagonal, $\mathbf{d}$:

$ \mathbf{d} = \mathbf{a} - \mathbf{b} = (2 - 1) \hat{\mathbf{i}} + (4 - 2) \hat{\mathbf{j}} + (-5 - 3) \hat{\mathbf{k}} = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 8 \hat{\mathbf{k}} $

Next, we calculate the cosine of the angle $\theta$ between the diagonals using the dot product formula:

$ \cos \theta = \frac{\mathbf{c} \cdot \mathbf{d}}{|\mathbf{c}| |\mathbf{d}|} $

Dot Product:

$ \mathbf{c} \cdot \mathbf{d} = (3)(1) + (6)(2) + (-2)(-8) = 3 + 12 + 16 = 31 $

Magnitudes of the diagonals:

$ |\mathbf{c}| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 $

$ |\mathbf{d}| = \sqrt{1^2 + 2^2 + (-8)^2} = \sqrt{1 + 4 + 64} = \sqrt{69} $

Substituting these into the formula for $\cos \theta$:

$ \cos \theta = \frac{31}{7 \sqrt{69}} $

Thus, the angle $\theta$ between the diagonals is:

$ \theta = \cos^{-1}\left(\frac{31}{7 \sqrt{69}}\right) $

Therefore, the angle between the diagonals of the parallelogram is $\cos^{-1}\left(\frac{31}{7 \sqrt{69}}\right)$.

To determine if the points with the given position vectors are coplanar, we need to ensure the determinant of the matrix formed by vectors $\mathbf{AB}$, $\mathbf{AC}$, and $\mathbf{AD}$ is zero.

Given:

$\mathbf{OA} = -\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 4\hat{\mathbf{k}}$

$\mathbf{OB} = 3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$

$\mathbf{OC} = -3\hat{\mathbf{i}} + 8\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$

$\mathbf{OD} = -3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + \lambda\hat{\mathbf{k}}$

Calculate the vectors:

$ \mathbf{AB} = \mathbf{OB} - \mathbf{OA} = (3 + 1)\hat{\mathbf{i}} + (2 - 4)\hat{\mathbf{j}} + (-5 + 4)\hat{\mathbf{k}} = 4\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - \hat{\mathbf{k}} $

$ \mathbf{AC} = \mathbf{OC} - \mathbf{OA} = (-3 + 1)\hat{\mathbf{i}} + (8 - 4)\hat{\mathbf{j}} + (-5 + 4)\hat{\mathbf{k}} = -2\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - \hat{\mathbf{k}} $

$ \mathbf{AD} = \mathbf{OD} - \mathbf{OA} = (-3 + 1)\hat{\mathbf{i}} + (2 - 4)\hat{\mathbf{j}} + (\lambda + 4)\hat{\mathbf{k}} = -2\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + (\lambda + 4)\hat{\mathbf{k}} $

The determinant for coplanarity is:

$ \left|\begin{array}{ccc} 4 & -2 & -1 \\ -2 & 4 & -1 \\ -2 & -2 & \lambda + 4 \end{array}\right| = 0 $

Expanding the determinant:

$ = 4((4)(\lambda + 4) - (-2)(-1)) - (-2)((-2)(\lambda + 4) - (-2)(-1)) - 1((-2)(-2) - (4)(-2)) $

$ = 4(4\lambda + 16 - 2) - 2(2\lambda + 8 - 2) - (4 + 8) $

Simplify:

$ = 16\lambda + 56 - 4\lambda - 20 - 12 = 0 $

Solving:

$ 12\lambda + 24 = 0 $

$ 12\lambda = -24 $

$ \lambda = -2 $

Therefore, the value of $\lambda$ is $-2$.

Given, $|\mathrm{f}|=10,|\mathrm{~g}|=14$ and

$ |f-g|=15 $

We know that

$ \begin{aligned} & |\mathbf{f}-\mathbf{g}|^2=|\mathbf{f}|^2+|\mathbf{g}|^2-2 R e|\mathbf{f}||\mathbf{g}| \\ & \Rightarrow 225=100+196-2 R e|\mathbf{f}||\mathbf{g}| \\ & \Rightarrow 2 R e|\mathbf{f} \| \mathbf{g}|=296-225=71 \\ & \begin{aligned} & \text { Now, }|\mathbf{f}+\mathbf{g}|^2=|\mathbf{f}|^2+|\mathbf{g}|^2+2 R e|\mathbf{f}||\mathbf{g}| \\ &=100+196+71 \\ &=367 \\ & \Rightarrow|\mathbf{f}+\mathbf{g}|=\sqrt{367} \end{aligned} \end{aligned} $

Given, $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=\sqrt{3}$ and

$ \begin{aligned} & (a+b-c)^2+(b+c-a)^2+(c+a-b)^2=36 \\ & |a|^2+|b|^2+|c|^2+2 a \cdot b-2 a \cdot c-2 b \cdot c \\ & +|b|^2+|\mathbf{c}|^2+|\mathbf{a}|^2+2 \mathbf{b} \cdot \mathbf{c}-2 \mathbf{b} \cdot \mathbf{a}-2 \mathbf{c} \cdot \mathbf{a} \\ & +|\mathbf{c}|^2+|\mathbf{a}|^2+|\mathbf{b}|^2+2 \mathbf{c} \cdot \mathbf{a}-2 \mathbf{a} \cdot \mathbf{b} \\ & -2 b \cdot c=36 \\ & 3\left(|a|^2+|b|^2+|c|^2\right)-2 a \cdot b-2 a \cdot c \\ & -2 b \cdot c=36 \\ & 3(3+3+3)-2 a \cdot b-2 a \cdot c-2 b \cdot c=36 \\ & (\mathbf{a} \cdot \mathbf{b}+\mathbf{a} \cdot \mathbf{c}+\mathbf{b} \cdot \mathbf{c})=\frac{27-36}{2}=\frac{-9}{2} \quad \ldots \text { (i) } \end{aligned} $

Now, $|2 \mathbf{a}-3 \mathrm{~b}+2 \mathbf{c}|^2$

$ =4|a|^2+4|b|^2+4|c|^2-12 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{a} \cdot \mathbf{c}-12 \mathbf{b} \cdot \mathbf{c} $

Also, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 \geq 0$

$ \begin{aligned} & \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \geq 0 \\ & \Rightarrow \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a} \geq-\frac{9}{2} \end{aligned} $

Since, $\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=-\frac{9}{2}$

$ \begin{aligned} & \text { So, }|\mathbf{a}+\mathbf{b}+\mathbf{c}|=0 \\ & \begin{aligned} & \Rightarrow \quad a+b+c \quad \ldots \text { (ii) } \\ & \text { So, }|2 a-3 b+2 c|=|(2 a+2 c)-3 b| \\ & \quad|-2 b-3 b|, d=|-5 b| \\ & \quad=5|b|=5 \sqrt{3} \end{aligned} \end{aligned} $

$ \therefore|2 a-3 b+2 c|^2=(5 \sqrt{3})^2=75 $

Given the non-coplanar vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$, we start with the equations:

$ \mathbf{a} + \mathbf{b} + \mathbf{c} = \alpha \mathbf{d} $

From this equation, we can express $\mathbf{b} + \mathbf{c}$ as:

$ \mathbf{b} + \mathbf{c} = \alpha \mathbf{d} - \mathbf{a} \quad \text{(Equation i)} $

We are also given:

$ \mathbf{b} + \mathbf{c} + \mathbf{d} = \beta \mathbf{a} $

From which we derive:

$ \mathbf{b} + \mathbf{c} = \beta \mathbf{a} - \mathbf{d} \quad \text{(Equation ii)} $

By equating both expressed forms of $\mathbf{b} + \mathbf{c}$ from Equations (i) and (ii), we have:

$ \alpha \mathbf{d} - \mathbf{a} = \beta \mathbf{a} - \mathbf{d} $

Rearranging terms gives:

$ \alpha \mathbf{d} + \mathbf{d} = \beta \mathbf{a} + \mathbf{a} $

Thus, we have:

$ (\alpha + 1) \mathbf{d} = (\beta + 1) \mathbf{a} $

Since $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are non-coplanar, they are linearly independent. Therefore, the coefficients of $\mathbf{a}$ and $\mathbf{d}$ must separately satisfy the equality:

$1 + \beta = 0$ implies $\beta = -1$

$1 + \alpha = 0$ implies $\alpha = -1$

Therefore:

$ \mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = 0 $

Thus, the magnitude is:

$ |\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d}| = 0 $

Given that $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are unit vectors, we define:

$\hat{\mathbf{p}} = \hat{\mathbf{u}} + \hat{\mathbf{v}} + \hat{\mathbf{w}}$

$\hat{\mathbf{q}} = \hat{\mathbf{u}} \times (\hat{\mathbf{v}} \times \hat{\mathbf{w}})$

Additionally, we know:

$\hat{\mathbf{p}} \cdot \hat{\mathbf{u}} = \frac{3}{2}$

$\hat{\mathbf{p}} \cdot \hat{\mathbf{v}} = \frac{7}{4}$

$|\hat{\mathbf{p}}| = 2$

Calculation Steps:

From the equation $\mathbf{p} \cdot \mathbf{u}$:

$ \mathbf{p} \cdot \mathbf{u} = |\mathbf{u}|^2 + \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} = \frac{3}{2} $

This implies:

$ 1 + \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} = \frac{3}{2} $

$ \Rightarrow \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} = \frac{1}{2} $

From the equation $\mathbf{p} \cdot \mathbf{v}$:

$ \mathbf{p} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^2 + \mathbf{v} \cdot \mathbf{w} = \frac{7}{4} $

Thus:

$ \mathbf{u} \cdot \mathbf{v} + 1 + \mathbf{v} \cdot \mathbf{w} = \frac{7}{4} $

$ \Rightarrow \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{w} = \frac{3}{4} $

Equating results:

Solve the system:

$ \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} = \frac{1}{2} $

$ \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{w} = \frac{3}{4} $

Subtract to find:

$ \mathbf{u} \cdot \mathbf{w} - \mathbf{v} \cdot \mathbf{w} = -\frac{1}{4} $

Vector subtraction and dot product:

$ \mathbf{w} \cdot (\mathbf{u} - \mathbf{v}) = -\frac{1}{4} $

Given also:

$ \mathbf{p} \cdot (\mathbf{u} - \mathbf{v}) = \frac{3}{2} - \frac{7}{4} = -\frac{1}{4} $

Conclude for $K$:

This means:

$ \mathbf{p} = \mathbf{w} \quad \Rightarrow \quad |\mathbf{p}| = |\mathbf{w}| \quad \Rightarrow \quad \mathbf{u} + \mathbf{v} = 0 $

For:

$ \mathbf{q} = \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) $

Expanding:

$ \mathbf{q} = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} $

Knowing $\mathbf{v} = K \mathbf{q}$:

$ \Rightarrow \mathbf{u} \cdot \mathbf{v} = 0 $

Then solve for $K$:

$ K = \frac{1}{\mathbf{u} \cdot \mathbf{w}} $

From above, $\mathbf{u} \cdot \mathbf{v} = \frac{1}{2}$, hence:

$ K = 2 $

The expression $|\mathbf{b}|\mathbf{a} + |\mathbf{a}|\mathbf{b}$ represents a vector parallel to the angle bisector of the two non-collinear vectors $\mathbf{a}$ and $\mathbf{b}$.

To understand why, consider the angle bisector of the angle between vectors $\mathbf{a}$ and $\mathbf{b}$:

$ \lambda (\hat{\mathbf{a}} + \hat{\mathbf{b}}) $

This can be rewritten as:

$ \lambda \left(\frac{\mathbf{a}}{|\mathbf{a}|} + \frac{\mathbf{b}}{|\mathbf{b}|}\right) $

Simplifying further, we have:

$ \frac{\lambda}{|\mathbf{a}||\mathbf{b}|} (|\mathbf{b}|\mathbf{a} + |\mathbf{a}|\mathbf{b}) = k(|\mathbf{b}|\mathbf{a} + |\mathbf{a}|\mathbf{b}) $

This derivation shows that $|\mathbf{b}|\mathbf{a} + |\mathbf{a}|\mathbf{b}$ is indeed parallel to the angle bisector of $\mathbf{a}$ and $\mathbf{b}$.

By mid-point 5 theorem,

$ L N=Q M=\frac{1}{2} Q R $

$\begin{gathered}L M=N R=\frac{1}{2} P R \\ M N=Q L=\frac{1}{2} P Q \\ \mathbf{Q M}+\mathbf{L N}+\mathbf{M L}+\mathbf{R N}-\mathbf{M N}-\mathbf{Q L} \\ = \\ =\frac{1}{2} \mathbf{Q R}+\frac{1}{2} \mathbf{Q R}-\frac{1}{2} \mathbf{P R} \\ \quad-\frac{1}{2} \mathbf{P R}+\frac{1}{2} \mathbf{P Q}+\frac{1}{2} \mathbf{P Q} \\ =\mathbf{Q R}-\mathbf{P R}+\mathbf{P Q}\end{gathered}$

The given information states that the projection of vector $\mathbf{b}$ onto vector $\mathbf{a}$ is $\frac{8}{\sqrt{14}}$. Let's go through the calculation step-by-step to find $|\mathbf{b}|$.

First, we express the projection of $\mathbf{b}$ onto $\mathbf{a}$:

$ \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|} = \frac{8}{\sqrt{14}} $

From this, we have:

$ \frac{\mathbf{a} \cdot \mathbf{b}}{\sqrt{14}} = \frac{8}{\sqrt{14}} \implies \mathbf{a} \cdot \mathbf{b} = 8 $

Next, the magnitudes of the vectors satisfy the following relation:

$ |\mathbf{a}|^2 |\mathbf{b}|^2 = (\mathbf{a} \cdot \mathbf{b})^2 + |\mathbf{a} \times \mathbf{b}|^2 $

Given that $\mathbf{a} \times \mathbf{b} = 7 \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}}$, the magnitude $|\mathbf{a} \times \mathbf{b}|$ can be calculated as:

$ |\mathbf{a} \times \mathbf{b}| = \sqrt{7^2 + (-5)^2 + (-4)^2} = \sqrt{49 + 25 + 16} = \sqrt{90} $

Now, substitute these values into the equation:

$ 14 |\mathbf{b}|^2 = 64 + 90 $

$ 14 |\mathbf{b}|^2 = 154 $

Therefore:

$ |\mathbf{b}|^2 = \frac{154}{14} = 11 $

Thus, the magnitude of $\mathbf{b}$ is:

$ |\mathbf{b}| = \sqrt{11} $

Consider point $ A $ as the origin, denoted as $ O $. Let $ \mathbf{b} $ and $ \mathbf{c} $ be the position vectors of points $ B $ and $ C $ respectively.

Since the vector $ \mathbf{AM} = \frac{\mathbf{AB}}{3} = \frac{\mathbf{b}}{3} $, the position vector of $ M $ is $\frac{\mathbf{b}}{3}$.

The position vector of $ N $ is given by $ \mathbf{AN} = K \mathbf{AC} = K \mathbf{c} $.

Since the vectors $ \mathbf{BN} $ and $ \mathbf{CM} $ are perpendicular, we use the dot product property for perpendicular vectors:

$ (K \mathbf{c} - \mathbf{b}) \cdot \left(\frac{\mathbf{b}}{3} - \mathbf{c}\right) = 0 $

Expanding this, we have:

$ \frac{K}{3} \mathbf{b} \cdot \mathbf{c} - \frac{|\mathbf{b}|^2}{3} - K|\mathbf{c}|^2 + \mathbf{b} \cdot \mathbf{c} = 0 $

Substituting, knowing that for an equilateral triangle, $\mathbf{b} \cdot \mathbf{c} = |\mathbf{b}||\mathbf{c}|\cos(60^\circ) = \frac{a^2}{2}$:

$ \frac{K}{3} \cdot \frac{a^2}{2} - \frac{a^2}{3} - K a^2 + \frac{a^2}{2} = 0 $

Simplifying the equation:

$ |\mathbf{a}|^2 \left[\frac{K}{6} - \frac{1}{3} - K + \frac{1}{2}\right] = 0 $

Since $ |\mathbf{a}| \neq 0 $ (as it represents side length $ a $), the term in square brackets must equal zero:

$ \frac{K}{6} - K - \frac{1}{3} + \frac{1}{2} = 0 $

This simplifies to:

$ \frac{-5K}{6} + \frac{1}{6} = 0 $

Thus:

$ -5K + 1 = 0 \quad \Rightarrow \quad K = \frac{1}{5} $

To clarify the problem, we start by considering two non-collinear unit vectors, $\mathbf{a}$ and $\mathbf{b}$.

We are given:

$\mathbf{u} = \mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{b}$

$\mathbf{v} = \mathbf{a} \times \mathbf{b}$

Let's analyze $|\mathbf{u}|$:

$ \mathbf{u} = \mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{b} $

The dot product of $\mathbf{u}$ with itself yields:

$ |\mathbf{u}|^2 = \mathbf{u} \cdot \mathbf{u} = \left[\mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{b}\right] \cdot \left[\mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{b}\right] $

Expanding gives:

$ |\mathbf{u}|^2 = \mathbf{a} \cdot \mathbf{a} - 2 (\mathbf{a} \cdot \mathbf{b}) (\mathbf{b} \cdot \mathbf{a}) + (\mathbf{a} \cdot \mathbf{b})^2 (\mathbf{b} \cdot \mathbf{b}) $

Given $ |\mathbf{a}| = |\mathbf{b}| = 1 $, this simplifies to:

$ |\mathbf{u}|^2 = 1 - (\mathbf{a} \cdot \mathbf{b})^2 $

Since $(\mathbf{a} \cdot \mathbf{b}) = \cos \theta$, we have:

$ |\mathbf{u}|^2 = 1 - \cos^2 \theta = \sin^2 \theta $

Thus, $|\mathbf{u}| = \sin \theta$.

Now, consider $|\mathbf{v}|$:

$ \mathbf{v} = \mathbf{a} \times \mathbf{b} $

The magnitude of the cross product is:

$ |\mathbf{v}| = |\mathbf{a}||\mathbf{b}|\sin \theta = \sin \theta $

Now let us verify the expression $|\mathbf{u}| + |\mathbf{u} \cdot \mathbf{v}|$:

$ \begin{aligned} |\mathbf{u} \cdot \mathbf{v}| &= \left| \left(\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{b} \right) \cdot \left(\mathbf{a} \times \mathbf{b}\right) \right| \\ &= \left| \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) - (\mathbf{a} \cdot \mathbf{b}) \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) \right| \end{aligned} $

Using vector triple product identities, we know:

$ \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0 $

$ \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 0 $

Therefore:

$ |\mathbf{u} \cdot \mathbf{v}| = 0 $

Thus, $|\mathbf{u}| + |\mathbf{u} \cdot \mathbf{v}| = \sin \theta + 0 = \sin \theta$.

Hence, $|\mathbf{v}| = |\mathbf{u}| + |\mathbf{u} \cdot \mathbf{v}|$.

$ \text { In } \triangle A B C, A B+B C=A C $

$ \Rightarrow \quad A C=a+b $

AD is parallel to BC and also

$ \begin{aligned} & A D=2 B C \\ & A D=2 b \\ & C D=A D-A C=2 b-(a+b)=b-a \end{aligned} $

FA is parallel to CD, but is opposite direction.

$ \mathrm{FA}=-\mathrm{CD}=-(\mathrm{b}-\mathrm{a})=\mathrm{a}-\mathrm{b} $

To find the angle between the vectors $\mathbf{f} + \mathbf{g} + \mathbf{h}$ and $\mathbf{h}$, we start by noting that $\mathbf{f}$, $\mathbf{g}$, and $\mathbf{h}$ are mutually orthogonal vectors of equal magnitude. This means:

$|\mathbf{f}| = |\mathbf{g}| = |\mathbf{h}|$.

$\mathbf{f} \cdot \mathbf{g} = \mathbf{g} \cdot \mathbf{h} = \mathbf{h} \cdot \mathbf{f} = 0$.

The dot product of $(\mathbf{f} + \mathbf{g} + \mathbf{h})$ with $\mathbf{h}$ can be calculated as:

$ (\mathbf{f} + \mathbf{g} + \mathbf{h}) \cdot \mathbf{h} = \mathbf{f} \cdot \mathbf{h} + \mathbf{g} \cdot \mathbf{h} + \mathbf{h} \cdot \mathbf{h} = |\mathbf{h}|^2 $

Let $\theta$ be the angle between $\mathbf{f} + \mathbf{g} + \mathbf{h}$ and $\mathbf{h}$. We use the formula for the dot product in terms of angle:

$ |\mathbf{f}+\mathbf{g}+\mathbf{h}||\mathbf{h}| \cos \theta = |\mathbf{h}|^2 $

Since $\mathbf{f}$, $\mathbf{g}$, and $\mathbf{h}$ are orthogonal and of equal magnitude, the magnitude of $\mathbf{f} + \mathbf{g} + \mathbf{h}$ is:

$ |\mathbf{f}+\mathbf{g}+\mathbf{h}| = \sqrt{|\mathbf{f}|^2 + |\mathbf{g}|^2 + |\mathbf{h}|^2} = \sqrt{3}|\mathbf{h}| $

Substitute this back into the equation:

$ \sqrt{3} |\mathbf{h}| |\mathbf{h}| \cos \theta = |\mathbf{h}|^2 $

Simplifying, we find:

$ \cos \theta = \frac{1}{\sqrt{3}} $

Therefore, the angle $\theta$ is:

$ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) $

Given two unit vectors $\mathbf{a}$ and $\mathbf{b}$, we know that:

$ |\mathbf{a}| = |\mathbf{b}| = 1 $

Consider the vectors $\mathbf{c} = \mathbf{a} + 2\mathbf{b}$ and $\mathbf{d} = 5\mathbf{a} - 4\mathbf{b}$. These vectors are perpendicular, meaning:

$ \mathbf{c} \cdot \mathbf{d} = 0 $

Let's compute the dot product:

$ (\mathbf{a} + 2\mathbf{b}) \cdot (5\mathbf{a} - 4\mathbf{b}) = \mathbf{a} \cdot (5\mathbf{a} - 4\mathbf{b}) + 2\mathbf{b} \cdot (5\mathbf{a} - 4\mathbf{b}) $

Expanding this, we have:

$ = \mathbf{a} \cdot 5\mathbf{a} - \mathbf{a} \cdot 4\mathbf{b} + 2\mathbf{b} \cdot 5\mathbf{a} - 2\mathbf{b} \cdot 4\mathbf{b} $

This simplifies to:

$ = 5|\mathbf{a}|^2 - 4\mathbf{a} \cdot \mathbf{b} + 10\mathbf{a} \cdot \mathbf{b} - 8|\mathbf{b}|^2 $

Simplifying further by substituting $|\mathbf{a}| = 1$ and $|\mathbf{b}| = 1$:

$ = 5(1) + 6\mathbf{a} \cdot \mathbf{b} - 8(1) = 0 $

Which simplifies to:

$ 5 - 8 + 6(\mathbf{a} \cdot \mathbf{b}) = 0 $

$ -3 + 6(\mathbf{a} \cdot \mathbf{b}) = 0 $

Solving for $\mathbf{a} \cdot \mathbf{b}$:

$ 6(\mathbf{a} \cdot \mathbf{b}) = 3 $

$ \mathbf{a} \cdot \mathbf{b} = \frac{3}{6} = \frac{1}{2} $

The dot product $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, so:

$ \cos\theta = \frac{1}{2} $

Thus, the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is:

$ \theta = \frac{\pi}{3} $

To determine the value of $ p $ for which the vectors $\mathbf{a} = 2 \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}, \mathbf{b} = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 3 \hat{\mathbf{k}},$ and $\mathbf{c} = 3 \hat{\mathbf{i}} + p \hat{\mathbf{j}} + 5 \hat{\mathbf{k}}$ are coplanar, we need to evaluate the condition of coplanarity. The vectors are coplanar if the determinant of the matrix formed by their components is zero:

$ \left|\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & p & 5 \end{array}\right| = 0 $

Calculate the determinant:

$ = 2 \left( 2 \cdot 5 - (-3) \cdot p \right) - (-1) \left( 1 \cdot 5 - (-3) \cdot 3 \right) + 1 \left( 1 \cdot p - 2 \cdot 3 \right) $

Simplify each term:

The first term: $ 2(10 + 3p) = 20 + 6p $

The second term: $-1(5 + 9) = -14$

The third term: $1(p - 6) = p - 6$

Combining these terms gives:

$ 20 + 6p - 14 + p - 6 = 0 $

Simplifying further:

$ 20 + 6p - 14 + p - 6 = 0 \implies 7p = -28 \implies p = -4 $

Thus, the value of $ p $ is $-4$.

Given the problem involves finding the value of $(\alpha + \beta + \gamma)^2$ when $(\alpha, \beta, \gamma)$ are the direction cosines of the angular bisector of two lines with direction ratios $(2, 2, 1)$ and $(2, -1, -2)$.

Finding Direction Cosines:

The direction ratios for line $L_1$ are $(2, 2, 1)$, and for line $L_2$ are $(2, -1, -2)$.

The direction cosines for $L_1$ are found as:

$ \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right) $

Similarly, for $L_2$ the direction cosines are:

$ \left(\frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}\right) $

Perpendicular Check:

The dot product of the two direction ratios confirms that the lines are perpendicular:

$ 2 \times 2 + 2 \times (-1) + 1 \times (-2) = 0 $

Angle Bisector Direction Cosines:

The formulas for the direction cosines of the angle bisectors are:

$ \left(\frac{l_1+l_2}{2 \cos \frac{\theta}{2}}, \frac{m_1+m_2}{2 \cos \frac{\theta}{2}}, \frac{n_1+n_2}{2 \cos \frac{\theta}{2}}\right) $

and

$ \left(\frac{l_1-l_2}{2 \sin \frac{\theta}{2}}, \frac{m_1-m_2}{2 \sin \frac{\theta}{2}}, \frac{n_1-n_2}{2 \sin \frac{\theta}{2}}\right) $

Calculate the Direction Cosines:

For an angle bisector direction, calculate:

$ \left(\frac{4}{3\sqrt{2}}, \frac{1}{3\sqrt{2}}, \frac{-1}{3\sqrt{2}}\right) $

Another possible set is:

$ \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $

Calculation of $(\alpha + \beta + \gamma)^2$:

Using $\alpha = \frac{4}{3\sqrt{2}}, \beta = \frac{1}{3\sqrt{2}}, \gamma = \frac{-1}{3\sqrt{2}}$:

Sum:

$ \alpha + \beta + \gamma = \frac{4}{3\sqrt{2}} + \frac{1}{3\sqrt{2}} - \frac{1}{3\sqrt{2}} = \frac{4}{3\sqrt{2}} $

Squaring:

$ (\alpha + \beta + \gamma)^2 = \left(\frac{4}{3\sqrt{2}}\right)^2 = \frac{16}{18} = \frac{8}{9} $

Alternatively, use $\alpha = 0, \beta = \frac{1}{\sqrt{2}}, \gamma = \frac{1}{\sqrt{2}}$:

Sum:

$ \alpha + \beta + \gamma = 0 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} $

Squaring:

$ (\alpha + \beta + \gamma)^2 = (\sqrt{2})^2 = 2 $

Based on calculations with valid assumptions, the value of $(\alpha + \beta + \gamma)^2$ considering a correct scenario for calculation comes out to be 2.

To compute the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$, we first use the determinant representation: $[a b c] = \left| \begin{array}{ccc} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{array} \right|$.

This can be expanded as follows:

$ [a b c] = 1 \left( (1 \cdot (1 + x - y)) - (x \cdot (1 - x)) \right) - 1 \left( x^2 - y \right) $

Breaking it down further:

$ = 1 \cdot (1 + x - y - x + x^2) - (x^2 - y) $

Simplifying the expression:

$ = (1 + x - y - x + x^2) - x^2 + y $

$ = 1 $

The result $1$ is independent of the variables $x$ and $y$. Therefore, the value of the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ does not depend on $x$ or $y$.

The projection of a vector $\mathbf{a}$ on another vector $\mathbf{b}$ is calculated using the formula:

$ |\mathbf{a}| \cos \theta = \frac{(\mathbf{a} \cdot \mathbf{b})}{|\mathbf{b}|} $

Consider the line joining the points $ (3, 4, 5) $ and $ (4, 6, 3) $ as $ L_1 $, and the line joining the points $ (-1, 2, 4) $ and $ (1, 0, 5) $ as $ L_2 $.

The direction vector $ L_1 $ is derived as follows:

$ L_1 = (4 - 3)\hat{\mathbf{i}} + (6 - 4)\hat{\mathbf{j}} + (3 - 5)\hat{\mathbf{k}} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}} $

Similarly, the direction vector $ L_2 $ is:

$ L_2 = (1 - (-1))\hat{\mathbf{i}} + (0 - 2)\hat{\mathbf{j}} + (5 - 4)\hat{\mathbf{k}} = 2\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}} $

The projection of $ L_1 $ onto $ L_2 $ is calculated by:

$ \text{Projection of } L_1 \text{ on } L_2 = \frac{L_1 \cdot L_2}{|L_2|} $

Calculating the dot product $ L_1 \cdot L_2 $:

$ L_1 \cdot L_2 = (1 \times 2) + (2 \times -2) + (-2 \times 1) = 2 - 4 - 2 = -4 $

The magnitude of $ L_2 $ is:

$ |L_2| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $

Thus, the projection is:

$ \frac{-4}{3} $

Therefore, the magnitude of the projection is:

$ \frac{4}{3} $

Step 1: Given condition for coplanarity

For three vectors to be coplanar, their scalar triple product must be zero. We have the vectors A, B, and C, and we know the given relation between λ and μ:

$A = \lambda \hat{i} - \hat{j} + \hat{k}$

$B = \hat{i} + 2 \hat{j} + \mu \hat{k}$

$C = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$

Scalar triple product condition:

$[A, B, C] = A \cdot (B \times C) = 0$

Step 2: Calculate the cross product of B and C

$B \times C = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & \mu \\ 3 & -4 & 5 \\ \end{vmatrix}$

$B \times C = (10 + 4 \mu) \hat{i} - (5 - 3 \mu) \hat{j} - 10 \hat{k}$

Step 3: Calculate the scalar triple product and apply the coplanarity condition

$[A, B, C] = A \cdot (B \times C) = \lambda(10 + 4 \mu) - 1(5 - 3 \mu) + 1(-10) = 0$

Step 4: Substitute the given relation between λ and μ Given that:

$\lambda - \mu = 5$

From the coplanarity condition, we have:

$4 \lambda \mu + 10 \lambda - 3 \mu = 5$

Now, substitute λ in terms of μ:

$4(5 + \mu) \mu + 10(5 + \mu) - 3 \mu = 5$

Step 5: Solve for μ and λ

Simplify the equation and solve for μ:

$4 \mu^2 + 27 \mu + 45 = 0$

Factor the equation:

$(4 \mu + 15)(\mu + 3) = 0$

So, the two possible values for μ are:

$\mu = -3, \frac{-15}{4}$

For each value of μ, find the corresponding value of λ using the given relation:

$\lambda = \mu + 5$

So, we get the values for λ:

$\lambda = 2, \frac{5}{4}$

Step 6: Calculate the sum

Now, we need to find the sum:

$\sum\limits_{(\lambda, \mu) \in S} 80\left(\lambda^2 + \mu^2\right)$

Substitute the values of λ and μ, and simplify:

$80\left[(2^2 + (-3)^2) + \left(\frac{5}{4}\right)^2 + \left(\frac{-15}{4}\right)^2\right] = 80\left[13 + \frac{25}{16} + \frac{225}{16}\right]$

$= 80\left[13 + \frac{250}{16}\right] = 10 \times 229$

$= 2290$

So, the correct answer is 2290 (Option D).

Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$, respectively.

Then the position vector of $E$ is:

$ \vec{E} = \frac{\vec{a} + \vec{c}}{2} $

And the position vector of $F$ is:

$ \vec{F} = \frac{\vec{b} + \vec{d}}{2} $

Now, we are given the equation:

$ (\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k\overrightarrow{FE} $

We can rewrite this equation using the position vectors:

$ (\vec{b} - \vec{a} - (\vec{c} - \vec{b})) + (\vec{d} - \vec{a} - (\vec{c} - \vec{d})) = k(\vec{E} - \vec{F}) $

Simplifying the equation, we get:

$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}(2\vec{E} - 2\vec{F}) $

Now substitute $\vec{E}$ and $\vec{F}$ expressions we found earlier:

$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}\left(2\left(\frac{\vec{a} + \vec{c}}{2}\right) - 2\left(\frac{\vec{b} + \vec{d}}{2}\right)\right) $

Simplifying the equation:

$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = -\frac{k}{2}(\vec{b} + \vec{d} - \vec{a} - \vec{c}) $

Since both sides of the equation are equal:

$ k = -4 $

$ \begin{aligned} & |\vec{a}|=2 \\\\ & |\vec{b}|=3 \\\\ & \vec{a} \cdot \vec{b}=\frac{\pi}{4} \end{aligned} $

$ \begin{aligned} & |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\ & = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\ & = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2 \\\\ & = |-7 \vec{a} \times \vec{b}|^2 \\\\ & = \left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2 \end{aligned} $

= $ 49 \times 4 \times 9 \times \frac{1}{2}=882 $

$ \begin{aligned} & C A+A B=C B \\\\ & \Rightarrow +2 i+4 j+(\delta+3) k=\alpha i+\beta j+\gamma k \\\\ & \Rightarrow \alpha=+2, \beta=4, \gamma=\delta+3 \end{aligned} $

$ \begin{aligned} \text { Area } & =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|=\left| {{1 \over 2}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 2} & 1 & 3 \cr { + 2} & 4 & \gamma \cr } } \right|} \right|=5 \sqrt{6} \\\\ & =(\gamma-12)^2+(6+2 \gamma)^2+100=(10 \sqrt{6})^2 \\\\ & \Rightarrow 5 \gamma^2=320 \end{aligned} $

$ \begin{aligned} \Rightarrow\gamma^2 =64 \\\\ \Rightarrow\gamma =8 \\\\ \Rightarrow\delta =5 \\\\ \overrightarrow{C B} \cdot \overrightarrow{C A} & =(2 i+4 j+8 k)(4 i+3 j+5 k) \\\\ & =8+12+40 \\\\ & =60 \end{aligned} $

Given that $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$, we can rewrite this as:

$(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$

This implies that the vector $\vec{d} - \vec{c}$ is a scalar multiple of $\vec{b}$:

$\vec{d} = \vec{c} + \lambda \vec{b}$

Also, we are given that $\vec{d} \cdot \vec{a} = 24$:

$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24$

Now, we can find the value of $\lambda$ :

$\lambda = \frac{24 - \vec{a} \cdot \vec{c}}{\vec{b} \cdot \vec{a}} = \frac{24 - 6}{9} = 2$

Therefore, we have :

$\vec{d} = \vec{c} + 2(\vec{b}) = 8\hat{i} - 5\hat{j} + 18\hat{k}$

Now, we can find the squared magnitude of $\vec{d}$ :

$|\vec{d}|^2 = 64 + 25 + 324 = 413$

$ \left|\begin{array}{lll} a & 1 & 1 \\\\ 1 & \mathrm{~b} & 1 \\\\ 1 & 1 & \mathrm{c} \end{array}\right|=0 $

$ \mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1 $

$ \begin{aligned} & \left|\begin{array}{lll} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{array}\right|=0 \\\\ & a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0 \\\\ & a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0 \end{aligned} $

$ \begin{aligned} & \frac{\mathrm{a}}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\ & \Rightarrow-1+\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\ & \Rightarrow \frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1 \end{aligned} $

The given vectors are :

$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$

$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$

We are given that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0$ which implies $(\vec{a} + \vec{b}) \times \vec{c} = 0$. So, $\vec{c}$ is in the direction of $\vec{a} + \vec{b}$.

Let's denote $\vec{c} = \alpha (\vec{a} + \vec{b})$.

Substituting values for $\vec{a}$ and $\vec{b}$, we get :

$\vec{c} = \alpha((\lambda + 3) \hat{i} + \hat{k})$

From the conditions $\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$, we can form two equations :

$\alpha \lambda(\lambda + 3) - \alpha = -17 \quad \text{and} \quad \alpha(3\lambda + 9 + 2) = -20.$

By solving the above equations, we find $\alpha = -1$ and $\lambda = 3$.

Substituting these values back into $\vec{c}$ gives $\vec{c} = -6 \hat{i} - \hat{k}$.

Next, we need to find $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$.

This simplifies to :

$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2$

Calculate the cross product $\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})$ which gives :

$ =\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1 \end{array}\right|=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} $

Finally, the square of the magnitude of this vector is:

$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2 = (1)^2 + 3^2 + (-6)^2 = 1 + 9 + 36 = 46.$

$ \begin{aligned} & {[\vec{b}-\vec{a} \,\,\,\,\,\vec{c}-\vec{a} \,\,\,\,\,\vec{d}-\vec{a}]=0} \\\\ & (\vec{b}-\vec{a}) \cdot[(\vec{c}-\vec{a}) \times(\vec{d}-\vec{a})]=0 \\\\ & (\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{d}-\vec{c} \times \vec{a}-\vec{a} \times \vec{d})=0 \\\\ & {[\vec{b}\,\,\,\,\, \vec{c} \,\,\,\,\,\vec{d}]-[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]-[\vec{b} \,\,\,\,\,\vec{a} \,\,\,\,\,\vec{d}]-[\vec{a} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]=0} \\\\ & \therefore [\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{c}]=[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{c}] \\\\ & \quad=[\vec{d} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]+[\vec{b} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{a}]+[\vec{c} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{b}] \end{aligned} $

We have,

$ \begin{aligned} & 10\left|a_i\right|<1, i=1,2,3 \\\\ & \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\ & |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\ & \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \text {. } \end{aligned} $

Hence, (A) is true.

$ \begin{array}{rlrl} & |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2} \leq \sqrt{a_1^2+a_1^2+a_1^2} \\\\ & =\sqrt{3}\left|a_1\right| \\\\ \therefore & |\vec{a}|=\sqrt{3}\left|a_1\right|<3\left|a_1\right| \\\\ \therefore & |\vec{a}|<3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \end{array} $

Hence, (B) is also true.

We have, $\vec{a}$ is non-zero vector parallel to the line of intersection of the two planes described by $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$.

Let $\mathbf{n}_1$ and $\mathbf{n}_2$ are the normal vector to the plane $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$, respectively.

$ \begin{aligned} & n_1=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ & n_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{array}\right|=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \vec{a}=\lambda\left|n_1 \times n_2\right| \end{aligned} $

[ $\because \mathbf{a}$ is parallel to line of intersection of both planes]

$ \begin{aligned} & =\lambda\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{array}\right| \\\\ & =\lambda(-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

$\vec{a} \cdot \vec{b} =6 [Given]$

$\lambda(0+4+2) =6 $

$\lambda =1$

$ \begin{aligned} \therefore \vec{a} & =-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ \cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{6}{\sqrt{4+4} \sqrt{4+4+1}} \\\\ & =\frac{1}{\sqrt{2}} \end{aligned} $

$ \therefore \theta=\frac{\pi}{4} $

$ \begin{aligned} \vec{a} \times \vec{b} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 2 \\ 2 & -2 & 1 \end{array}\right| \\\\ & =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\ |\vec{a} \times \vec{b}| & =\sqrt{4+16+16}=6 \\\\ \text { Hence, }(\theta,|\vec{a} \times \vec{b}|) & =\left(\frac{\pi}{4}, 6\right) \end{aligned} $

If $\vec{d}$ is $\perp$ to both $\vec{a}$ and $\vec{b}$ then

$ \vec{d}=\lambda(\vec{a} \times \vec{b})=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{array}\right|=(35 \hat{i}-13 \hat{j}-21 \hat{k}) \lambda $

$ \begin{aligned} & \text { but } \vec{c} \cdot \vec{d}=12 \Rightarrow \lambda(35 \times 1+13 \times 1-21 \times 2)=12 \\\\ & \Rightarrow \lambda(6)=12 \Rightarrow \lambda=2 \\\\ & \vec{\lambda}=2(35 \hat{i}-13 \hat{j}-21 \hat{k}) \end{aligned} $

$ \begin{aligned} & \text { Now, }(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d}) \\\\ & =\left|\begin{array}{ccc} -1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{array}\right| \\\\ & =-94+182-44=44 \end{aligned} $

1. **Circumcenter $ P $**:

The circumcenter of a triangle is equidistant from the vertices of the triangle. It is the center of the circumcircle, the circle that passes through all three vertices of the triangle.

2. **Orthocenter $ Q $**:

The orthocenter of a triangle is the point of intersection of its three altitudes. An altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side (or its extension).

3. **Centroid $ G $**:

The centroid of a triangle is the point of intersection of its medians. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The centroid always divides each median in a 2:1 ratio, with the larger segment being closer to the vertex.

With the above definitions, it's known that the centroid divides the line segment joining the circumcenter and orthocenter in the ratio 2 : 1, meaning :

$ \overrightarrow{PG} = \frac{2}{3} \overrightarrow{PQ} $

$ \overrightarrow{PQ} = 3\overrightarrow{PG} $

The position vector of the centroid $G$ in terms of the vertices $A, B,$ and $C$ is :

$ \overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} $

Because $P$ is the circumcenter and is at the origin in this problem:

$ \overrightarrow{PA} = \overrightarrow{A} $

$ \overrightarrow{PB} = \overrightarrow{B} $

$ \overrightarrow{PC} = \overrightarrow{C} $

Substituting these into the equation for $G$ :

$ \overrightarrow{PG} = \frac{\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}}{3} $

Now, using the relationship between $PQ$ and $PG$ established earlier :

$ \overrightarrow{PQ} = 3\overrightarrow{PG} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} $

Given, the position vector of point P is : $ \overrightarrow{OP} = -\widehat{i} - 2\widehat{j} + 3\widehat{k} $

Position vectors of points A, B, and C are :

$ \overrightarrow{OA} = -2\widehat{i} + \widehat{j} - 3\widehat{k} $

$ \overrightarrow{OB} = 2\widehat{i} + 4\widehat{j} - 2\widehat{k} $

$ \overrightarrow{OC} = -4\widehat{i} + 2\widehat{j} - \widehat{k} $

Now, vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ can be calculated as :

$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} $

$ = (2 + 2)\widehat{i} + (4 - 1)\widehat{j} - (-2 + 3)\widehat{k} $

$ = 4\widehat{i} + 3\widehat{j} - \widehat{k} $

$ \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} $

$ = (-4 + 2)\widehat{i} + (2 - 1)\widehat{j} - (-1 + 3)\widehat{k} $

$ = -2\widehat{i} + \widehat{j} - 2\widehat{k} $

$ \begin{aligned} \text { Now, } \overrightarrow{A B} \times \overrightarrow{A C} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(5)+\hat{\mathbf{j}}(-2-8)+\hat{\mathbf{k}}(4+6) \\\\ & =5 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \\\\ &\overrightarrow{O P} =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

To find the projection of $ \overrightarrow{OP} $ onto $ \overrightarrow{AB} \times \overrightarrow{AC} $, we need to find the dot product between $ \overrightarrow{OP} $ and the normalized vector $ \overrightarrow{AB} \times \overrightarrow{AC} $.

First, find the magnitude of $ \overrightarrow{AB} \times \overrightarrow{AC} $ :

$ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{5^2 + (-10)^2 + 10^2} $

$ = \sqrt{25 + 100 + 100} $

$ = \sqrt{225} $

$ = 15 $

The projection of vector $ \overrightarrow{OP} $ onto a vector perpendicular to both $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ (which is $ \overrightarrow{AB} \times \overrightarrow{AC}$) is given by :

$ \frac{\overrightarrow{OP} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})}{| \overrightarrow{AB} \times \overrightarrow{AC} |} $

= $ \frac{(-\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (5\hat{i} - 10\hat{j} + 10\hat{k})}{\sqrt{25 + 100 + 100}} $

$ = \frac{-5 + 20 + 30}{15} $

$ = \frac{45}{15} = 3 $

An arc $P Q$ of a circle subtends a right angle at its centre $O$. The mid-point of an $\operatorname{arc} P Q$ is $R$. So, $P R=R Q$

Also given that, $\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $, and $\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \overrightarrow v $



Let $\overrightarrow {OP} = \overrightarrow u$

$ \begin{aligned} & \overrightarrow{OQ}=\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v} \\\\ & \overrightarrow{O R}=\overrightarrow{v} \end{aligned} $

Clearly, $\overrightarrow{q}=\hat{\mathbf{j}}$

$ \overrightarrow{u}=\hat{\mathbf{i}}, \overrightarrow{v}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}} $

Now, given, $\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v}$

$ \begin{array}{ll} \Rightarrow & \hat{\mathbf{j}}=\alpha(\hat{\mathbf{i}})+\beta\left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}\right) \\\\ \Rightarrow & \hat{\mathbf{j}}=\mathbf{i}\left(\alpha+\frac{\beta}{\sqrt{2}}\right)+\frac{\beta}{\sqrt{2}} \hat{\mathbf{j}} \end{array} $

On comparing both sides, we get

$ \begin{array}{rlrl} \alpha+\frac{\beta}{\sqrt{2}} =0 \text { and } \frac{\beta}{\sqrt{2}}=1 \\\\ \Rightarrow \alpha =-\frac{\beta}{\sqrt{2}} \quad \therefore \beta =\sqrt{2} \\\\ \Rightarrow \alpha =-1, \beta^2=2 \end{array} $

Now, $\alpha$ and $\beta^2$ are roots of the quadratic equation is $x^2-x-2=0$

Since, $\vec{u}_1, \vec{u}_2, \vec{u}_3$ are coplanar.

So, $\left[\begin{array}{lll}\vec{u}_1 & \vec{u}_2 & \vec{u}_3\end{array}\right]=0$

$ \begin{aligned} & \Rightarrow\left|\begin{array}{lll} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{array}\right|=0 \\\\ & \Rightarrow 1(b-1)-1(1-c)+a(1-b c)=0 \\\\ & \Rightarrow b-1-1+c+a-a b c=0 \\\\ & \Rightarrow a+b+c-2=a b c ........... (i) \end{aligned} $

Also, $\left[\begin{array}{lll}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{array}\right]=0$

$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} a+b & c & c \\ a & b+c & a \\ b & b & c+a \end{array}\right|=0 \\ & \Rightarrow(a+b)\left[b c+b a+c^2+c a-a b\right]-c\left[a c+a^2-a b\right] \\\\ & \quad+c\left[a b-b^2-b c\right]=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow a b c+a c^2+a^2 c+b^2 c+b c^2+a b c-a c^2-a^2 c \\\\ & +a b c+a b c-b^2 c-b c^2=0 \\\\ & \Rightarrow 4 a b c=0 \Rightarrow a b c=0 \\\\ & \begin{array}{lr} \text { So, } a+b+c-2=0 [from (i)]\\\\ \Rightarrow a+b+c=2 \end{array} \\\\ & \Rightarrow 6(a+b+c)=12 \end{aligned} $

$ \begin{aligned} & \text { Here } \overrightarrow{\mathrm{AC}}=(-2-2) \hat{i}+(-3-1) \hat{j}+(5-1) \hat{k} \\\\ & =-4 \hat{i}-4 \hat{j}+4 \hat{k} \\\\ & \overrightarrow{\mathrm{BD}}=(1-1) \hat{i}+(-6-2) \hat{j}+(-7-5) \hat{k} \\\\ & =-8 \hat{j}-12 \hat{k} \end{aligned} $

So, area of quadrilateral $=\frac{1}{2}| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} \mid$

$ \begin{aligned} & =\frac{1}{2}\left\|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{array}\right\| \\\\ & =\frac{1}{2}|(48+32) \hat{i}-(48-0) \hat{j}+(32-0) \hat{k}| \\\\ & =\frac{1}{2}|80 \hat{i}-48 \hat{j}+32 \hat{k}| \\\\ & =\frac{1}{2} 16|15 \hat{i}-3 \hat{j}+2 \hat{k}| \\\\ & =8 \sqrt{25+9+4}=8 \sqrt{38} \text { sq units. } \end{aligned} $

Given : Points with position vectors

$ \alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k} $

and $\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$ are collinear.

So, $\frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11+8}$

$ \begin{aligned} & \Rightarrow \frac{2(\alpha-6)}{3}=\frac{-1}{11-\beta}=\frac{2}{19} \\\\ & \Rightarrow \frac{2}{3}(\alpha-6)=\frac{2}{19} \end{aligned} $

$ \begin{aligned} & \Rightarrow 19 \alpha-114=3 \Rightarrow 19 \alpha=117 \\\\ & \Rightarrow \alpha=\frac{117}{19} \end{aligned} $

And, $\frac{-1}{11-\beta}=\frac{2}{19}$

$ \begin{aligned} & \Rightarrow-19=22-2 \beta \\\\ & \Rightarrow 2 \beta=41 \\\\ & \Rightarrow \beta=\frac{41}{2} \end{aligned} $

$ \begin{aligned} & \therefore(19 \alpha-6 \beta)^2=\left(19 \times \frac{117}{19}-\frac{6 \times 41}{2}\right)^2 \\\\ & =(117-123)^2=36 \end{aligned} $

Given that the volume $V$ of the parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is represented by the scalar triple product $[\vec{a},\vec{b},\vec{c}]$, which is the determinant of the 3 x 3 matrix with vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as its rows (or columns).

When the vectors representing the coterminous edges of the parallelepiped are $\vec{a}$, $\vec{b} + \vec{c}$, and $\vec{a} + 2\vec{b} + 3\vec{c}$, the volume $V$ of the parallelepiped is represented by :

$\begin{aligned} & V=[\vec{a}, \vec{b}+\vec{c}, \vec{a}+2 \vec{b}+3 \vec{c}] \\\\ & =\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3\end{array}\right|\left[\begin{array}{ll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] \\\\ & =1(3-2)[\vec{a}, \vec{b}, \vec{c}] \\\\ & =\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=V \end{aligned}$

Let $\overrightarrow{O A}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

$ \begin{aligned} & \overrightarrow{O B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\ & \overrightarrow{O C}=(a+1) \hat{\mathbf{i}}+2 \hat{\mathbf{k}} \end{aligned} $

and $ \overrightarrow{O D}=9 \hat{\mathbf{i}}+(a-8) \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

$ \begin{array}{ll} &\therefore \overrightarrow{A B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \overrightarrow{A C}=a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ &\text { and } \overrightarrow{A D}=8 \hat{\mathbf{i}}+(a-6) \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{array} $

Since, given point are coplanar

$ \begin{aligned} & \therefore \quad[\overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D}]=0 \\\\ & \Rightarrow\left|\begin{array}{ccr} 1 & -1 & 1 \\ a & 2 & -1 \\ 8 & a-6 & 3 \end{array}\right|=0 \\\\ & \Rightarrow 1(6+a-6)+1(3 a+8)+1\left(a^2-6 a-16\right)=0 \\\\ & \Rightarrow a+3 a+8+a^2-6 a-16=0 \Rightarrow a^2-2 a-8=0 \\\\ & \Rightarrow a^2-4 a+2 a-8=0 \Rightarrow a(a-4)+2(a-4)=0 \\\\ & \Rightarrow(a-4)(a+2)=0 \Rightarrow a=4,-2 \\\\ & \therefore \text { Sum of all values of } a=4-2=2 \end{aligned} $

Given, position vectors of the points $A, B, C$ and $D$ be

$5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

$ \begin{aligned} \overrightarrow{A B} & =(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}})=-4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+(3-2 \lambda) \hat{\mathbf{k}} \\\\ \overrightarrow{A C} & =(-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\ & =-7 \hat{\mathbf{i}}+(\lambda-5) \hat{\mathbf{j}}+(4-2 \lambda) \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \text { and } \overrightarrow{A D} & =(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\ & =-6 \hat{\mathbf{i}}+(6-2 \lambda) \hat{\mathbf{k}} \end{aligned} $

Since, points $A, B, C$ and $D$ are coplanar

$ \begin{aligned} & \therefore [ { \overrightarrow{AB}~ \overrightarrow{AC}~ \overrightarrow{AD}] }=0 \\\\ & \Rightarrow \left|\begin{array}{ccc} -4 & -3 & (3-2 \lambda) \\ -7 & (\lambda-5) & (4-2 \lambda) \\ -6 & 0 & 6-2 \lambda \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow -6\left\{(-12+6 \lambda)-\left(13 \lambda-15-2 \lambda^2\right)\right\} \\\\ & +(6-2 \lambda)\{-4 \lambda+20-21\} =0 \\\\ & \Rightarrow -6\left(2 \lambda^2-7 \lambda+3\right)+(6-2 \lambda)(-4 \lambda-1) =0 \\\\ & \Rightarrow -12 \lambda^2+42 \lambda-18+8 \lambda^2-22 \lambda-6 =0 \\\\ & \Rightarrow -4 \lambda^2+20 \lambda-24 =0 \\\\ & \Rightarrow \lambda^2-5 \lambda+6 =0 \\\\ & \Rightarrow (\lambda-2)(\lambda-3) =0 \\\\ & \Rightarrow \lambda =2,3 \end{aligned} $

$ \therefore \sum\limits_{\lambda \varepsilon S}(\lambda+2)^2=(2+2)^2+(3+2)^2=16+25=41 $

Given vectors :

$ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} $

$ \vec{b} = \hat{i} - 2\hat{j} - 2\hat{k} $

$ \vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k} $

Since $ \vec{d} $ is perpendicular to both $ \vec{b} $ and $ \vec{c} $, its direction is given by their cross product :

$ \vec{d} = \lambda(\vec{b} \times \vec{c}) $

$ \begin{aligned} \vec{b} \times \vec{c} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(-6+8)-\hat{\mathbf{j}}(3-2)+\hat{\mathbf{k}}(4-2)=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

Thus, $ \vec{b} \times \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $

Given this, $ \vec{d} $ can be expressed as :

$ \vec{d} = \lambda(2\hat{i} - \hat{j} + 2\hat{k}) $

Using the given condition that $ \vec{a} \cdot \vec{d} = 18 $ :

$ (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda(2\hat{i} - \hat{j} + 2\hat{k}) = 18 $

$ \Rightarrow 2\lambda(2) - 3\lambda(1) + 4\lambda(2) = 18 $

$ \Rightarrow \lambda(4 + 8 - 3) = 18 $

$ \Rightarrow 9\lambda = 18 $

$ \Rightarrow \lambda = 2 $

So, $ \vec{d} = 4\hat{i} - 2\hat{j} + 4\hat{k} $

Now, using the identity :

$ |\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2 $

Given $ |\vec{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29} $

and $ |\vec{d}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Using your corrected calculations :

$ |\vec{a} \times \vec{d}|^2 = 29 \times 36 - 18^2 = 1044 - 324 = 720 $

$\begin{aligned} & \text { Projection of }\vec{a} \text { on } \vec{b} =\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\\\ & = \frac{(5 \hat{i}-\hat{j}-3 \hat{k}) \cdot(\hat{i}+3 \hat{j}+5 \hat{k})}{\sqrt{1^2+3^2+5^2}}=\frac{5-3-15}{\sqrt{35}} \\\\ & = \frac{-13}{\sqrt{35}}\end{aligned}$

Negative sign indicates that direction of the projection vector is opposite to the direction of $\vec{b}$.

$\begin{aligned} & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \Rightarrow \vec{r}-\vec{c}=\lambda \vec{a}((\vec{r}-\vec{c} ) \text{and} \overrightarrow{a} \text { are parallel }) \\\\ & \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a} \\\\ & \Rightarrow \vec{r} \cdot \vec{b}=\vec{c} \cdot \vec{b}+\lambda \vec{a} \cdot \vec{b} \\\\ & 0=(1-3)+\lambda(2+5) \Rightarrow \lambda=\frac{2}{7} \\\\ & \text { Hence, } \vec{r}=\vec{c}+\frac{2 \vec{a}}{7} \\\\ & \vec{r} \Rightarrow \frac{11}{7} \hat{i}-\frac{11}{7} \hat{k} \\\\ & |\vec{r}|=\sqrt{\left(\frac{11}{7}\right)^2+\left(-\frac{11}{7}\right)^2} \Rightarrow r=\frac{11 \sqrt{2}}{7}\end{aligned}$

$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{c}}=\hat{5 \mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \vec{b}=0, \quad \vec{r} \cdot \vec{a}=0$

$\Rightarrow \vec{r}-\vec{c}=\lambda \vec{b}$

Also, $(\vec{c}+\lambda \vec{b}) \cdot \vec{a}=0$

$\Rightarrow \vec{a} \cdot \vec{c}+\lambda(\vec{a} \cdot \vec{b})=0$

$\therefore \lambda=\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}=\frac{-8}{5}$

$\overrightarrow{\mathrm{r}}=\frac{5(5 \hat{\mathrm{i}}-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})-8(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{5}$

$ \Rightarrow $ $\overrightarrow{\mathrm{r}}=\frac{17 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{5}$

$ \Rightarrow $ $|\overrightarrow{\mathrm{r}}|^{2}=\frac{1}{25}(289+50)$

$ \Rightarrow $ $25|\overrightarrow{\mathrm{r}}|^{2}=339$