Vector Algebra

Given $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$

$\vec{\mathrm{c}}=2(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-3 \vec{\mathrm{b}}$

Dot product with $\overrightarrow{\mathrm{a}}$ on both sides

$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$ ..... (1)

Dot product with $\vec{b}$ on both sides

$\begin{aligned} & \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 \quad \text{... (2)}\\ & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) \\ & |\overrightarrow{\mathrm{c}}|^2=4[12]+144 \\ & |\overrightarrow{\mathrm{c}}|^2=48+144 \\ & |\overrightarrow{\mathrm{c}}|^2=192 \\ & \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{c}} \mid} \\ & \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} \\ & \therefore \cos \theta=\frac{-48}{8 \sqrt{3} .4} \\ & \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} \\ & \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right) \end{aligned}$

Unit vector $\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$

$\begin{aligned} & \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \end{aligned}$

Now angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$

$\Rightarrow \mathrm{x}+\mathrm{z}=0$ ...... (i)

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}$

$\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \Rightarrow y+z=\frac{1}{\sqrt{2}}$ ...... (ii)

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3}$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}}$ ..... (iii)

from equation (i), (ii) and (iii) we get

$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$

$\begin{aligned} & \text { Thus } \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \\ & \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2} \end{aligned}$

Area of parallelogram, $S=|\vec{a} \times \vec{b}|$

Area of quadrilateral $=\operatorname{Area}(\triangle \mathrm{OAB})+\operatorname{Area}(\triangle \mathrm{OBC})$

$\begin{aligned} & =\frac{1}{2}\{|\vec{a} \times(12 \vec{a}+4 \vec{b})|+|\vec{b} \times(12 \vec{a}+4 \vec{b})|\} \\ & =8|(\vec{a} \times \vec{b})| \end{aligned}$

$\text { Ratio }=\frac{8|(\vec{a} \times \vec{b})|}{|(\vec{a} \times \vec{b})|}=8$

$\begin{aligned} & \vec{a}+5 \vec{b}=\lambda \vec{c} \\ & \vec{b}+6 \vec{c}=\mu \vec{a} \end{aligned}$

Eliminating $\vec{a}$

$\begin{aligned} & \lambda \overrightarrow{\mathrm{c}}-5 \overrightarrow{\mathrm{b}}=\frac{6}{\mu} \overrightarrow{\mathrm{c}}+\frac{1}{\mu} \overrightarrow{\mathrm{b}} \\ & \therefore \mu=\frac{-1}{5}, \lambda=-30 \\ & \alpha=5, \beta=30 \end{aligned}$

$\triangle \mathrm{ABC}$ is equilateral

Orthocentre and centroid will be same

$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$

Mid-point of $\mathrm{AB}$ is $\mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right)$

$\begin{aligned} & \therefore \ell_1=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}} \\ & \ell_1=\sqrt{\frac{1}{6}}=\ell_2=\ell_3 \\ & \therefore \ell_1^2+\ell_2^2+\ell_3^2=\frac{1}{2} \end{aligned}$

$\begin{aligned} & \mathrm{AB}=5 \\ & \mathrm{AC}=5 \end{aligned}$

$\therefore \mathrm{D}$ is midpoint of $\mathrm{BC}$

$\begin{aligned} & \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) \\ & \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} \\ & l=\sqrt{\frac{45}{2}} \\ & \therefore 2 l^2=45 \end{aligned}$

$\begin{aligned} & \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\ & \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)} \end{aligned}$

$\begin{aligned} & \text { given } \vec{a} \times \vec{c}=\vec{b} \\ & \Rightarrow(\vec{a} \times \vec{c}) \cdot \vec{b}=\vec{b} \cdot \vec{b}=|\vec{b}|^2=27 \\ & \Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=[\vec{a} \quad \vec{c} \quad \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27 \quad \text{.... (ii)} \end{aligned}$

$\begin{aligned} & \text { Now } \vec{a} \cdot \vec{b}=3-6+3=0 \quad \text{.... (iii)}\\ & \vec{a} \cdot \vec{c}=3 \quad \text{.... (iv) (given)} \end{aligned}$

$\begin{gathered} \text { By (i), (ii), (iii) & (iv) } \\ 27-0-3=24 \end{gathered}$

$ C=\left(\frac{344}{5}, \frac{6-6}{5}, \frac{-9+2}{5}\right) $

$ \begin{array}{l} C\left(\frac{7}{5}, 0,-\frac{7}{5}\right) \\\\ O C=\frac{7}{5} \hat{\mathbf{i}}-\frac{7}{5} \hat{\mathbf{k}} \\\\ O D=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ \mathbf{C D}=\frac{8}{5} \hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{17}{5} \hat{\mathbf{k}} \\\\ \begin{aligned} \mathbf{C D} & =\sqrt{\frac{64}{25}+1+\frac{289}{25}}=\sqrt{\frac{378}{25}}=\frac{3 \sqrt{42}}{5} \\\\ \mathbf{C D} & =\frac{\mathbf{C D}}{|\mathbf{C D}|}=\frac{\frac{1}{5}(8 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+17 \hat{\mathbf{k}})}{\frac{3 \sqrt{42}}{5}} \\\\ & =\frac{1}{3 \sqrt{42}}(8 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+17 \hat{\mathbf{k}}) \end{aligned} \end{array} $

To solve for the expression $2a^2 + 3b^2 + 4c^2$, follow these steps:

Given that $\hat{\mathbf{e}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} + c \hat{\mathbf{k}}$ is a unit vector, we have:

$ a^2 + b^2 + c^2 = 1 $

The vector $\hat{\mathbf{e}}$ is coplanar with the vectors $\hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 5 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}} + \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}$. The condition for coplanarity implies the determinant formed by these vectors should be zero:

$ \left|\begin{array}{ccc} a & b & c \\ 1 & -3 & 5 \\ 3 & 1 & -5 \end{array}\right| = 0 $

Calculating the determinant gives:

$ a(10) - b(-20) + c(10) = 0 \quad \Rightarrow \quad 10a + 20b + 10c = 0 $

Additionally, since $\hat{\mathbf{e}}$ is perpendicular to $\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$, we have:

$ a + b + c = 0 $

Solving the linear equations, we use the relation derived from coplanarity:

$ 10a + 20b + 10c = 0 \quad \Rightarrow \quad a + 2b + c = 0 $

Together with $ a + b + c = 0 $, it follows:

$ \frac{a}{1} = \frac{-b}{0} = \frac{c}{-1} = \lambda $

Thus:

$ a = \lambda, \quad b = 0, \quad c = -\lambda $

Substituting into the unit vector condition:

$ a^2 + b^2 + c^2 = 1 \quad \Rightarrow \quad \lambda^2 + 0 + (-\lambda)^2 = 1 \quad \Rightarrow \quad 2\lambda^2 = 1 $

Hence:

$ \lambda^2 = \frac{1}{2} \quad \Rightarrow \quad \lambda = \pm \frac{1}{\sqrt{2}} $

Therefore, the values for $a$, $b$, and $c$ are:

$ a = \frac{1}{\sqrt{2}}, \quad b = 0, \quad c = -\frac{1}{\sqrt{2}} $

Calculating $2a^2 + 3b^2 + 4c^2$:

$ 2\left(\frac{1}{2}\right) + 3(0) + 4\left(\frac{1}{2}\right) = 1 + 2 = 3 $

Thus, $2a^2 + 3b^2 + 4c^2 = 3$.

To find the magnitude of vector $\hat{\mathbf{d}}$, which is normal to the plane formed by vectors $\mathbf{a}$ and $\mathbf{b}$, and also satisfies $\hat{\mathbf{d}} \cdot \hat{\mathbf{c}} = 2$, we perform the following calculations:

Firstly, calculate $\mathbf{a} \times \mathbf{b}$, the cross product of vectors $\mathbf{a}$ and $\mathbf{b}$:

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

The calculation proceeds as follows:

$ \begin{aligned} \mathbf{d} = & \ \lambda \left(\hat{\mathbf{i}}(1 \cdot 1 - (-2) \cdot (-2)) - \hat{\mathbf{j}}(1 \cdot 1 - (-2) \cdot 1) + \hat{\mathbf{k}}(1 \cdot (-2) - 1 \cdot 1)\right) \\ = & \ \lambda (\hat{\mathbf{i}}(1 - 4) - \hat{\mathbf{j}}(1 + 2) + \hat{\mathbf{k}}(-2 - 1)) \\ = & \ \lambda (-3\hat{\mathbf{i}} - 3\hat{\mathbf{j}} - 3\hat{\mathbf{k}}) \\ = & \ -3\lambda(\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) \end{aligned} $

Next, use the condition $\hat{\mathbf{d}} \cdot \hat{\mathbf{c}} = 2$:

$ \mathbf{d} \cdot \mathbf{c} = -3\lambda [2 \hat{\mathbf{i}} + 1 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}}] $

Solves as:

$ \begin{aligned} -3\lambda[2 + 1 - 1] & = 2 \\ -3\lambda \times 2 & = 2 \\ -6\lambda & = 2 \\ \lambda & = -\frac{1}{3} \end{aligned} $

Substitute $\lambda$ back to determine $\mathbf{d}$:

$ \mathbf{d} = \lambda (-3(\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}})) = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} $

Finally, the magnitude of $\hat{\mathbf{d}}$ is:

$ |\mathbf{d}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} $

The position vectors and direction vectors for two lines are given:

Line $ L_1 $: $\mathbf{r} = \mathbf{a} + t\mathbf{b}$, where $\mathbf{a} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + 3\hat{\mathbf{k}}$ and $\mathbf{b} = 2\hat{\mathbf{i}} - \hat{\mathbf{j}} + \lambda \hat{\mathbf{k}}$.

Line $ L_2 $: $\mathbf{r} = \mathbf{c} + s\mathbf{d}$, where $\mathbf{c} = -\hat{\mathbf{k}}$ and $\mathbf{d} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} - \hat{\mathbf{k}}$.

To determine whether the lines are coplanar or skew, we set up equations based on their parametric forms:

$ L_1 = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ \lambda \end{pmatrix} $

$ L_2 = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} + s\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} $

By equating the components of the parametric equations, we get:

$ 1 + 2t = s $

$-1 - t = 2s$

$3 + \lambda t = -1 - s$

Solving the first two equations simultaneously gives:

$ 2 + 4t = 2s $

$ -1 - t = 2s $

$ \frac{-1 - t}{2s} = \frac{3 + 5t}{0} $

From these equations, solve for $ t $ and $ s $:

$\Rightarrow t = -\frac{3}{5}$

$\Rightarrow s = -\frac{1}{5}$ after substituting $ t $.

For the third equation:

$ 3 - \frac{3\lambda}{5} = -1 + \frac{1}{5} $

From this, solve for $\lambda$:

$ 21 - \frac{1}{5} = \frac{3\lambda}{5} $

$ \Rightarrow \lambda = \frac{19}{3} $

Thus, the lines are skew when $\lambda \neq \frac{19}{3}$. If $\lambda = \frac{19}{3}$, the lines would be coplanar.

Given that the vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ each have a magnitude of $\sqrt{2}$ and the angle between any two vectors is $\frac{\pi}{3}$, we can analyze the vectors $\mathbf{x}$ and $\mathbf{y}$ defined as follows:

$ |\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = \sqrt{2} $

First, let's express $\mathbf{x}$:

$ \mathbf{x} = \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) $

Using the vector triple product identity:

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

Given that the angle between each pair of vectors is $\frac{\pi}{3}$, the dot products are:

$ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\left(\frac{\pi}{3}\right) = \sqrt{2} \times \sqrt{2} \times \frac{1}{2} = 1 $

Therefore, the expression for $\mathbf{x}$ becomes:

$ \mathbf{x} = 1 \cdot \mathbf{b} - 1 \cdot \mathbf{c} = \mathbf{b} - \mathbf{c} $

Now, consider $\mathbf{y}$:

$ \mathbf{y} = \mathbf{b} \times (\mathbf{c} \times \mathbf{a}) $

Using the vector triple product identity again:

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

This simplifies similarly to:

$ \mathbf{y} = 1 \cdot \mathbf{c} - 1 \cdot \mathbf{a} = \mathbf{c} - \mathbf{a} $

Hence, the magnitudes of $\mathbf{x}$ and $\mathbf{y}$ are:

$ |\mathbf{x}| = |\mathbf{b} - \mathbf{c}| $

$ |\mathbf{y}| = |\mathbf{c} - \mathbf{a}| $

Since both expressions have the same form, it follows that:

$ |\mathbf{x}| = |\mathbf{y}| $

Given that the vector $\mathbf{a}$ is perpendicular to the plane formed by the non-zero vectors $\mathbf{b}$ and $\mathbf{c}$, we have:

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

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

From the problem statement, we know:

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

Expanding both sides, we get:

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

This simplifies to:

$ 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 0 $

Hence, we conclude:

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

Now, considering $|(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}| + |(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|$, we find:

$ \left|\left(|\mathbf{a}||\mathbf{b}| \sin 90^\circ \right) \cdot \mathbf{c}\right| + |(\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{a}| $

Since $\sin 90^\circ = 1$, this becomes:

$ |\mathbf{a}||\mathbf{b}||\mathbf{c}| + 0 - 0 \quad \text{(because $\mathbf{a} \cdot \mathbf{c} = 0$ and $\mathbf{b} \cdot \mathbf{c} = 0$)} $

Thus, the result is:

$ |\mathbf{a}||\mathbf{b}||\mathbf{c}| $

Given the relationships $ \mathbf{a} \times \mathbf{c} = \mathbf{b} $ and $ \mathbf{a} \cdot \mathbf{c} = 3 $, we need to find $ \mathbf{a} \cdot (\mathbf{c} \times \mathbf{b} - \mathbf{b} - \mathbf{c}) $.

First, consider:

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

We are given $ |\mathbf{b}|^2 = (\mathbf{a} \times \mathbf{c})^2 $. Thus:

$ |\mathbf{b}|^2 + 9 = |\mathbf{a}|^2 \cdot |\mathbf{c}|^2 $

Start with:

$ 27 + 9 = 6 \cdot |\mathbf{c}|^2 \implies 36 = 6 \cdot |\mathbf{c}|^2 \implies |\mathbf{c}|^2 = 6 $

We then know:

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

Substituting:

$ 6\mathbf{a} - 3\mathbf{c} = \mathbf{c} \times \mathbf{b} $

Now compute:

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

Breaking it down:

$ 6 \mathbf{a}^2 = 6 \times |\mathbf{a}|^2 = 6 \times 6 $

$ -4 (\mathbf{a} \cdot \mathbf{c}) = -4 \times 3 = -12 $

$ \mathbf{a} \cdot \mathbf{b} = 0 $ because $ \mathbf{a} \times \mathbf{c} = \mathbf{b} $ implies orthogonality of $ \mathbf{a} $ and $ \mathbf{b} $.

Substituting:

$ = 6 \times 6 - 12 - 0 = 36 - 12 - 0 = 24 $

Thus, the final result is:

$ 24 $

Position vector of

$ P=\frac{4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-6 \hat{\mathbf{k}}+4 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{3}=\frac{8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}}{3} $

Position vector of

$ \begin{aligned} Q & =\frac{8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}+2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{3} \\\\ & =\frac{10 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}}{3} \end{aligned} $

Let the point which divides $ P Q $ in ratio $ 2: 3 $ be $ M $.

Position vector of $ M $

$ \begin{array}{l} =\frac{2\left(\frac{10 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}}{3}\right)+3\left(\frac{8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}}{3}\right)}{5} \\\\ =\frac{20 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}+24 \hat{\mathbf{i}}-27 \hat{\mathbf{j}}}{15} \\\\ =\frac{44 \hat{\mathbf{i}}-33 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}}{15} \end{array} $

Line joining the points ( $ 1,-1,1 $ ) and $ (1,1,-2) $ be

$ \frac{x-1}{0}=\frac{y+1}{2}=\frac{z-1}{-3} $

Line joining the points $ (2,1,-6) $ and $ (3,-1,-7) $ is

$ \frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+6}{-1} $

For intersection of points,

$ \frac{x-1}{0}=\frac{y+1}{2}=\frac{z-1}{-3}=K $

$ \Rightarrow x=1, y=2 K-1 \text { and } Z=-3 K+1 $

$ \text { and } \frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+6}{-1}=\lambda $

$ x=\lambda+2, y=1-2 \lambda, z=-\lambda-6 $

$ \therefore \lambda=-1 $ and $ K=2 $

$ \therefore $ Positivion vector of points of intersection is $ \mathbf{i}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} $

$ \mathbf{a}=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $ and $ \mathbf{b}=6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} $

$ \begin{array}{l} \text { Required magnitude }=\frac{\mathbf{a} \cdot \mathbf{b}}{|a|} \\\\ \qquad \begin{aligned} & =\frac{24-10+6}{\sqrt{16+25+9}}=\frac{20}{\sqrt{50}}=\frac{20}{5 \sqrt{2}} \\\\ & =2 \sqrt{2} \end{aligned} \end{array} $

Let $ \mathbf{d}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} $

Given, $ |\mathbf{d}|=1 \Rightarrow x^{2}+y^{2}+z^{2}=1 \quad \ldots \text { (i) } $

$ d \cdot \mathbf{c}=0 \Rightarrow-x+2 z=0 \Rightarrow x=2 z \quad \ldots \text { (ii) } $

$ \mathbf{a , b} $ and $ \mathbf{d} $ are coplanar

$ \left|\begin{array}{ccc}2 & -1 & 0 \\\\ 0 & 2 & -1 \\\\ x & y & z\end{array}\right|=0 $

$ \begin{array}{l} 2(2 z+y)+x(l)=0 \\\\ 4 z+2 y+2 z=0 \\\\ 6 z+2 y=0 \Rightarrow y=-3 z \end{array} $

$ [\because x=2 z] $

From Eq. (i), $ z^{2}+4 z^{2}+9 z^{2}=1 $

$ \begin{aligned} 14 z^{2} & =1 \\\\ z^{2} & =\frac{1}{14} \Rightarrow z=\frac{ \pm 1}{\sqrt{14}} \end{aligned} $

If $ z=\frac{1}{\sqrt{14}}, x=\frac{2}{\sqrt{14}}, y=\frac{-3}{\sqrt{14}} $

If $ z=\frac{-1}{\sqrt{14}}, x=\frac{-2}{\sqrt{14}}, y=\frac{3}{\sqrt{14}} $

$ |d \cdot b|=\left(\frac{6}{\sqrt{14}}+\frac{1}{\sqrt{14}}\right)=\frac{7}{\sqrt{14}}=\sqrt{\frac{7}{2}} $

Given,

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors and the three points $\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c}, 2 \mathbf{a}+\lambda \mathbf{b}-2 \mathbf{c}$, $4 a+7 b-8 c$ are collinear.

$ \begin{aligned} \text { Let } P & =\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c} \\\\ Q & =2 \mathbf{a}+\lambda \mathbf{b}-2 \mathbf{c} \\\\ R & =4 \mathbf{a}+7 \mathbf{b}-8 \mathbf{c} \\\\ \mathbf{P Q} & =(2 \mathbf{a}+\lambda \mathbf{b}-2 \mathbf{c})-(\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c}) \\\\ & =(2-\lambda) \mathbf{a}+(\lambda+2) \mathbf{b}+(-3) \mathbf{c} \\\\ \mathbf{P R} & =(4 a+7 b-8 \mathbf{c})-(\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c}) \\\\ & =(4-\lambda) \mathbf{a}+9 \mathbf{b}-9 \mathbf{c} \end{aligned} $

$\because P, Q$ and $R$ collinear.

$ \begin{aligned} & \therefore \mathbf{P Q}=k \mathbf{P R} \\\\ & \begin{array}{c} \Rightarrow(2-\lambda) \mathbf{a}+(\lambda+2) \mathbf{b}+(-3) \mathbf{c} \\\\ \quad=k(4-\lambda) \mathbf{a}+9 k \mathbf{b}-9 k \mathbf{c} \\\\ \therefore \quad 2-\lambda=k(4-\lambda) \Rightarrow \lambda+2=9 k \\\\ \quad-3=-9 k \Rightarrow k=\frac{1}{3} \end{array} \end{aligned} $

Now, $\lambda+2=9\left(\frac{1}{3}\right)=3$

$ \begin{array}{lrl} \Rightarrow & 6-3 \lambda & =4-\lambda \Rightarrow 2=2 \lambda \\\\ \Rightarrow & & \lambda=1 \end{array} $

We have,

$ \begin{aligned} |\mathbf{a}| & =3,|\mathbf{b}|=4 \\\\ |\mathbf{a}+\mathbf{b}| & =\sqrt{37},|\mathbf{a}-\mathbf{b}|=k \text { and } \\\\ (\mathbf{a}, \mathbf{b}) & =0 \end{aligned} $

Now, $|\mathbf{a}+\mathbf{b}|=\sqrt{37}$

$ \begin{array}{rr} \Rightarrow & (\mathbf{a}+\mathbf{b}) \cdot(\mathbf{a}+\mathbf{b})=37 \\\\ \Rightarrow & |\mathbf{a}|^2+|\mathbf{b}|^2+2 \mathbf{a} \cdot \mathbf{b}=37 \\\\ \Rightarrow & 9+16+2|\mathbf{a}||\mathbf{b}| \cos \theta=37 \\\\ \Rightarrow & 2(3)(4) \cos \theta=12 \\\\ \Rightarrow & \cos \theta=\frac{1}{2} \\\\ \Rightarrow & \theta=60^{\circ} \end{array} $

Now, $|\mathbf{a}-\mathbf{b}|=k$

$ \begin{array}{lr} \Rightarrow (\mathbf{a}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})=k^2 \\\\ \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2-2|\mathbf{a}||\mathbf{b}| \cos (\theta)=k^2 \\\\ \Rightarrow 9+16-2(3)(4) \cos \left(60^{\circ}\right)=k^2 \end{array} $

$ \begin{aligned} & \Rightarrow 25-24\left(\frac{1}{2}\right)=k^2 \\\\ & \Rightarrow k^2=13 \\\\ & \therefore \frac{4}{13}(k \sin \theta)^2 \\\\ & =\frac{4}{13} k^2 \sin ^2\left(60^{\circ}\right) \\\\ & =\frac{4}{13}(13)\left(\frac{\sqrt{3}}{2}\right)^2=3 \end{aligned} $

Given,

$\mathbf{r}$ is a vector perpendicular to the plane determined by the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

$ \therefore \quad \mathbf{r}=\lambda\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ 2 & -1 & 0 \\\\ 0 & 1 & 2 \end{array}\right|=\lambda(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) $

Also, given that

The magnitude of the projection of $\mathbf{r}$ on the vector $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is 1

$ \begin{aligned} & \therefore \quad\left|\frac{\mathbf{r} \cdot(2 \hat{i}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{\sqrt{2^2+1^2+2^2}}\right|=1 \\\\ & \Rightarrow \quad|\lambda(-4-4+4)|=3 \Rightarrow \lambda=\frac{3}{4} \\\\ & \therefore \quad \mathbf{r}=\frac{3}{4}(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

Now,

$ \begin{aligned} |r| & =\frac{3}{4} \sqrt{(-2)^2+(4)^2+(2)^2}=\frac{3}{4} \sqrt{24} \\\\ & =\frac{3}{4}(2 \sqrt{6})=\frac{3 \sqrt{6}}{2} \end{aligned} $

Given,

$ \begin{aligned} & \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ & \cos (\mathbf{a}, \mathbf{b} \times \mathbf{c})=\sqrt{\frac{2}{3}} \text { and } \mathbf{a} \text { is a unit vectors. } \\\\ & \text { Now, } \mathbf{b} \times \mathbf{c}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ 1 & -1 & 2 \\\\ 1 & 2 & -1 \end{array}\right|=-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\\\ & \therefore|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})| \\\\ & =|\mathbf{a}||\mathbf{b} \times \mathbf{c}| \sin (\mathbf{a}, \mathbf{b} \times \mathbf{c}) \\\\ & =(1) \sqrt{(-3)^2+(3)^2+(3)^2} \sqrt{1-(\cos (\mathbf{a}, \mathbf{b} \times \mathbf{c}))^2} \\\\ & =3 \sqrt{3} \sqrt{1-\left(\sqrt{\frac{2}{3}}\right)^2}=3 \sqrt{3}\left(\frac{1}{\sqrt{3}}\right)=3 \end{aligned} $

$A B=\sqrt{(4-3)^2+(1-2)^2+(0+1)^2}$

$ \begin{aligned} & =\sqrt{1+1+1}=\sqrt{3} \\\\ A C & =\sqrt{(2-3)^2+(1-2)^2+(4+1)^2} \\\\ & =\sqrt{1+1+25} \\\\ & =\sqrt{27}=3 \sqrt{3} \end{aligned} $

By angle bisector theorem,

$ \frac{B D}{C D}=\frac{A B}{A C}=\frac{\sqrt{3}}{3 \sqrt{3}}=\frac{1}{3}=1: 3 $

The point $D$ divides the segment $B C$ internally in the ratio $1: 3$

$\therefore$ Coordinate of $D$ will be

$ \left(\frac{4(3)+2(1)}{1+3}, \frac{1(3)+1(1)}{1+3}, \frac{0(3)+4(1)}{1+3}\right) $

i.e. $\left(\frac{7}{2}, 1,1\right)$

Given, $D \equiv(p, q, r)$

$\therefore \quad p=\frac{7}{2}, q=1$ and $r=1$

Now, $\sqrt{2 p+q+r}=\sqrt{2\left(\frac{7}{2}\right)+1+1}=\sqrt{9}=3$

Given, $(3,0,2)$ and $(0,2, k)$ are the direction ratios of two lines and $\theta$ is the angle between them.

$ \begin{aligned} & \therefore \cos \theta=\frac{3 \times 0+0 \times 2+2 \times k}{\sqrt{3^2+0^2+2^2} \cdot \sqrt{0^2+2^2+k^2}} \\\\ & \Rightarrow|\cos \theta|=\left|\frac{2 k}{\sqrt{13\left(4+k^2\right)}}\right| \\\\ & \Rightarrow\left|\frac{2 k}{\sqrt{13\left(4+k^2\right)}}\right|=\frac{6}{13}\left[\text { given, }|\cos \theta|=\frac{6}{13}\right] \\\\ & \Rightarrow \quad \frac{4(k)^2}{13\left(4+k^2\right)}=\left(\frac{6}{13}\right)^2 \Rightarrow \frac{4 k^2}{4+k^2}=\frac{36}{13} \\\\ & \Rightarrow \quad 52 k^2=144+36 k^2 \\\\ & \Rightarrow \quad 16 k^2=144 \Rightarrow k^2=9 \\\\ & \Rightarrow \quad k= \pm 3 \end{aligned} $

We have,

$ A(\hat{i}-2 \hat{j}+\hat{k}) $

$ B(2 \hat{i}-\hat{j}+\hat{k}) \quad C(\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) $

Position vector of

$ \begin{aligned} D & =\frac{2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{2} \\\\ & =\frac{3}{2}(\hat{\mathbf{i}}-\hat{\mathbf{k}}) \end{aligned} $

Position vector of

$ \begin{aligned} E & =\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{2} \\\\ & =\frac{1}{2}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\\\ \therefore \mathrm{DE}= & \frac{1}{2}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}})-\frac{3}{2}(\hat{\mathbf{i}}-\hat{\mathbf{k}}) \\\\ = & -\frac{1}{2} \hat{\mathbf{i}}-\frac{3}{2} \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

$\Rightarrow$ Unit vector along

$\begin{aligned} \mathrm{DE} & =\frac{\left(-\frac{1}{2} \hat{\mathbf{i}}-\frac{3}{2} \hat{\mathbf{j}}+\hat{\mathbf{k}}\right)}{\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{-3}{2}\right)^2+(1)^2}} \\ & =\frac{1}{\sqrt{14}}(-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\end{aligned}$

$ \begin{aligned} & \text { Let } \mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ & \therefore \quad|a|=3,|b|=3 \end{aligned} $

A vector along bisectors $=\frac{\mathbf{a}}{|\mathbf{a}|} \pm \frac{b}{|b|}$

$ \begin{aligned} & =\frac{2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}}{3} \pm \frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{3} \\\\ & =\frac{3 \hat{\mathbf{i}}+3 \hat{\mathbf{k}}}{3}, \frac{\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{3} \end{aligned} $

$\begin{aligned} \therefore \text { Required vector }=\hat{\mathbf{i}} & +\hat{\mathbf{k}} \\\\ & (\because \text { Magnitude }=\sqrt{2})\end{aligned}$

$ \begin{aligned} & \cos \theta=\frac{(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})}{\sqrt{4^2+(-1)^2+2^2} \cdot \sqrt{1^2+3^2+\left(-2^2\right.}} \\\\ & \begin{aligned} \cos \theta & =\frac{4-3-4}{\sqrt{21} \cdot \sqrt{14}}=\frac{-3}{7 \sqrt{6}} \Rightarrow \sin \theta=\sqrt{\frac{285}{294}} \\\\ \text { So, } \sin 2 \theta & =2 \sin \theta \cos \theta \\\\ & =2 \cdot \frac{\sqrt{285}}{7 \sqrt{6}}\left(\frac{-3}{7 \sqrt{6}}\right)=\frac{-\sqrt{285}}{49} \end{aligned} \end{aligned} $

We have, $|a|=3,|b|=2 \sqrt{2}$ and $|c|=5$

$\because \mathrm{c}$ is perpendicular to the plane of a and b .

$ \begin{aligned} & \begin{aligned} & \therefore \quad k(\mathbf{a} \times \mathbf{b}) \Rightarrow \mathbf{c} \perp \mathbf{a}, \mathbf{c} \perp \mathbf{b} \\\\ & \text { Now, }|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=|\mathbf{a}|^2+|b|^2+|c|^2+2 \end{aligned} \\\\ & \quad(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\\\ & =9+8+25+2\left(3 \times 2 \sqrt{2} \cdot \sin \frac{\pi}{4}+0+0\right) \\\\ & |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=42+12=54 \\\\ & \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}|=3 \sqrt{6} \end{aligned} $

Let $A(\lambda \mathbf{a}+3 \mathbf{b}-\mathbf{c}), B(\mathbf{a}-\lambda \mathbf{b}+3 \mathbf{c})$

$ \begin{aligned} & \mathbf{c}(3 \mathbf{a}+4 \mathbf{b}-\lambda \mathbf{c}) \text { and } D(\mathbf{a}-6 \mathbf{b}+6 \mathbf{c}) \\\\ & \mathbf{A D}=(1-\lambda) \mathbf{a}-9 \mathbf{b}+7 \mathbf{c} \\\\ & \mathbf{B D}=(\lambda-6) \mathbf{b}+3 \mathbf{c} \\\\ & \mathbf{C D}=-2 \mathbf{a}-10 \mathbf{b}+(6+\lambda) \mathbf{c} \end{aligned} $

$\because A, B, C$ and $D$ are coplanar

$ \Rightarrow\left|\begin{array}{ccc} 1-\lambda & -9 & 7 \\\\ 0 & \lambda-6 & 3 \\\\ -2 & -10 & 6+\lambda \end{array}\right|=0 $

$\begin{aligned} & \Rightarrow \quad(1-\lambda)\left[\lambda^2-36+30\right] \\\\ & -2[-27-7 \lambda+42]=0 \\\\ & \Rightarrow \quad(1-\lambda)\left(\lambda^2-6\right)-2(15-7 \lambda)=0 \\\\ & \Rightarrow \quad \lambda^2-6-\lambda^3+6 \lambda-30+14 \lambda=0 \\\\ & \Rightarrow \quad-\lambda^3+\lambda^2+20 \lambda-36=0 \\\\ & \Rightarrow \quad \lambda^3-\lambda^2-20 \lambda+36=0 \\\\ & \Rightarrow \quad(\lambda-2)\left(\lambda^2+\lambda-18\right)=0 \Rightarrow \lambda=2\end{aligned}$

Given the vectors $ a \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}, \hat{\mathbf{i}} + b \hat{\mathbf{j}} + \hat{\mathbf{k}} $, and $ \hat{\mathbf{i}} + \hat{\mathbf{j}} + c \hat{\mathbf{k}} $, we need to verify that they are coplanar. When vectors are coplanar, it means they can be expressed as a linear combination of the other vectors.

Thus, we can write:

$ a \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} = x (\hat{\mathbf{i}} + b \hat{\mathbf{j}} + \hat{\mathbf{k}}) + y (\hat{\mathbf{i}} + \hat{\mathbf{j}} + c \hat{\mathbf{k}}) $

where $ x $ and $ y $ are scalars, and not both zero.

Expanding both sides, we get:

$ a \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} = (x+y) \hat{\mathbf{i}} + (bx + y) \hat{\mathbf{j}} + (x + cy) \hat{\mathbf{k}} $

Equating the coefficients of $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$, we derive:

$ a = x + y \tag{1} $

$ 1 = bx + y \tag{2} $

$ 1 = x + cy \tag{3} $

Now, from Equation (1), we have:

$ 1-a = 1 - (x + y) $

Substitute the rearranged form of Equation (2):

$ 1-b = 1 - (bx + y) = \frac{x-1+y}{x} $

Substitute from Equation (3):

$ 1-c = 1 - (x + cy) = \frac{y-1+x}{y} $

Calculate the desired expression:

$ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{1}{1-(x+y)} + \frac{x}{x-1+y} + \frac{y}{y-1+x} $

By simplifying:

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

Thus, the expression simplifies to $1$.

Given, $\mathbf{A B}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$

$ \mathrm{BC}=6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} $

We know that $\mathbf{A B}+\mathbf{B C}+\mathbf{C A}=0$

$ \begin{array}{ll} \Rightarrow & (8 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})=\mathbf{A C} \\ \therefore & |\mathbf{A B}|=\sqrt{4+9+36}=7 \\ & |\mathbf{B C}|=\sqrt{36+4+9}=7 \\ \text { and } & |\mathbf{A C}|=\sqrt{64+1+9}=\sqrt{74} \end{array} $

$\therefore$ Perimeter of $\triangle A B C$ is

$ 7+7+\sqrt{74}=14+\sqrt{74} $

To find the orthogonal projection of vector $\mathbf{a} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}$ on vector $\mathbf{b} = \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}$, we use the formula for the projection of $\mathbf{a}$ onto $\mathbf{b}$:

$ \text{Projection}_{\mathbf{b}}(\mathbf{a}) = \frac{(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}}{|\mathbf{b}|^2} $

Step-by-step, this involves:

Calculate the dot product $\mathbf{a} \cdot \mathbf{b}$:

$ \mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(-2) + (3)(1) = 2 - 6 + 3 = -1 $

Find the magnitude squared of $\mathbf{b}$, $ |\mathbf{b}|^2 $:

$ |\mathbf{b}|^2 = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6 $

Substitute these values into the projection formula:

$ \text{Projection}_{\mathbf{b}}(\mathbf{a}) = \frac{-1(\hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + \hat{\mathbf{k}})}{6} = \frac{-\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - \hat{\mathbf{k}}}{6} $

Thus, the orthogonal projection vector of $\mathbf{a}$ on $\mathbf{b}$ is:

$ \frac{-\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - \hat{\mathbf{k}}}{6} $

Given vectors are:

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

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

First, we compute $2 \mathbf{a}$ and $\frac{\mathbf{b}}{2}$:

$ 2 \mathbf{a} = -8 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} + 8 \hat{\mathbf{k}} $

$ \frac{\mathbf{b}}{2} = \frac{1}{\sqrt{2}} \hat{\mathbf{i}} - \frac{1}{\sqrt{2}} \hat{\mathbf{j}} $

To find the angle between $2 \mathbf{a}$ and $\frac{\mathbf{b}}{2}$, we use the dot product formula:

$ \cos \theta = \frac{2 \mathbf{a} \cdot \frac{\mathbf{b}}{2}}{|2 \mathbf{a}|\left|\frac{\mathbf{b}}{2}\right|} $

This simplifies to:

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

Calculating the dot product $\mathbf{a} \cdot \mathbf{b}$:

$ \begin{aligned} \mathbf{a} \cdot \mathbf{b} &= (-4)(\sqrt{2}) + (2)(-\sqrt{2}) + (4)(0) \\ &= -4\sqrt{2} - 2\sqrt{2} \\ &= -6\sqrt{2} \end{aligned} $

Next, we find the magnitudes of $\mathbf{a}$ and $\mathbf{b}$:

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

$ |\mathbf{b}| = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 $

Substitute these into the cosine formula:

$ \cos \theta = \frac{-6\sqrt{2}}{6 \times 2} = \frac{-6\sqrt{2}}{12} = -\frac{1}{\sqrt{2}} $

The angle $\theta$ is calculated as:

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

Since $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ corresponds to $\pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$:

$ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} = 135^{\circ} $

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}}) $