Vector Algebra

Given,

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

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

and $\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$

Also, $\left| {OB \times OC} \right| = 9/2$ ..... (i)

Now, $\overrightarrow {OB} \times \overrightarrow {OC} = \overrightarrow {OB} \times {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$

$ = {{ - \lambda } \over 2}\overrightarrow {OB} \times \overrightarrow {OA} = {\lambda \over 2}(\overrightarrow {OA} \times \overrightarrow {OB} )$

We need to find $\overrightarrow {OA} \times \overrightarrow {OB} $ for this,

($\because$ $\overrightarrow a \times \overrightarrow b = - \overrightarrow b \times \overrightarrow a $ and $\overrightarrow a \times \overrightarrow a $ = 0)

$\overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 2 & 1 \cr 1 & { - 2} & 2 \cr } } \right|$

$ = 6\widehat i - 3\widehat j - 6\widehat k = 3(2\widehat i - \widehat j - 2\widehat k)$

So, $\overrightarrow {OB} \times \overrightarrow {OC} = {{3\lambda } \over 2}(2\widehat i - \widehat j - 2\widehat k)$

From Eq. (i), $\left| {{{3\lambda } \over 2}\sqrt {4 + 1 + 4} } \right| = 9/2$

$ \Rightarrow \left| {{{9\lambda } \over 2}} \right| = 9/2 \Rightarrow {9 \over 2}\left| \lambda \right| = {9 \over 2}$

$ \Rightarrow \left| \lambda \right| = 1 \Rightarrow \lambda = \pm 1$

But $\lambda$ > 0

$\therefore$ $\lambda$ = 1

So, $\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \overrightarrow {OA} )$ (By putting $\lambda$ = 1)

$ \Rightarrow \overrightarrow {OC} = {1 \over 2}( - \widehat i - 4\widehat j + \widehat k) = - {1 \over 2}\widehat i - 2\widehat j + {1 \over 2}\widehat k$

Option (a)

Projection of ${\overrightarrow {OC} }$ and ${\overrightarrow {OA} }$ $ = {{\overrightarrow {OC} \,.\,\overrightarrow {OA} } \over {\left| {\overrightarrow {OA} } \right|}}$

$ = {{{1 \over 2}( - 2 - 8 + 1)} \over 3} = {{ - 3} \over 2}$

Option (b)

Area of $\Delta OAB = {1 \over 2}\left| {\overrightarrow {OA} \times \overrightarrow {OB} } \right| = {1 \over 2}\left| {\overrightarrow {OA} } \right|\left. {\overrightarrow {OB} } \right|\sin 90^\circ $

$ = {1 \over 2}\left| {3 \times 3} \right| = {9 \over 2}$

Option (c)

Area of the $\Delta ABC = {1 \over 2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right|$

$ = {1 \over 2}\left\| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & { - 4} & 1 \cr { - 5/2} & { - 4} & {{{ - 1} \over 2}} \cr } } \right\|$

$ = {1 \over 2}\left| {6\widehat i - 3\widehat j - 6\widehat k} \right| = {1 \over 2} \times 3 \times 3 = {9 \over 2}$

Option (d)

The acute angle between the diagonals of the parallelogram with adjacent sides ${\overrightarrow {OA} }$ and ${\overrightarrow {OC} }$ = $\theta$

$ \Rightarrow {{(\overrightarrow {OA} + \overrightarrow {OC} )\,.\,(\overrightarrow {OA} - \overrightarrow {OC} )} \over {\left| {\overrightarrow {OA} + \overrightarrow {OC} } \right|\left| {\overrightarrow {OA} - \overrightarrow {OC} } \right|}} = \cos \theta $

$ \Rightarrow \cos \theta = {{\left( {{3 \over 2}\widehat i + {3 \over 2}\widehat k} \right)\,.\,\left( {{5 \over 2}\widehat i + 4\widehat j + {1 \over 2}\widehat k} \right)} \over {{3 \over 2}\sqrt 2 \times \sqrt {{{90} \over 4}} }}$

$ = {{18} \over {3\sqrt 2 \times \sqrt {90} }} = {1 \over {\sqrt 5 }} \ne {1 \over 2}$ [$\therefore$ $\theta$ $\ne$ $\pi$/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

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

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

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 $=\frac{1}{2}|(\mathbf{a}+\mathbf{b}) \times(\mathbf{b}+\mathbf{c})|$

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

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

To determine if the points with position vectors $10\widehat{i} + 3\widehat{j}$, $12\widehat{i} - 5\widehat{j}$, and $a\widehat{i} + 11\widehat{j}$ are collinear, we need to check if the vectors between these points are scalar multiples of each other. This is equivalent to checking if the determinant of the matrix formed by these points is zero.

Let's consider the vectors formed by the points:

• Vector 1, $\mathbf{AB}$: from $10\widehat{i} + 3\widehat{j}$ to $12\widehat{i} - 5\widehat{j}$
• Vector 2, $\mathbf{AC}$: from $10\widehat{i} + 3\widehat{j}$ to $a\widehat{i} + 11\widehat{j}$

We can find these vectors as follows:

$\mathbf{AB} = (12 - 10)\widehat{i} + (-5 - 3)\widehat{j}$

$\mathbf{AB} = 2\widehat{i} - 8\widehat{j}$

$\mathbf{AC} = (a - 10)\widehat{i} + (11 - 3)\widehat{j}$

$\mathbf{AC} = (a - 10)\widehat{i} + 8\widehat{j}$

For the points to be collinear, $\mathbf{AB}$ and $\mathbf{AC}$ must be linearly dependent, such that:

$\mathbf{AB} = k \cdot \mathbf{AC}$ for some scalar $k$

Equating the components:

$2\widehat{i} - 8\widehat{j} = k \cdot \left((a - 10)\widehat{i} + 8\widehat{j}\right)$

Separating the components gives us two equations:

1. $2 = k(a - 10)$

2. $-8 = k \cdot 8$

From the second equation:

$k = -1$

Substituting $k = -1$ into the first equation:

$2 = -1 \cdot (a - 10)$

$2 = -a + 10$

Solving for $a$:

$a = 10 - 2$

$a = 8$

Thus, the points will be collinear if $a = 8$.

We have, | a | = 3, | b | = 4 and | c | = 5.

It is given that

a $ \bot $ (b + c), b $ \bot $ (c + a) and c $ \bot $ (a + b)

$\Rightarrow$ a . (b + c) = 0, b . (c + a) = 0 and c . (a + b) = 0

$\Rightarrow$ a . b + a . c = b . c + b . a = c . a = c . b = 0

or, a . b + b . c + c . a = 0

(adding all the above equations)

Now, | a + b + c |² = | a |² + | b |² + | c |² + 2 (a . b + b . c + c . a)

= 3² + 4² + 5² = 50

$\therefore$ | a + b + c | = 5$\sqrt2$

We have, $\left| {(a \times b).\,c} \right| = \left| a \right|\left| b \right|\left| c \right|$

$ \Rightarrow \left\| a \right\|\left\| b \right\|\left| c \right|\left. {\sin \theta \cos \alpha } \right| = \left| a \right|\left| b \right|\left| c \right|$

$ \Rightarrow \left| {\sin \theta } \right|\left| {\cos \alpha } \right| = 1 \Rightarrow \theta = {\pi \over 2}$ and $\alpha = 0$

$ \Rightarrow a \bot b$ and $c||\widehat n \Rightarrow a \bot b$ and c $\bot$ both a and b.

$\Rightarrow$ a, b, c are mutually perpendicular.

$\Rightarrow$ a . b = b . c = c . a = 0