Vector Algebra

Given vectors $ \mathbf{a} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}} $ and $ \mathbf{b} = 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} $, we need to find a unit vector that is perpendicular to both $ \mathbf{a} $ and $ \mathbf{b} $. This can be found by determining the cross product $ \mathbf{a} \times \mathbf{b} $.

First, let's compute the cross product $ \mathbf{a} \times \mathbf{b} $:

$ \begin{aligned} \mathbf{a} \times \mathbf{b} & = \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 4 \\ 0 & 3 & 2 \end{array}\right| \\ & = \hat{\mathbf{i}}(3 \cdot 2 - 4 \cdot 3) - \hat{\mathbf{j}}(2 \cdot 2 - 0 \cdot 4) + \hat{\mathbf{k}}(2 \cdot 3 - 0 \cdot 3) \\ & = \hat{\mathbf{i}}(6 - 12) - \hat{\mathbf{j}}(4 - 0) + \hat{\mathbf{k}}(6 - 0) \\ & = -6 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 6 \hat{\mathbf{k}} \end{aligned} $

Next, we need to find the magnitude of the cross product $ \mathbf{a} \times \mathbf{b} $:

$ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-6)^2 + (-4)^2 + 6^2} = \sqrt{36 + 16 + 36} = \sqrt{88} = 2 \sqrt{22} $

Thus, the unit vector perpendicular to both $ \mathbf{a} $ and $ \mathbf{b} $ is derived by normalizing the cross product:

$ \begin{aligned} \text{Unit vector} & = \pm \frac{-6 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 6 \hat{\mathbf{k}}}{2 \sqrt{22}} \\ & = \frac{-3 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}}{\sqrt{22}} \quad \text{or} \quad \frac{3 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 3 \hat{\mathbf{k}}}{\sqrt{22}} \end{aligned} $

Therefore, the required unit vectors are $ \frac{-3 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}}{\sqrt{22}} $ or $ \frac{3 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 3 \hat{\mathbf{k}}}{\sqrt{22}} $.

Given that the vectors $a \hat{\mathbf{i}} + \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}$, $4 \hat{\mathbf{i}} + 5 \hat{\mathbf{j}} + \hat{\mathbf{k}}$, and $4 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 6 \hat{\mathbf{k}}$ are coplanar, we can determine the value of $a$ by setting the determinant of their coefficients to zero.

The determinant is:

$ \begin{vmatrix} a & 1 & 3 \\ 4 & 2 & 6 \\ 4 & 5 & 1 \\ \end{vmatrix} = 0 $

Expanding the determinant, we have:

$ a(2 \times 1 - 6 \times 5) - 1(4 \times 1 - 6 \times 4) + 3(4 \times 5 - 2 \times 4) = 0 $

Calculating each term gives:

$ a(2 - 30) - 1(4 - 24) + 3(20 - 8) = 0 $

$ a(-28) + 20 + 36 = 0 $

This simplifies to:

$ -28a + 56 = 0 $

Solving for $a$, we get:

$ 28a = 56 \quad \Rightarrow \quad a = 2 $

Thus, the value of $a$ is $2$.

Given:

$|\mathbf{a}| = 2$

$|\mathbf{b}| = 3$

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

The parallelogram has adjacent sides:

$\mathbf{p} = 2\mathbf{a} + 3\mathbf{b}$

$\mathbf{q} = \mathbf{a} - \mathbf{b}$

The diagonals of the parallelogram are described by $\mathbf{p} + \mathbf{q}$ and $\mathbf{p} - \mathbf{q}$.

Calculate $\mathbf{p} + \mathbf{q}$:

$ \mathbf{p} + \mathbf{q} = (2\mathbf{a} + 3\mathbf{b}) + (\mathbf{a} - \mathbf{b}) = 3\mathbf{a} + 2\mathbf{b} $

Calculate $|\mathbf{p} + \mathbf{q}|^2$:

$ \begin{aligned} |\mathbf{p} + \mathbf{q}|^2 &= |3\mathbf{a} + 2\mathbf{b}|^2 \\ &= 9|\mathbf{a}|^2 + 4|\mathbf{b}|^2 + 12 (\mathbf{a} \cdot \mathbf{b}) \\ &= 9 \times 4 + 4 \times 9 + 12 \times |\mathbf{a}| |\mathbf{b}| \cos \theta \\ &= 36 + 36 + 12 \times 2 \times 3 \times \frac{1}{2} \\ &= 36 + 36 + 36 \\ &= 108 \end{aligned} $

Therefore, the length of the shorter diagonal, $|\mathbf{p} + \mathbf{q}|$, is:

$ \sqrt{108} = 6 \sqrt{3} $

Given,

and

$ \begin{aligned} & \mathbf{v}_1=x^2 \hat{\mathbf{i}}+2 x \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{v}_2=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+x \hat{\mathbf{k}} \end{aligned} $

The angle between $\mathbf{v}_1$ and $\mathbf{v}_2$ is obtuse.

Thus, $\mathbf{v}_1 \cdot \mathbf{v}_2<0$

$ \begin{aligned} & \Rightarrow\left(x^2-4 x+x\right)<0 \Rightarrow x^2-3 x<0 \\ & \Rightarrow \quad x(x-3)<0 \\ & \end{aligned} $

$ x \in(0,3) $

To find the projection of the sum of vectors $\mathbf{a}$ and $\mathbf{b}$ on a vector perpendicular to their plane, follow these steps:

Calculate $\mathbf{a} + \mathbf{b}$:

$ \mathbf{a} = 3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}, \quad \mathbf{b} = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $

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

$ = 5 \hat{\mathbf{i}} + 5 \hat{\mathbf{j}} - 7 \hat{\mathbf{k}} $

Find $\mathbf{a} \times \mathbf{b}$ (the cross product):

$ \mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & -5 \\ 2 & 1 & -2 \end{array} \right| $

$ = \hat{\mathbf{i}}(-8+9) - \hat{\mathbf{j}}(-6+10) + \hat{\mathbf{k}}(3-8) $

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

Normalize the perpendicular vector $\mathbf{n} = -3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}$:

First, find the magnitude $|\mathbf{n}|$:

$ |\mathbf{n}| = \sqrt{(-3)^2 + (-4)^2 + (-5)^2} = \sqrt{9+16+25} = \sqrt{50} = 5 \sqrt{2} $

The unit vector $\hat{\mathbf{n}}$ is:

$ \hat{\mathbf{n}} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{-3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}}{5 \sqrt{2}} $

$ = \frac{-3}{5 \sqrt{2}} \hat{\mathbf{i}} - \frac{4}{5 \sqrt{2}} \hat{\mathbf{j}} - \frac{5}{5 \sqrt{2}} \hat{\mathbf{k}} $

Projection of $\mathbf{a} + \mathbf{b}$ on $\hat{\mathbf{n}}$:

$ (\mathbf{a} + \mathbf{b}) \cdot \hat{\mathbf{n}} = (5 \hat{\mathbf{i}} + 5 \hat{\mathbf{j}} - 7 \hat{\mathbf{k}}) \cdot \left(\frac{-3}{5 \sqrt{2}} \hat{\mathbf{i}} - \frac{4}{5 \sqrt{2}} \hat{\mathbf{j}} - \frac{5}{5 \sqrt{2}} \hat{\mathbf{k}}\right) $

Calculate the dot product, which results in 0. Thus, the projection of the sum of the vectors $\mathbf{a}$ and $\mathbf{b}$ on the vector perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$ is 0.

In triangle $ \triangle PQR $, the vertices have the following position vectors:

$ P(4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 6 \hat{\mathbf{k}}) $

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

$ R(3 \hat{\mathbf{i}} + \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) $

The angle bisector of a vertex in a triangle divides the opposite side into segments proportional to the lengths of the other two sides. To find the position vector $ I $ of the intersection of the angle bisector from vertex $ P $ with side $ QR $, we use the formula:

$ I = \frac{|PR| \cdot Q + |PQ| \cdot R}{|PR| + |PQ|} $

First, calculate the lengths of $ PQ $ and $ PR $:

$ |PQ| = \sqrt{(4-2)^2 + (3-2)^2 + (6-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} $

$ |PR| = \sqrt{(4-3)^2 + (3-1)^2 + (6-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} $

Since $ |PQ| = |PR| $, the intersection point $ I $ is the midpoint of $ Q $ and $ R $:

$ I = \frac{Q + R}{2} = \frac{(2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) + (3 \hat{\mathbf{i}} + \hat{\mathbf{j}} + 3 \hat{\mathbf{k}})}{2} $

$ I = \frac{5 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 6 \hat{\mathbf{k}}}{2} $

$ I = \frac{5}{2} \hat{\mathbf{i}} + \frac{3}{2} \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} $

To find the projection vector of $\hat{\mathbf{f}}$ onto $\hat{\mathbf{g}}$, we start with the given vectors:

$ \hat{\mathbf{f}} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} $

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

The formula for the orthogonal projection of $\hat{\mathbf{f}}$ on $\hat{\mathbf{g}}$ is:

$ \text{Projection of } \hat{\mathbf{f}} \text{ on } \hat{\mathbf{g}} = \frac{\hat{\mathbf{f}} \cdot \hat{\mathbf{g}}}{|\hat{\mathbf{g}}|^2} \cdot \hat{\mathbf{g}} $

First, calculate the dot product $\hat{\mathbf{f}} \cdot \hat{\mathbf{g}}$:

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

Next, find the magnitude squared of $\hat{\mathbf{g}}$:

$ |\hat{\mathbf{g}}|^2 = (2)^2 + (-1)^2 + (3)^2 = 4 + 1 + 9 = 14 $

Now use these results to find the projection:

$ \text{Projection of } \hat{\mathbf{f}} \text{ on } \hat{\mathbf{g}} = \frac{4}{14}(2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 3\hat{\mathbf{k}}) $

Simplify the fraction:

$ = \frac{2}{7}(2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 3\hat{\mathbf{k}}) $

Thus, the projection vector of $\hat{\mathbf{f}}$ on $\hat{\mathbf{g}}$ is:

$ \frac{2}{7}(2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 3\hat{\mathbf{k}}) $

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.

$\mathbf{a}+3 \mathbf{b}+4 \mathbf{c}=\mathbf{a}(x+y+6 z)+\mathbf{b}(-2 x+5 y+14 z) +\mathbf{c}(3 x-2 y+4 z)$

(because $a, b$ and $c$ are non-coplanar vectors)

Comparing both sides,

$\begin{aligned} & x+y+6 z=1 \quad \text{.... (i)}\\ & -2 x+5 y+14 z=3 \quad \text{.... (ii)}\\ & 3 x-2 y+4 z=4 \quad \text{.... (iii)} \end{aligned}$

By solving Eqs. (i), (ii) and (iii), we get

$\begin{aligned} & x=-2, y=-3, z=1 \\ & x+y+z=-4 \end{aligned}$

Let the vectors of magnitude $a, 2 a, 3 a$ are along $O P, O Q, O R$, respectively.

Then, vectors are $O P, O Q, O R$ are $a\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right), 2 a\left(\frac{\hat{j}+\hat{k}}{\sqrt{2}}\right), 3 a\left(\frac{\hat{k}+\hat{i}}{\sqrt{2}}\right)$, respectively.

Their resultant say $R$ is given by

$\begin{aligned} & \mathbf{R}=a\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)+2 a\left(\frac{\hat{j}+\hat{k}}{\sqrt{2}}\right)+3 a\left(\frac{\hat{k}+\hat{i}}{\sqrt{2}}\right) \\ & =\frac{a}{\sqrt{2}}(4 \hat{i}+3 \hat{j}+5 \hat{k}) \\ & \therefore|\mathbf{R}|=\sqrt{\frac{a^2}{2}(16+9+25)}=5 a \\ & \end{aligned}$

$\begin{aligned} \text { Let } \mathbf{a} & =x \hat{i}+y \hat{j}+z \hat{k}, \mathbf{b}=3 \hat{i}+6 \hat{j}+6 \hat{k} \\ \mathbf{a} \cdot \mathbf{b} & =27 \end{aligned}$

According to question, a is a collinear with b, then

$\begin{array}{rlrl} & \quad \mathbf{a} =\lambda \mathbf{b} \quad \text{... (i)}\\ \Rightarrow & \mathbf{a} \cdot \mathbf{b} =27 \\ \Rightarrow & \lambda \mathbf{b} \cdot \mathbf{b} =27 \quad \text{[from Eq. (i)]}\\ \Rightarrow & \lambda|\mathbf{b}|^2 =27 \end{array}$

$\begin{aligned} & \lambda=\frac{27}{\left(\sqrt{(3)^2+(6)^2+(6)^2}\right)^2}=\frac{27}{(9)^2} \\ & \lambda=\frac{1}{3} \\ & \mathrm{a}=\frac{1}{3}(3 \hat{i}+6 \hat{j}+6 \hat{k}) \Rightarrow \mathbf{a}=\hat{i}+2 \hat{j}+2 \hat{k} \end{aligned}$

$\begin{aligned} \text { So, } |\mathbf{a}| & =\sqrt{1+(2)^2+(2)^2} \\ & =\sqrt{9}=3 \text { units } \end{aligned}$

As given $\mathbf{a}$ is perpendicular to $\mathbf{b}$ and $\mathbf{c}$

$\Rightarrow \mathbf{a} \cdot \mathbf{b}=0$ and $\mathbf{a} \cdot \mathbf{c}=0$ and angle between $\mathbf{b}$ and $\mathbf{c}=\frac{\pi}{3}$

$\therefore \mathbf{b} \cdot \mathbf{c}=|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3}=1 \cdot 1 \cdot \frac{1}{2}$

$=\frac{1}{2}$ [$\because$ b and c are unit vectors]

Now, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=|a|^2+|b|^2+|c|^2 +2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})$

$\begin{aligned} & =1+1+1+2\left(0+\frac{1}{2}+0\right)=1+1+1+1=4 \\ & |a+b+c|=\sqrt{4}=2 \text { units } \end{aligned}$

Given,

$\begin{aligned} \mathbf{F} & =2 i+2 j+5 k, A=(1,2,5) \\ B & =(-1,-2,-3) \\ \mathbf{B A} & =\mathbf{A}-\mathbf{B}=(\hat{i}+2 \hat{j}+5 \hat{k})-(-\hat{i}-2 \hat{j}-3 \hat{k}) \\ \mathbf{B A} & =2 \hat{i}+4 \hat{j}+8 \hat{k} \\ \mathbf{B A} & \times \mathbf{F}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 8 \\ 2 & 2 & 5 \end{array}\right| \\ & =\hat{i}(20-16)-\hat{j}(10-16)+\hat{k}(4-8) \\ & =4 \hat{i}+6 \hat{j}-4 \hat{k} \end{aligned}$

So, $4 \hat{i}+6 \hat{j}-4 \hat{k}=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$

$\begin{array}{ll} \because \quad 2 \lambda=-4 \\ \qquad\lambda=-2 \end{array}$

$O A B C$ is a tetrahedron. If $D$ and $E$ are the mid-points of $O A$ and $B C$ respectively.

Thus, take $O A=a, O B=b, O C=c$

Thus, $O D=\frac{\mathbf{a}}{2}$

$\mathrm{OE}=\frac{\mathrm{b}+\mathbf{c}}{2}$

Thus, $\mathrm{DE}=\mathrm{OE}-\mathrm{OD}$

$\begin{aligned} & =\frac{b+c}{2}-\frac{a}{2}=\frac{b+c-a}{2} \\ & =\frac{O B+O C-O A}{2} \end{aligned}$

$\begin{aligned} & \text { Given, } \mathbf{a}+\mathbf{b}+\mathbf{c}=0 \text { and } \\ &|\mathbf{a}|=7,|\mathbf{b}|=5,|\mathbf{c}|=3 \\ \Rightarrow & \mathbf{b}+\mathbf{c}=-\mathbf{a} \Rightarrow|\mathbf{b}+\mathbf{c}|^2=|-\mathbf{a}|^2 \\ \Rightarrow & |\mathbf{b}|^2+|\mathbf{c}|^2+2 \cdot \mathbf{b} \cdot \mathbf{c}=|\mathbf{a}|^2 \\ \Rightarrow & 25+9+2 \cdot \mathbf{b} \cdot \mathbf{c}=49 \\ \Rightarrow & 2 \mathbf{b}=15 \Rightarrow 2|\mathbf{b}| \cdot|\mathbf{c}| \cos \theta=15 \\ \Rightarrow & \cos \theta=\frac{1}{2}=\cos 60^{\circ} \\ \therefore & \theta=60^{\circ} \end{aligned}$

Given, OP $\cdot \hat{i}=-1 \Rightarrow P_x=-1$

$P$ lies on $y=2^{x+2}$

$\therefore \quad y=2^{-1+2}=2$

Thus, $(P x, P y)=(-1,2)$

and also given

$O Q \cdot \hat{i}=2 \Rightarrow Q_x=2$

$\begin{array}{rlrl} \text { Qlies on } y =2^{x+2} \\ \therefore & y =16 \\ \therefore \quad\left(Q_x, Q_y\right) & =(2,16) \end{array}$

Thus $\mathbf{O P}=-\hat{i}+2 \hat{j}$ and $\mathbf{O Q}=2 \hat{i}+16 \hat{j}$

$\text { OQ }-4 \mathbf{O P}=2 \hat{i}+16 \hat{j}+4 \hat{i}-8 \hat{j}=6 \hat{i}+8 \hat{j}$

$\mathrm{AB}=\mathrm{a}, \mathrm{BC}=\mathbf{b}, \mathrm{DA}=\mathbf{a}-\mathbf{b}$

In $\triangle A B C$, using vector concept, we have

$\begin{aligned} & & \mathbf{A B}+\mathbf{B C} & =\mathbf{A C} \\ \Rightarrow & & \mathbf{A C} & =\mathbf{a}+\mathbf{b} \quad \text{.... (i)} \end{aligned}$

In $,$\triangle A D C$,

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

In $\triangle M D C$,

$\begin{aligned} & M D+D C=M C \\ \Rightarrow & M D=\frac{b}{2}-2 a \end{aligned}$

[$\because M$ is mid-point of $BC$]

In $\triangle A D X$,

$\begin{aligned} & \mathbf{A D}+\mathbf{D X}=\mathbf{A X} \\ & \mathbf{A D}+\frac{4}{5} \mathbf{D M}=\mathbf{A X} \\ \Rightarrow \quad \mathbf{A X} & =\mathbf{b}-\mathbf{a}+\frac{4}{5}\left(2 \mathbf{a}-\frac{\mathbf{b}}{2}\right) \\ & =\mathbf{b}-\mathbf{a}+\frac{8}{5} \mathbf{a}-\frac{2}{5} \mathbf{b}=\frac{3}{5}(\mathbf{a}+\mathbf{b}) \\ \Rightarrow \quad \mathbf{A X} & =\frac{3}{5} \mathbf{A C} \quad \text{[from Eq. (i)]} \end{aligned}$

which says that $\mathbf{A X} \| \mathbf{A C}$ and one point $A$ is common there.

Therefore, $A, X$ and $C$ are collinear.

$\begin{aligned} &\begin{array}{rrr} \text { } & (3 a-5 b) \perp(2 a+b) \\ \Rightarrow & \quad(3 a-5 b)(2 a+b)=0 \\ \Rightarrow & 6 a \cdot a+3 a \cdot b-10 b \cdot a-5 b \cdot b=0 \\ \Rightarrow & 6|a|^2+3 a \cdot b-10 a \cdot b-5|b|^2=0 \\ \Rightarrow & 6|a|^2-5|b|^2=7 a \cdot b \quad \text{.... (i)} \end{array}\\ &\text { Also, }(a+4 b) \perp(-a+b) \end{aligned}$

$\begin{array}{rr} \Rightarrow & (a+4 b) \cdot(-\mathbf{a}+\mathbf{b})=0 \\ \Rightarrow & -\mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}-4 \mathbf{b} \cdot \mathbf{a}+4 \mathbf{b} \cdot \mathbf{b}=0 \\ \Rightarrow & -|\mathbf{a}|^2+\mathbf{a} \cdot \mathbf{b}-4 \mathbf{a} \cdot \mathbf{b}+4|\mathbf{b}|^2=0 \\ \Rightarrow & -|\mathbf{a}|^2+4|\mathbf{b}|^2=3 \mathbf{a} \cdot \mathbf{b} \quad \text{... (ii)} \end{array}$

From Eqs. (i) and (ii), we have $ \begin{aligned} & \frac{1}{7}\left(6|\mathbf{a}|^2-5|\mathbf{b}|^2\right)=\frac{1}{3}\left(-|\mathbf{a}|^2+4|\mathbf{b}|^2\right) \\ \Rightarrow & 18|\mathbf{a}|^2-15|\mathbf{b}|^2=-7|\mathbf{a}|^2+28|\mathbf{b}|^2 \\ \Rightarrow & 25|\mathbf{a}|^2=43|\mathbf{b}|^2 \\ \because \quad & 3 \mathbf{a} \cdot \mathbf{b}=-|\mathbf{a}|^2+4|\mathbf{b}|^2 \quad \text{[from Eq. (ii)]}\\ & 3|\mathbf{a}||\mathbf{b}| \cos \theta=-\frac{43}{25}|\mathbf{b}|^2+4|\mathbf{b}|^2 \\ \Rightarrow & 3|\mathbf{a}||\mathbf{b}| \cos \theta=\frac{57}{25}|\mathbf{b}|^2 \\ \Rightarrow & \left.3 \sqrt{\frac{43}{25}}|| \mathbf{b}\right|^2 \cos \theta=\frac{57}{25}|\mathbf{b}|^2 \\ \Rightarrow & \frac{\sqrt{43}}{5} \cos \theta=\frac{19}{25} \Rightarrow \cos \theta=\frac{19}{5 \sqrt{43}} \\ \Rightarrow & \theta=\cos ^{-1}\left(\frac{19}{5 \sqrt{43}}\right) \end{aligned}$

$\begin{aligned} & \mathbf{a}=x \hat{i}+y \hat{j}+z \hat{k}, \\ & \quad x=2 y \text { and }|\mathbf{a}|=5 \sqrt{2} \\ & \therefore \quad x=|a| \cos \alpha \\ & \Rightarrow \quad x=5 \sqrt{2} \cos \alpha, \\ & y=5 \sqrt{2} \cos \beta \end{aligned}$

$\begin{aligned} & \text { and } z=5 \sqrt{2} \cos \gamma=5 \sqrt{2} \cos \left(135^{\circ}\right) \\ & =5 \sqrt{2}\left(-\frac{1}{\sqrt{2}}\right)=-5 \end{aligned}$

Now, $x=2 y$

$\begin{aligned} & \text { Now, } x=2 y \\ & \Rightarrow \quad 5 \sqrt{2} \cos \alpha=2 \times 5 \sqrt{2} \cos \beta \\ & \Rightarrow \quad \cos \alpha=2 \cos \beta \\ & \text { As, } \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad 4 \cos ^2 \beta+\cos ^2 \beta+\frac{1}{2}=1 \\ & \Rightarrow \quad \cos ^2 \beta=\frac{1}{10} \\ & \Rightarrow \quad \cos \beta=\frac{1}{\sqrt{10}} \\ & \therefore \quad x=2 y=2(5 \sqrt{2} \cos \beta)=2\left(5 \sqrt{2} \times \frac{1}{\sqrt{10}}\right)=2 \sqrt{5} \\ & \therefore \quad y=\frac{x}{2}=\sqrt{5} \\ & \therefore \quad a=2 \sqrt{5} \hat{i}+\sqrt{5} \hat{j}-5 \hat{k} \end{aligned}$

Position vectors of the vertices of the $\triangle A B C$ are $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$.

$\therefore$ Centroid of $\triangle A B C$ is $\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}$

We know, when through the vertices, lines are drawn parallel to the sides to form a new triangle, then obtained such triangle will have the same centroid as the original triangle, which is also clear from the above graphical figure.

$\therefore$ Centroid of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}$.

$\begin{aligned} \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \mathbf{c} & =x \hat{\mathbf{i}}+(x-2) \hat{\mathbf{j}}-\hat{\mathbf{k}}\end{aligned}$

Given, $\mathbf{c}$ lies in the plane of $\mathbf{a}$ and $\mathbf{b}$.

Then, $[\mathbf{a}, \mathbf{b}, \mathbf{c}]=0$

$\begin{aligned} & \Rightarrow \quad[\mathbf{a}, \mathbf{b}, \mathbf{c}]=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{array}\right|=0 \\ & 1(1-2(x-2)-\mathrm{l}(-1-2 x)+\mathrm{l}(x-2+x)=0 \\ & \Rightarrow \quad 1-2 x+4+1+2 x+x-2+x=0 \\ & \Rightarrow \quad 2 x+4=0 \\ & \Rightarrow \quad x=-2 \\ \end{aligned}$

The equation of line passing through $u$ and $v$ is $=\vec{u}+\lambda(\vec{v}-\vec{u})$

$\begin{aligned} \mathbf{r} & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}) \\ & =(2+\lambda) \hat{\mathbf{i}}+(1-6 \lambda) \hat{\mathbf{j}} \end{aligned}$

For the point $P$,

$\begin{aligned} & \frac{5}{2} \hat{\mathbf{i}}-2 \hat{\mathbf{j}}=(2+\lambda) \hat{\mathbf{i}}+(1-6 \lambda) \hat{\mathbf{j}} \\ & \Rightarrow \quad 2+\lambda=\frac{5}{2} \Rightarrow \lambda=\frac{1}{2} \\ & \text { and } \quad 1-6 \lambda=1-6 \times \frac{1}{2}=-2 \\ \end{aligned}$

Thus, point $P$ lie in the line through $u$ and $v$.

For the point $Q$

$\begin{aligned} & 2+\lambda=7 / 3 \Rightarrow \lambda=1 / 3 \\ & \text{and} \quad 1-6 \lambda=-1 \end{aligned}$

Thus, point $Q$ also lie in the line passes through $u$ and $v$.

Points $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}),(2 \hat{\mathbf{i}}-\hat{\mathbf{j}})$ i.e. $(1,2)$ and $(2,-1)$.

Line joining points $(1,2)$ and $(2,-1)$ is

$y-2=\frac{-1-2}{2-1}(x-1)$

or $\quad y-2=-3 x+3$

$y+3 x=5$ ..... (i)

Points $(-\hat{\mathbf{i}}),(2 \hat{\mathbf{i}}) \text { i.e. }(-1,0),(2,0)$

Line joining points $(-1,0)(2,0)$ is

$y-0 =\frac{0-0}{2+1}(x+1)$

$\Rightarrow \quad y =0$ ..... (ii)

Solve Eqs. (i) and (ii) for $x$ and $y$, we get

$y=0, x=5 / 3$

$\therefore$ Point of intersection $\left(\frac{5}{3}, 0\right)$,

i.e. $\frac{5}{3} \hat{\mathbf{i}}+0 \hat{\mathbf{j}}=\frac{5}{3} \hat{\mathbf{i}}$

$\begin{aligned} & \frac{(\mathbf{a} \times \mathbf{b})^2+(\mathbf{a} \cdot \mathbf{b})^2}{2|\mathbf{a}|^2|\mathbf{b}|^2} \\ & =\frac{(|\mathbf{a}| \cdot|\mathbf{b}| \sin \theta \hat{\mathbf{n}})^2+(|\mathbf{a}||\mathbf{b}| \cos \theta)^2}{2|\mathbf{a}|^2|\mathbf{b}|^2} \\ & =\frac{|\mathbf{a}|^2|\mathbf{b}|^2 \sin ^2 \theta \hat{\mathbf{n}} \cdot \hat{\mathbf{n}}+|\mathbf{a}|^2|\mathbf{b}|^2 \cos ^2 \theta}{2|\mathbf{a}|^2|\mathbf{b}|^2} \\ & =\frac{\sin ^2 \theta+\cos ^2 \theta}{2}=\frac{1}{2}\end{aligned}$

$\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}$

Let $\mathbf{d}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}$

Given, $\mathbf{a} \cdot \mathbf{d}=0$

$ \begin{aligned} \Rightarrow & (\hat{\mathbf{i}}-\hat{\mathbf{j}}) \cdot(p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}) =0 \\ \Rightarrow & p-q =0 \Rightarrow p=q \end{aligned}$

Given, $[\mathbf{b} \mathbf{c} \mathbf{d}]=0$

$\begin{array}{ll} \Rightarrow \left|\begin{array}{ccc} 0 & \mathrm{l} & -\mathrm{l} \\ -\mathrm{l} & 0 & \mathrm{l} \\ p & p & r \end{array}\right|=0 \\ \Rightarrow(-1)(-r-p)-1(-p)=0 \\ \Rightarrow r+p+p=0 \\ \Rightarrow r=-2 p \\ \therefore \mathbf{d}=p \hat{\mathbf{i}}+p \hat{\mathbf{j}}-2 p \hat{\mathbf{k}} \end{array}$

Also, $|\mathbf{d}|=1$

$\begin{aligned} & \Rightarrow \quad \sqrt{p^2+p^2+(-2 p)^2}=1 \\ & \Rightarrow \quad 6 p^2=1 \\ & \Rightarrow \quad p= \pm \frac{1}{\sqrt{6}} \\ & \therefore \quad d=\frac{ \pm \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{6}} \\ \end{aligned}$

$\begin{aligned}|\mathbf{u} \times \mathbf{v}| & =\| \mathbf{u}|\cdot| \mathbf{v}|\sin \theta \mathbf{n}| \\ & =\| \mathbf{u}|\cdot||\mathbf{v}| \sin 45^{\circ} \hat{\mathbf{n}} \mid \\ & =|| \mathbf{u}|\cdot| \mathbf{v}\left|\cdot \frac{1}{\sqrt{2}} \hat{\mathbf{n}}\right| \\ & =|\mathbf{u}| \cdot|\mathbf{v}| \cos 45^{\circ} \cdot 1=\mathbf{u} . \mathbf{v}\end{aligned}$

$ \begin{aligned} & \text { } \mathbf{a}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ & \text { Let } \left.\mathbf{b}_1=k(3 \hat{\mathbf{i}}-\hat{\mathbf{j}})\right], \mathbf{b}_2=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}} \\ & \quad \mathbf{b}=\mathbf{b}_1+\mathbf{b}_2 \\ & 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}=k(3 \hat{\mathbf{i}}-\hat{\mathbf{j}})+(p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}) \\ & 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}=(3 k+p) \hat{\mathbf{i}}+(-k+q) \hat{\mathbf{j}}+r \hat{\mathbf{k}} \\ & \Rightarrow \quad 3 k+p=2-k+q=1 \\ & \text { and } r=-3 \quad .....(\text{i})\\ & \text { Given that, } \mathbf{a} \cdot \mathbf{b}_2=0 \\ & \Rightarrow \quad 3 p-q=0 \quad .....(\text{ii}) \end{aligned}$

Using Eq. (ii) in Eq. (i) $3 p=q$, we obtain

$ \begin{aligned} & 3 k+p=2 \\ & 3 p-k=1 \end{aligned}$

Solve for $k$ and $p$, we get

$\begin{aligned} k & =\frac{1}{2}, p=\frac{1}{2} \\ \because \quad q & =3 p \Rightarrow q=\frac{3}{2} \\ \quad \mathbf{b}_2 & =\frac{1}{2} \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned}$