Vector Algebra

$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right) + \overrightarrow b \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right) \times \left( {\overrightarrow a - \overrightarrow b } \right)$

$ = \left( {\overrightarrow a \left( {\overrightarrow a .\,\overrightarrow b } \right) - \overrightarrow b \left( {\overrightarrow a .\,\overrightarrow a } \right) + \overrightarrow a \left( {\overrightarrow b .\,\overrightarrow b } \right) - \overrightarrow b \left( {\overrightarrow a .\,\overrightarrow b } \right)} \right) \times \left( {\overrightarrow a - \overrightarrow b } \right)$

$ = \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow a - \overrightarrow a \times \overrightarrow b } \right) - \left( {\overrightarrow a .\,\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right) + \left( {\overrightarrow b .\,\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow a - \overrightarrow a \times \overrightarrow b } \right) - \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right)$

$ = \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) - \left( {\overrightarrow a .\,\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) + \left( {\overrightarrow b .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) - \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)$

$ = \left( {\overrightarrow b \times \overrightarrow a } \right)\left( {\overrightarrow b .\,\overrightarrow b - \overrightarrow a .\,\overrightarrow a } \right)$

$ = \left( {5\overrightarrow b \times \overrightarrow a } \right)\left( {5 + {\lambda ^2} - 13 - {\lambda ^2}} \right)$

$ = 8\left( {\overrightarrow a \times \overrightarrow b } \right)$

$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \lambda & 2 & { - 3} \cr 1 & { - \lambda } & 2 \cr } } \right|$

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

$ \Rightarrow \lambda = 1$

${\left| {\overrightarrow a \times \left( {\overrightarrow a - \overrightarrow b } \right) + \overrightarrow b \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right|^2}$

$ = {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} = 4\,.\,35 = 140$

$\overrightarrow b .\overrightarrow c = \overrightarrow b .(2\overline a \times \overrightarrow b ) - 3\overline b .\overrightarrow b $

$ = 0 - 3|\overline b {|^2} = - 48$

$\widehat n = \alpha \overrightarrow b - \overrightarrow a $

$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = \left( {\overrightarrow c \,.\,\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c \,.\,\overrightarrow a } \right)\overrightarrow b $

$ = 12\overrightarrow a - \left( {\overrightarrow c \,.\,\left( {\alpha \overrightarrow b - \widehat n} \right)} \right)\overrightarrow b $

$ = 12\overrightarrow a - (12\alpha - 0)\overrightarrow b $

$ = 12\left( {\overrightarrow a - \alpha \overrightarrow b } \right)$

$\therefore$ $\left| {\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right| = 12$

Let $A \equiv (1,2,3),B \equiv (2,3,4),C \equiv (1,5,7)$

$\overrightarrow n = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{ i & j & k \cr 1 & 1 & 1 \cr 0 & 3 & 4 \cr } } \right|$

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

$\widehat {OP} = {{ \pm (\widehat i - 4\widehat j + 3\widehat k)} \over {\sqrt {26} }}$

Since $\cos \beta > 0$, take $-$ sign

$\widehat {OP} = {{\widehat i - 4\widehat j + 3\widehat k} \over {\sqrt {26} }}$

$ \Rightarrow \cos \alpha < 0,\cos \gamma < 0$

$\alpha ,\gamma \in \left( {{\pi \over 2},\pi } \right)$

$(\overrightarrow r - \overrightarrow c ) \times \overrightarrow b = 0$

$\overrightarrow r = \lambda \overrightarrow b + \overrightarrow c $

$ \Rightarrow \lambda \overrightarrow b \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a = 0$

$ \Rightarrow \lambda (3) + (7 + 8) = 0$

$ \Rightarrow \lambda = - 5$

$\overrightarrow r = 5\overrightarrow b + \overrightarrow c $

$ = - 5\widehat i - 5\widehat j - 5\widehat k + (7\widehat i + 3\widehat j + 4\widehat k)$

$ = 2\widehat i - 8\widehat j - \widehat k$

$\therefore$ $\overrightarrow r \,.\,\overrightarrow c = 17 + 24 - 4 = 34$

$[\matrix{ {\overrightarrow c } & {\overrightarrow a } & {\overrightarrow b } \cr } ] = - 25$

Let $\overrightarrow c = l\widehat i + n\widehat j + n\widehat k$

$\left| {\matrix{ l & m & n \cr 4 & 3 & 0 \cr 3 & { - 4} & 5 \cr } } \right| = - 25$

$ \Rightarrow 3l - 4m - 5n = - 5$ ..... (i)

$\overrightarrow c \,.\,(\widehat i + \widehat j + \widehat k) = 4$

$ \Rightarrow l + m + n = 4$ ..... (ii)

${{\overrightarrow c \,.\,\overrightarrow a } \over {|\overrightarrow a |}} = 1 \Rightarrow \overrightarrow c \,.\,\overrightarrow a = 5$

$ \Rightarrow 4l + 3m = 5$ ...... (iii)

Using (i), (ii) and (iii)

$l = 2,m = - 1,n = 3$

Now, ${{\overrightarrow c \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = {{25} \over {5\sqrt 2 }} = {5 \over {\sqrt 2 }}$

$\therefore$ Option (2) is correct.

First, we write $\overrightarrow{b}$ as a linear combination of $\overrightarrow{a}$ and $\overrightarrow{j}$ since $\overrightarrow{b}$ is a rotation of $\overrightarrow{a}$ about the y-axis.

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

$\overrightarrow{b}$ is orthogonal to $\overrightarrow{a}$ due to the right angle rotation, so $\overrightarrow{b} \cdot \overrightarrow{a} = 0$.

This implies :

$\begin{aligned} & (-\hat{i}+2 \hat{j}+\hat{k}) \cdot(-\lambda \hat{i}+(2 \lambda+\mu) \hat{j}+\lambda \hat{k})=0 \\\\ & \lambda+2(2 \lambda+\mu)+\lambda=0 \Rightarrow 6 \lambda+2 \mu=0 \Rightarrow \mu+3 \lambda=0\end{aligned}$

$ \therefore $ $\vec{b}=\lambda \vec{a}-3 \lambda \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})-3 \lambda \hat{j}=\lambda(-\hat{i}-\hat{j}+\hat{k})$

The magnitude of $\overrightarrow{b}$ is the same as the magnitude of $\overrightarrow{a}$ because a rotation doesn't change the magnitude of a vector. This gives us :

$ |\vec{b}|=\sqrt{3}|\lambda|=\sqrt{6}[\because|a|=\sqrt{6}] \Rightarrow|\lambda|=\sqrt{2} \Rightarrow \lambda \neq \sqrt{2} $

as for this value of $\lambda$ angle between $b$ and $y$-axis is not acute.

Therefore $\lambda=-\sqrt{2}$

Thus, we have :

$\overrightarrow{b} = -\sqrt{2}(-\hat{i} - \hat{j} + \hat{k}) = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$

Then, we find the vector $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ :

$3\overrightarrow{a} + \sqrt{2}\overrightarrow{b} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}) $

$= -3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} - 2\hat{k} = -\hat{i} + 8\hat{j} + \hat{k}$

The projection of this vector onto $\overrightarrow{c}$ is given by the dot product divided by the magnitude of $\overrightarrow{c}$ :

$\frac{(-\hat{i} + 8\hat{j} + \hat{k}) \cdot (5\hat{i} + 4\hat{j} + 3\hat{k})}{\sqrt{5^2 + 4^2 + 3^2}} = \frac{-5 + 32 + 3}{\sqrt{50}} = \frac{30}{5\sqrt{2}} = 3\sqrt{2}$

So the projection of $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ onto $\overrightarrow{c}$ is $3\sqrt{2}$ which is option D.

Let $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$

Given that, projection of $\mathbf{r}$ on $\mathbf{c}=\frac{9}{\sqrt{6}}$

$ \begin{aligned} & \text { Since, } \quad \mathbf{r} \cdot \mathbf{c}=\frac{9}{\sqrt{6}} \\ & \Rightarrow \quad \frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{9}{\sqrt{6}} \\ & \Rightarrow \frac{\mathbf{a} \cdot \mathbf{c}+\lambda \mathbf{b} \cdot \mathbf{c}}{\sqrt{1+1+4}}=\frac{9}{\sqrt{6}} \\ & \Rightarrow(2-1-2)+\lambda(1+2+2)=9 \\ & \Rightarrow \quad \lambda=2 \end{aligned} $

Hence, $\mathbf{r}=\mathbf{a}+\mathbf{2} \mathbf{b}=\mathbf{4} \mathbf{i}+3 \mathbf{j}-\mathbf{j}$

$ |\mathrm{r}|=\sqrt{16+9+1}=\sqrt{26} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } O A=a, O B=b, O C=c \\ & \therefore O G=\frac{a+b+c}{3}, O G=\frac{2 \times O S-1 \times O O}{2+1} \\ & O G=\frac{2 O S}{3} \Rightarrow O S=\frac{3}{2} O G \\ & O S=\frac{a+b+c}{2} \\ & \text { and } O A+O B+O C=a+b+c \\ & =2\left(\frac{a+b+c}{2}\right)=2 O S \\ & S A+S B+S C=O A-O S+O B-O S +O C-O S \end{aligned} \end{aligned} $

$ \begin{aligned} & =\mathbf{O A}+\mathbf{O B}+\mathbf{O C}-3 \mathbf{O B} \\ & =2 \mathbf{O S}-3 \mathbf{O S}=-\mathbf{O S}=\mathbf{S O} \\ \mathbf{G A} & +\mathbf{G B}+\mathbf{G C} \\ & =\mathbf{O A}-\mathbf{O G}+\mathbf{O B}-\mathbf{O G}+\mathbf{O C}-\mathbf{O G} \\ & =(\mathbf{a}+\mathbf{b}+\mathbf{c})-3(\mathbf{O G}) \\ & =\mathbf{a}+\mathbf{b}+\mathbf{c}-3\left(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\right) \\ & =\mathbf{0} \end{aligned} $

$ \begin{aligned} & \mathbf{a} \cdot \mathbf{a}=|\mathbf{a}|^2=5 \\ & \mathbf{b} \cdot \mathbf{b}=|\mathbf{b}|^2=5 \\ & \mathbf{c} \cdot \mathbf{c}=|\mathbf{c}|^2=5 \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &|\mathbf{a}+\mathbf{b}-\mathbf{c}|^2+|\mathbf{b}+\mathbf{c}-\mathbf{a}|^2+|\mathbf{c}+\mathbf{a}-\mathbf{b}|^2 \end{aligned} $

$ \begin{aligned} =3\left(|\mathbf{a}|^2+|\mathbf{b}|^2\right. & \left.+|\mathbf{c}|^2\right) -2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})=50 \end{aligned} $

$ \begin{aligned} & \Rightarrow \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=\frac{3(5+5+5)-50}{2} \\ & \Rightarrow \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=\frac{-5}{2} \end{aligned} $

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

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

Now, $[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-0=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

$ \begin{aligned} \Rightarrow \quad\{(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}\} \mathbf{c} & =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \Rightarrow \quad|\mathbf{c}| & =\frac{3}{|(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}|} \\ & =\frac{3}{\sin \frac{\pi}{6}}=6 \end{aligned} $

Given,

$ \begin{aligned} |a| & =4 \\ |b| & =5 \\ |a-b| & =3 \\ a \cdot b & =|a||b| \cos \theta \\ a \cdot b & =4 \times 5 \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

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

$ 9=16+25-40 \cos \theta $

[From Eq. (i)]

$ \therefore \quad \cos \theta=\frac{32}{40}=\frac{4}{5} $

Now, from $\triangle A B C$,

$ \begin{aligned} & A B=\sqrt{25-16}=3 \\ & B C=4 \Rightarrow A C=5 \end{aligned} $

$ \therefore \quad \cot \theta=\frac{\text { Base }}{\text { Perpendicular }}=\frac{4}{3} $

Hence, $\cot ^2 \theta=\frac{16}{9}$

(by squaring on both sides)

$|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7$

$ \begin{aligned} & \because a+b+c=0 \\ & \Rightarrow a+b=-c \\ & |a+b|=|-c| \end{aligned} $

[taking mod on both sides]

$|a+b|^2=|-c|^2$ [squaring on both sides]

$ \begin{gathered} |\mathbf{a}|^2+|\mathbf{b}|^2+2 \mathbf{a} \cdot \mathbf{b}=|\mathbf{c}|^2 \\ 9+25+2 \mathbf{a} \cdot \mathbf{b}=49 \end{gathered} $

Also, $\mathbf{a b}=3 \times 5$

Hence, $9+25+2|\mathbf{a}||\mathbf{b}| \cos \theta=49$

$ \begin{aligned} & \Rightarrow 30 \cos \theta=15 \Rightarrow \cos \theta=1 / 2 \\ & \therefore \theta=60^{\circ}=\pi / 3 \end{aligned} $

So, angle is $\pi / 3$.

Given,

$ A(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}), B(-12 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}), $

$C(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$ and $D(\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$ are position vector of four points.

$ \begin{aligned} & \mathbf{A B}=14 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \\ & \mathbf{B C}=-11 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathrm{CD}=(-1-\lambda) \hat{\mathbf{i}}+0 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

∴ For coplanar

$ \begin{array}{rlrl} \left|\begin{array}{ccc} 14 & 0 & 6 \\ -11 & -3 & 1 \\ -(\lambda+1) & 0 & -3 \end{array}\right| & =0 \\ 14(9)+6(-3(\lambda+1)) & =0 \\ 126-18 \lambda-18 & =0 \\ 18 \lambda & =108 \\ \Rightarrow & & \lambda & =6 \end{array} $

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

We know that, The orthogonal projection of $\mathbf{a}$ on

$ \begin{aligned} & \mathbf{b}(\mathbf{x})=\frac{(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}}{|\mathbf{b}|^2} \\ & \Rightarrow \mathbf{x}=\frac{[(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}})]}{\left(\sqrt{(2)^2+(-1)^2+(-2)^2}\right)^2} \\ & \quad(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}})) \end{aligned} $

$ \begin{aligned} & \Rightarrow x=\frac{(1 \times 2)+2 \times(-1)+(-2) \cdot(-2)}{9}(\hat{\mathbf{2i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \therefore x=\frac{4}{9}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Orthogonal projection of $\mathbf{b}$ on

$ \begin{aligned} & \mathbf{a}(\mathbf{y})=\frac{(\mathbf{b} \cdot \mathbf{a}) \mathbf{a}}{|\mathbf{a}|^2} \\ & \quad \mathbf{y}=\frac{4}{9}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

So, $\quad \mathbf{x}-\mathbf{y}=\frac{4}{9}(\hat{\mathbf{i}}-3 \hat{\mathbf{j}})$

$ \begin{aligned} & \text { Hence, }|x-y|=\frac{4}{9} \sqrt{(1)^2+(-3)^2} \\ & \therefore \quad|x-y|=\frac{4}{9} \sqrt{10} \end{aligned} $

Given, $\mathbf{O A}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

$ \mathbf{O B}=2 \hat{\mathbf{i}}-\hat{\mathbf{k}} \Rightarrow \mathbf{O C}=\hat{\mathbf{j}}+2 \hat{\mathbf{k}} $

$and\,\,\, \mathrm{OD}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} $

$ \begin{aligned} So,\,\,& \mathrm{AB}=\mathrm{OB}-\mathrm{OA}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathrm{AC}=\mathrm{OC}-\mathrm{OA}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathrm{AD}=\mathrm{OD}-\mathrm{OA}=2 \hat{\mathbf{j}}+\hat{\mathbf{k}}(\lambda-1) \end{aligned} $

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

$ \begin{aligned} & \because \quad\left|\begin{array}{ccc} 1 & 1 & -2 \\ -1 & 2 & 1 \\ 0 & 2 & \lambda-1 \end{array}\right|=0 \\ & \Rightarrow 1(2 \lambda-2-2)-1(-\lambda+1-0)-2(-2)=0 \\ & \Rightarrow 2 \lambda-4+\lambda-1+4=0 \\ & \Rightarrow \quad 3 \lambda=1 \\ & \Rightarrow \quad \lambda=\frac{1}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\text {} \end{aligned} $

Magnitude of vector $6 \lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

$ \begin{aligned} & =\sqrt{(6 \lambda)^2+(-3)^2+(6)^2} \\ & =\sqrt{\left(6 \times \frac{1}{3}\right)^2+9+36} \text { [from Eq. (i), } \lambda=\frac{1}{3} \text { ] } \\ & =\sqrt{4+9+36}=7 \text { units } \end{aligned} $

Clearly, the point $(\mathbf{a}-\mathbf{b}+\mathbf{c})$ lies on the line because line is passing through this point.

∴ Equation of plane is

$ \begin{array}{r} {[\mathbf{r}-(2 \mathbf{a}-\mathbf{b})][(2 \mathbf{a}-\mathbf{b})-(2 \mathbf{b}-\mathbf{c})] \times[(2 c-\mathbf{a})} -(2 \mathbf{b}-\mathbf{c})]=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

Now, we have to show that the point. $(\mathbf{a}-\mathbf{b}+\mathbf{c})$ satisfying the equation plane.

Taking LHS of Eq. (i),

$ \begin{aligned} &(\mathbf{r}-2 \mathbf{a}+\mathbf{b}) \cdot[(2 \mathbf{a}-3 \mathbf{b}+\mathbf{c}) \times(-\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})] \\ &=(\mathbf{r}-2 \mathbf{a}+\mathbf{b})[-4 \mathbf{a} \times \mathbf{b}+6 \mathbf{a} \times \mathbf{c}+ \\ &+3 \mathbf{b} \times \mathbf{a}-9 \mathbf{b} \times \mathbf{c}-\mathbf{c} \times \mathbf{a}-2 \mathbf{c} \times \mathbf{b}] \\ &=(\mathbf{r}-2 \mathbf{a}+\mathbf{b})[-7 \mathbf{a} \times \mathbf{b}+7 \mathbf{a} \times \mathbf{c}-7 \mathbf{b} \times \mathbf{c}] \\ &=(\mathbf{r}-2 \mathbf{a}+\mathbf{b})(-\mathbf{a} \times \mathbf{b}+\mathbf{a} \times \mathbf{c}-\mathbf{b} \times \mathbf{c}) \end{aligned} $

On putting $\mathbf{r}=\mathbf{a}-\mathbf{b}+\mathbf{c}$,

$ \begin{aligned} & =(-\mathbf{a}+\mathbf{c})(-\mathbf{a} \times \mathbf{b}+\mathbf{a} \times \mathbf{c}-\mathbf{b} \times \mathbf{c}) \\ & \Rightarrow[\mathrm{a}: \mathbf{b} \mathbf{c}]-[\mathrm{c} \mathbf{a} \mathbf{b}]=0 \end{aligned} $

Hence, the point of intersection of line and plane is $\mathbf{a}-\mathbf{b}+\mathbf{c}$.

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

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\, \mathbf{b}=6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ &and\,\,\, \mathbf{c}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-12 \hat{\mathbf{k}} \end{aligned} $

The projection of $\mathbf{b}$ on $\mathbf{a}=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \mathbf{a}$

$ \begin{aligned} &\Rightarrow \mathbf{p}=\frac{6-4-6}{\sqrt{\left(1^2+(-2)^2+2^2\right)^2}} \mathbf{a} \Rightarrow \mathbf{p}=-\frac{4}{9} \mathbf{a}\\ &\text { The projection of } \mathbf{c} \text { on } \mathbf{a}=\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}|^2} \mathbf{a}\\ &\begin{array}{ll} \Rightarrow & \mathbf{q}=\frac{3+8-24}{\sqrt{\left(1^2+(-2)^2+2^2\right)^2}} \mathbf{a} \\ \Rightarrow & \mathbf{q}=-\frac{13}{9} \mathbf{a} \\ \therefore & 13 \mathbf{p}=13\left(-\frac{4}{9} \mathbf{a}\right)=4\left(-\frac{13}{9} \mathbf{a}\right)=4 \mathbf{q} \end{array} \end{aligned} $

Given, $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

$ \begin{aligned} & \mathbf{b}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}} \\ & \mathbf{c}=\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Since, $\mathbf{r}$ is perpendicular to both $\mathbf{b}$ and $\mathbf{c}$.

$ \begin{aligned} & \therefore \mathbf{r}=t\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -1 & 5 \\ 1 & -4 & -2 \end{array}\right| \\ & =[\hat{\mathbf{i}}(2+20)-\hat{\mathbf{j}}(-6-5)+\hat{\mathbf{k}}(-12+1)] t \\ & =(22 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}-11 \hat{\mathbf{k}}) t=22 t \hat{\mathbf{i}}+11 \hat{\mathbf{j}}-11 \hat{\mathbf{k}} \\ & \because \mathbf{r} \cdot \mathbf{a}=11 \\ & \therefore(22 t \hat{\mathbf{i}}+11 t \hat{\mathbf{j}}-11 t \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=11 \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & 22 t+22 t-33 t & =11 \\ \Rightarrow & 11 t & =11 \\ \Rightarrow & t & =1 \end{array} $

Thus, $\mathbf{r}=22 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}-11 \hat{\mathbf{k}}$

Since, $\mathbf{r} \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=22-11-11=0$

$\therefore \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is perpendicular to $\mathbf{r}$.

$ \begin{aligned} &\text { Volume of a tetrahedron }\\ &=\frac{1}{6}\left[\begin{array}{lll} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{array}\right] \Rightarrow 2=\frac{1}{6}\left[\begin{array}{ccc} 1 & -\lambda & 1 \\ \lambda & -1 & -1 \\ 1 & 1 & \lambda \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2=\frac{1}{6}\left[1(-\lambda+1)+\lambda\left(\lambda^2+1\right)+1(\lambda+1)\right] \\ & \Rightarrow \quad 12=-\lambda+1+\lambda^3+\lambda+\lambda+1 \\ & \quad 12=\lambda^3+\lambda+2 \\ & \Rightarrow \quad \lambda^3+\lambda-10=0 \\ & \quad \therefore \quad \lambda=2 \\ & \text { Hence, }|\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}}+3 \hat{\mathbf{k}}| \\ & \begin{aligned} =|2 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}| & =\sqrt{(2)^2+(-6)^2+(3)^2} \\ & =\sqrt{4+36+9}=\sqrt{49}=7 \end{aligned} \end{aligned} $

$ \begin{aligned} & \mathrm{OA}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}=\mathbf{a} \\ & \mathrm{OB}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}=\mathbf{b} \\ & \mathrm{OC}=4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}=\mathbf{c} \end{aligned} $

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

$ \begin{aligned} \mathbf{P C} & =\mathbf{O C}-\mathbf{O P} \\ & =(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})-(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ \mathbf{P C} & =3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \end{aligned} $

Here, $l=3, m=-4, n=6$

Then, $l+3 m+2 n=3-12+12=3$

$ \begin{aligned} & B C=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & C D=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Diagonal $\mathbf{B D}=\mathbf{B C}+\mathbf{C D}$

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

Diagonal $\mathbf{A C}=\mathbf{A B}+\mathbf{B C}$

$ \begin{aligned} & =-\mathbf{C D}+\mathbf{B C} \\ & =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}+2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

Angle between $\mathbf{A C}$ and $\mathbf{B D}$

$ \begin{aligned} \cos \theta & =\frac{\mathbf{A C} \cdot \mathbf{B D}}{|\mathbf{A C}||\mathbf{B D}|}=\frac{-3}{\sqrt{19} \sqrt{11}}=\frac{-3}{\sqrt{209}} \\ \Rightarrow \tan \theta & =\frac{-10 \sqrt{2}}{3} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} \mathbf{a} & =\lambda \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \\ \mathbf{b} & =3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} \end{aligned} \\ & \text { and } \quad \mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

Volume of parallelopiped = 61

$ \begin{aligned} & {[\mathbf{a b c}]=61 } \\ \Rightarrow & \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=61 \\ \Rightarrow & \left|\begin{array}{ccc} \lambda & 3 & 4 \\ 3 & -1 & \lambda \\ \lambda & 1 & -3 \end{array}\right|=61 \\ \Rightarrow & \lambda(3-\lambda)-3\left(-9-\lambda^2\right)+4(3+\lambda)=61 \\ \Rightarrow & 3 \lambda-\lambda^2+27+3 \lambda^2+12+4 \lambda-61=0 \\ \Rightarrow & 2 \lambda^2+7 \lambda-22=0 \\ \Rightarrow & 2 \lambda^2+11 \lambda-4 \lambda-22=0 \\ \Rightarrow & 2 \lambda(\lambda+11)-4(\lambda+11)=0 \\ \Rightarrow & \lambda=2,-11 \end{aligned} $

$ \begin{aligned} & \text { Given, } \mathbf{a} \cdot \mathbf{b}=0 \text { and }|\mathbf{a}|=8,|\mathbf{b}|=3 \\ & \text { } \begin{aligned}Now, |\mathbf{a}-2 \mathbf{b}|^2 & =(\mathbf{a}-2 \mathbf{b}) \cdot(\mathbf{a}-2 \mathbf{b}) \\ & =|\mathbf{a}|^2+4|\mathbf{b}|^2 \end{aligned} \end{aligned} $

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

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=(8)^2+4(3)^2 $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =64+4 \times 9=64+36=100 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \therefore \quad|a-2 b|=\sqrt{100}=10 \end{aligned} $

Given, $\mathbf{a}$ and $\mathbf{b}$ be non-collinear vectors.

If the vectors $(\lambda-1) \mathbf{a}+2 \mathbf{b}$ and $3 \mathbf{a}+\lambda \mathbf{b}$ are collinear.

Now, $(\lambda-1) \mathbf{a}+2 \mathbf{b}=P(3 \mathbf{a}+\lambda \mathbf{b})$

$ \Rightarrow\{(\lambda-1)-3 P\} \mathbf{a}+\{2-\lambda P\} \mathbf{b}=0 $

Since, $\mathbf{a}$ and $\mathbf{b}$ are non-collinear,

So, $\quad \lambda-1-3 P=0$ and $2-\lambda P=0$

$ \begin{aligned} & \Rightarrow \quad \lambda-1=3 \cdot P \text { and } P=\frac{2}{\lambda} \\ & \Rightarrow \quad \lambda-1 \Rightarrow \frac{6}{\lambda}=\lambda^2-\lambda-6=0 \\ & \Rightarrow \quad \lambda=-2 \text { or } 3 \end{aligned} $

Thus, the set of all possible value of $\lambda$ is $\{-2,3\}$.

Vectors, $\mathbf{p}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$, $\mathbf{q}=d \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{r}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ forming a $\triangle A B C$ are such that, $\mathbf{p}=\mathbf{q}+\mathbf{r}$.

where, $a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$

$ \begin{aligned} & =(d \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})+(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \Rightarrow a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}=(d+3) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

Comparing coefficient of $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$, we get

$ a=d+3, b=4 \text { and } c=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, area of $\triangle A B C$ is $5 \sqrt{6}$ sq units

$ \Rightarrow \quad \frac{1}{2}|\mathbf{q} \times \mathbf{r}|=5 \sqrt{6} $

$ \begin{aligned} & \Rightarrow \frac{1}{2}\left\|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ d & 3 & 4 \\ 3 & 1 & -2 \end{array}\right\|=5 \sqrt{6} \\ & \Rightarrow \frac{1}{2} \sqrt{(-10)^2+(2 d+12)^2+(d-9)^2}=5 \sqrt{6} \\ & \Rightarrow \quad \frac{1}{4}\left(5 d^2+30 d+325\right)=150 \\ & \Rightarrow \quad \quad 5 d^2+30 d+325=600 \\ & \Rightarrow \quad \quad d^2+6 d-55=0 \\ & \therefore d=-11,5 \quad a=-8,8 \quad \text { [from Eq. (i)] } \\ & \quad b=4, \text { and } c=2 \quad|a|+|b|+|c| \quad \text { Now, } \\ & =| \pm 8|+|4|+|2|=14 \end{aligned} $

we have

$ \begin{aligned} & (\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b} \\ & =(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c} \end{aligned} $

Comparing on both side, we get

$ \begin{aligned} \mathbf{a} \cdot \mathbf{c}+\mathbf{a} \cdot \mathbf{b} & =4-2 \beta-\sin \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } \quad-\mathbf{a} \cdot \mathbf{b} & =\left(\beta^2-1\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Given, $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$

$ \begin{array}{lc} \Rightarrow & (\mathbf{c} \cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{c})=(\mathbf{c} \cdot \mathbf{c}) \\ \Rightarrow & \mathbf{a} \cdot \mathbf{c}=1 \end{array} $

Now, from Eq. (i),

$ 1+\mathbf{a} \cdot \mathbf{b}=4-2 \beta-\sin \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Using Eq. (ii), we get

$ \begin{aligned} & 1+1-\beta^2=4-2 \beta-\sin \alpha \\ \Rightarrow & \beta^2-2 \beta+2=\sin \alpha \\ \Rightarrow & (\beta-1)^2+1=\sin \alpha \end{aligned} $

It is possible only when $\beta=1$ and $\sin \alpha=1$

$ \Rightarrow \quad \beta=1 \text { and } \alpha=\frac{\pi}{2} $

Now, $\sin (\alpha+\beta)=\sin \left(\frac{\pi}{2}+1\right) \equiv \cos (1)$

Given the position vectors of $\mathbf{P}$ and $\mathbf{Q}$ are:

$ \begin{aligned} \mathbf{P} &= \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 7 \hat{\mathbf{k}} \\ \mathbf{Q} &= 5\hat{\mathbf{i}} - 3\hat{\mathbf{j}} + 4\hat{\mathbf{k}} \end{aligned} $

To find the vector $\mathbf{PQ}$, we calculate:

$ \mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (5\hat{\mathbf{i}} - 3\hat{\mathbf{j}} + 4\hat{\mathbf{k}}) - (\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 7\hat{\mathbf{k}}) $

Simplifying the components, we get:

$ \mathbf{PQ} = (5 - 1)\hat{\mathbf{i}} + (-3 - 2)\hat{\mathbf{j}} + (4 + 7)\hat{\mathbf{k}} = 4\hat{\mathbf{i}} - 5\hat{\mathbf{j}} + 11\hat{\mathbf{k}} $

The direction ratios of $\mathbf{PQ}$ are $a = 4$, $b = -5$, and $c = 11$.

To find the cosine of the angle $\gamma$ between the vector $\mathbf{PQ}$ and the $Z$-axis, we use the relation:

$ \cos \gamma = \frac{c}{\sqrt{a^2 + b^2 + c^2}} $

Substituting the values, we have:

$ \cos \gamma = \frac{11}{\sqrt{4^2 + (-5)^2 + 11^2}} = \frac{11}{\sqrt{16 + 25 + 121}} = \frac{11}{\sqrt{162}} $

To solve this problem, we need to evaluate the given vector conditions.

We are given that $|\mathbf{a}+\mathbf{b}+\mathbf{c}| = 1$, where $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors, meaning that $|\mathbf{a}|^2 = |\mathbf{b}|^2 = |\mathbf{c}|^2 = 1$. Additionally, $\mathbf{a}$ is perpendicular to $\mathbf{b}$, which implies $\mathbf{a} \cdot \mathbf{b} = 0$.

To proceed, we expand and square the magnitude of the vector sum:

$ | \mathbf{a} + \mathbf{b} + \mathbf{c} |^2 = 1 $

$ | \mathbf{a} |^2 + | \mathbf{b} |^2 + | \mathbf{c} |^2 + 2(\mathbf{a} \cdot \mathbf{b}) + 2(\mathbf{b} \cdot \mathbf{c}) + 2(\mathbf{c} \cdot \mathbf{a}) = 1 $

Substituting the known values:

$ 1 + 1 + 1 + 0 + 2(\mathbf{b} \cdot \mathbf{c}) + 2(\mathbf{c} \cdot \mathbf{a}) = 1 $

This simplifies to:

$ 3 + 2 \cos \beta + 2 \cos \alpha = 1 $

Therefore:

$ 2 \cos \alpha + 2 \cos \beta = -2 $

Dividing the entire equation by 2 gives:

$ \cos \alpha + \cos \beta = -1 $

To determine the equation of the line through a given point and parallel to a vector $\mathbf{a}$, consider the following:

Given:

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

$\mathbf{a} \cdot \hat{\mathbf{i}} = 1$

First, find the vector $\mathbf{a}$. Assume $\mathbf{a}$ can be expressed as $x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}$. We have:

$ (x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}) \times \hat{\mathbf{i}} = \hat{\mathbf{j}} + \hat{\mathbf{k}} $

Breaking down the cross product:

$x \hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0$

$y \hat{\mathbf{j}} \times \hat{\mathbf{i}} = -y \hat{\mathbf{k}}$

$z \hat{\mathbf{k}} \times \hat{\mathbf{i}} = z \hat{\mathbf{j}}$

Combining, we get:

$ -y \hat{\mathbf{k}} + z \hat{\mathbf{j}} = \hat{\mathbf{j}} + \hat{\mathbf{k}} $

Equating components:

$z = 1$

$-y = 1 \Rightarrow y = -1$

Given $\mathbf{a} \cdot \hat{\mathbf{i}} = 1$:

$ (x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}) \cdot \hat{\mathbf{i}} = x = 1 $

Thus, we find:

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

Now, the equation of the line passing through the point $\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$ and parallel to the vector $\mathbf{a}$ is:

$ \mathbf{r} = (\text{Passing through point}) + t (\text{Parallel vector}) $

Substitute:

$ \mathbf{r} = (\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) + t (\hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}) $

Simplify:

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

This is the equation of the line.