Vector Algebra

$\begin{aligned} A O & =\sqrt{2^2+2^2+1^2}=3 \\\\ O B & =\sqrt{2^2+4^2+4^2}=6 \\\\ \frac{A O}{B O} & =\frac{A D}{B D}=\frac{3}{6} \quad \text { [Angle bisector theorem] } \\\\ D & =\frac{2(2,2,1)+1(2,4,4)}{2+1}=\frac{(6,8,6)}{3} \\\\ & =\left(2, \frac{8}{3}, 2\right) \\\\ (O D) & =\sqrt{2^2+\left(\frac{8}{3}\right)^2+2^2}=\frac{\sqrt{136}}{3}\end{aligned}$

Given,

$\begin{aligned} \mathbf{u} & =<2,3,1> \\ \mathbf{v} & =<-3,2,0> \\ \mathbf{w} & =< 1,-1,4> \\ \mathbf{u} \cdot \mathbf{v} & =-6+6+0=0 \\ \mathbf{u} \cdot \mathbf{w} & =2-3+4=3 \neq 0 \end{aligned}$

$\mathbf{u} \perp \mathbf{v}$ but $\mathbf{u}$ is not perpendicular to $\mathbf{w}$.

$\begin{aligned} & \mathbf{a}=(1,1,0) \text {, } \\ & \mathbf{b}=(1,1,1) \Rightarrow 0=(\mathbf{a}+\lambda \mathbf{b}) \cdot \mathbf{a} \\ & \Rightarrow \quad \mathbf{a} \cdot \mathbf{a}+\lambda \mathbf{a} \cdot \mathbf{b}=0 \\ & \Rightarrow \quad(1+1+0)+\lambda(1+1+0)=0 \\ & \Rightarrow \quad \lambda=-1 \\ & \mathbf{c}=\mathbf{a}+\lambda \mathbf{b}=(0,0,1)=\mathbf{k} \\ & \end{aligned}$

$\begin{aligned} & \mathbf{a}=\hat{\mathbf{i}}, \mathbf{b}=\hat{\mathbf{j}} \\ & (\mathbf{r} \times \mathbf{a})=(\mathbf{b} \times \mathbf{a}) \Rightarrow(\mathbf{r} \times \mathbf{a})-(\mathbf{b} \times \mathbf{a})=0 \\ & (\mathbf{r}-\mathbf{b}) \times \mathbf{a}=0 \Rightarrow \mathbf{r}-\mathbf{b}=\lambda \mathbf{a} \\ & \mathbf{r}=\mathbf{b}+\lambda \mathbf{a} \\ & =\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}} \quad ..........\text{(i)}\\ & \mathbf{r} \times \mathbf{b}=\mathbf{a} \times \mathbf{b} \\ & \Rightarrow \quad(\mathbf{r}-\mathbf{a}) \times \mathbf{b}=0 \\ & \mathbf{r}=\mathbf{a}+\alpha \mathbf{b} \\ & =\hat{\mathbf{i}}+\propto \hat{\mathbf{j}} \quad .........\text{(ii)}\\ \end{aligned}$

Comparing Eqs. (i) and (ii), we get

$\begin{aligned} \lambda \hat{\mathbf{i}}+\hat{\mathbf{j}} & =\hat{\mathbf{i}}+\alpha \hat{\mathbf{j}} \Rightarrow \lambda=1, \alpha=1 \\ \mathbf{r} & =\hat{\mathbf{i}}+\hat{\mathbf{j}} \end{aligned}$

In option (d) it is clearly that $\mathrm{dr}^{\prime}$ s of $4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ is $4,4,4$

Which makes equal angles with coordinate axes.

Position vector of $A=\mathbf{O A}=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

Position vector of $B=\mathbf{O B}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

Position vector of $C=\mathbf{O C}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$

Position vector of $D=\mathbf{O D}=$ mid point of $\mathbf{O B}$ and $\mathbf{O C}=\frac{\mathbf{O B}+\mathbf{O C}}{2}$

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

Position vector of $E=\mathbf{O E}=$ mid point of $\mathbf{O A}$

$\text { and } \begin{aligned} \mathbf{O C} & =\frac{\mathbf{O A}+\mathbf{O C}}{2} \\ & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \mathbf{D E} & =\mathbf{O E}-\mathbf{O D}=\hat{\mathbf{j}} \end{aligned}$

$\begin{array}{ll} \text { } \because \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot|\mathbf{b}|}<0 \\ \Rightarrow \mathbf{a} \cdot \mathbf{b}<0 \end{array}$

Let $\theta$ be the angle between $\overline{\mathbf{a}}$ and $\overline{\mathbf{b}}$, then

$\begin{aligned} & & |\mathbf{a} \cdot \mathbf{b}| & =|\mathbf{a} \times \mathbf{b}| \\ \Rightarrow & & -|\mathbf{a}| \cdot|\mathbf{b}| \cos \theta & =|\mathbf{a}| \mathbf{b} \mid \sin \theta \\ \Rightarrow & & \tan \theta & =-1 \\ \Rightarrow & & \sec \theta & =-\sqrt{2} \\ \Rightarrow & & \theta & =\sec ^{-1}(-\sqrt{2}) \end{aligned}$

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors

$\begin{array}{ll} \Rightarrow & |\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1 \\ \because & \mathbf{a} \cdot \mathbf{b}=0 \text { and } \mathbf{a} \cdot \mathbf{c}=0 \\ \Rightarrow & \mathbf{a} \perp \mathbf{b} \text { and } \mathbf{a} \perp \mathbf{c} \end{array}$

Angle between $\mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$

$\therefore \quad \mathbf{b} \times \mathbf{c}=|\mathbf{b}| \cdot|\mathbf{c}| \sin \frac{\pi}{3} \cdot \hat{\mathbf{a}}$

$[\because \mathbf{a}$ is perpendicular to $\mathbf{b}$ and $\mathbf{c}$ and $\hat{\mathbf{a}}=1]$

$=1 \cdot 1 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$

Now, $[\mathbf{a} \mathbf{b} \mathbf{c}]^2=[\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})]^2$

$\begin{aligned} & =\left[\mathbf{a} \cdot \frac{\sqrt{3}}{2}\right]^2=|\mathbf{a}|^2 \cdot\left(\frac{\sqrt{3}}{2}\right)^2 \\ & =1 \cdot \frac{3}{4}=\frac{3}{4} \end{aligned}$

$\begin{aligned} \text { } \mathbf{a} & =(\sin x) \hat{\mathbf{i}}+(\sin y) \hat{\mathbf{j}} \\ \mathbf{b} & =(\cos x) \hat{\mathbf{i}}+(\cos y) \hat{\mathbf{j}} \\ \mathbf{a} \times \mathbf{b} & =\left[\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \sin x & \sin y & 0 \\ \cos x & \cos y & 0\end{array}\right] \\ & =\hat{\mathbf{k}}(\sin x \cos y-\cos x \sin y) \\ \mathbf{a} \times \mathbf{b} & =\hat{\mathbf{k}} \sin (x-y) \\ \text { or } \mid \mathbf{a} & \times \mathbf{b} \mid=\sin (x-y) \leq 1 \quad[\because-1<\sin x \leq 1, \forall x]\end{aligned}$

$l =\cos \alpha$

$m =\cos \alpha$

$n =\cos 90 \gamma=0$

$\because l^2+m^2+n^2 =1$

$\Rightarrow 2 \cos ^2 \alpha =1$

$\Rightarrow \cos ^2 \alpha=\frac{1}{2}$

$\Rightarrow \cos \alpha = \pm \frac{1}{\sqrt{2}}$

$\Rightarrow \alpha =45 \Upsilon$ or $135^{\circ}$

Let $A(5,-4,-3)$

Direction ratios of $\mathbf{O A}=5,-4,-3$

Direction ratios of $X$-axis $=1,0,0$

$\begin{aligned} \text { Angle, } \cos \theta & =\frac{5 \cdot 1+(-4) \cdot 0+(-3) \cdot 0}{\sqrt{5^2+(-4)^2+(-3)^2} \sqrt{1^2+0^2+0^2}} \\ & =\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}} \\ \therefore \quad \theta & =\frac{\pi}{4} \end{aligned}$

Let $\mathbf{p}=\mathbf{i}+a \mathbf{j}+\mathbf{k}$

$\mathbf{q}=\hat{\mathbf{j}}+a \hat{\mathbf{k}} \Rightarrow \mathbf{r}=a \hat{\mathbf{i}}+\hat{\mathbf{k}}$

$\begin{aligned} & =\mathbf{p} \cdot(\mathbf{q} \times \mathbf{r}) \\ & V=\left|\begin{array}{lll} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{array}\right| \\ & \Rightarrow \quad V=a^3-a+1 \\ \end{aligned}$

Differentiating $V$ with respect to $a$

$\frac{d V}{d a}=3 a^2-1$

Again, differentiating w.r.t. $a$

$\frac{d^2 V}{d a^2}=6 a$

For stationary points

$\begin{aligned} \frac{d V}{d a} & =0 \\ \Rightarrow \quad 3 a^2-1 & =0 \end{aligned}$

or

$a^2=\frac{1}{3} \text { or } a= \pm \frac{1}{\sqrt{3}} $

At,

$\begin{gathered} a=\frac{1}{\sqrt{3}}, \\ \frac{d^2 V}{d a^2}=6 \cdot \frac{1}{\sqrt{3}}>0 \end{gathered}$

$\therefore$ Volume is minimum at $a=\frac{1}{\sqrt{3}}$

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

Let $\theta$ be the angle between $(\mathbf{a}+\mathbf{b})$ and $(\mathbf{a}-\mathbf{b})$ then

$\begin{aligned} \cos \theta & =\frac{(\mathbf{a}+\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})}{|\mathbf{a}+\mathbf{b}||\mathbf{a}-\mathbf{b}|} \\ & =\frac{1(-1)+1(-1)+1(2)}{\sqrt{1^2+1^2+1^2} \sqrt{(-1)^2+(-1)^2+2^2}}=0 \\ \therefore \quad \theta & =90 \Upsilon \end{aligned}$

$\begin{aligned} & \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{b}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \\ & \mathbf{c}=7 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+11 \hat{\mathbf{k}} \end{aligned}$

Then, $\quad \mathbf{a}+\mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\mathbf{b}+\mathbf{c}=8 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}=4(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})$ Area of parallelogram $=\frac{1}{2}|(\mathbf{a}+\mathbf{b}) \times(\mathbf{b}+\mathbf{c})|$

$\begin{aligned} & =\frac{1}{2} \mid 2(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times 4(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\ & =\frac{8}{2}|(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})| \\ & =4\left\|\begin{array}{|lll} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{array}\right\|=4|-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}| \\ & =4 \sqrt{1+4+1}=4 \sqrt{6} \text { sq units } \\ & \end{aligned}$

$\begin{aligned} & \text { }|\mathbf{a}|=2,|\mathbf{b}|=3 \\ & \because \quad(\mathbf{a}+t \mathbf{b}) \perp(\mathbf{a}-t \mathbf{b}) \\ & \Rightarrow \quad(\mathbf{a}+t \mathbf{b}) \cdot(\mathbf{a}-t \mathbf{b})=0 \\ & \Rightarrow \quad \mathbf{a} \cdot \mathbf{a}-t(\mathbf{a} \cdot \mathbf{b})+t(\mathbf{b} \cdot \mathbf{a})-t^2(\mathbf{b} \cdot \mathbf{b})=0 \\ & \Rightarrow \quad|\mathbf{a}|^2-t^2|b|^2=0 \\ & \Rightarrow \quad t^2=\frac{|\mathbf{a}|^2}{|\mathbf{b}|^2} \\ & \Rightarrow \quad t=\frac{|\mathbf{a}|}{|\mathbf{b}|}=\frac{2}{3}\end{aligned}$