Vector Algebra

$|a|=\sqrt{2^2+2^2+p^2}=\sqrt{8+p^2}$ and $|b|=7$

$\mathbf{a} \cdot \mathbf{b}=4 \,\,\,\,[given]$

$ \begin{aligned} & \Rightarrow \quad|\mathbf{a}||\mathbf{b}| \cos \theta=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Again, } \quad|\mathbf{a} \times \mathbf{b}|=5 \sqrt{17} \\ & \Rightarrow \quad|\mathbf{a}||\mathbf{b}| \sin \theta=5 \sqrt{17} \\ & \Rightarrow \quad \frac{4}{\cos \theta} \times \sin \theta=5 \sqrt{17} \quad \text { [From Eq. (i)] } \\ & \Rightarrow \quad \tan \theta=\frac{5 \sqrt{17}}{4} \Rightarrow \cos \theta=\frac{4}{21} \end{aligned} $

$ \begin{aligned} & & \frac{4}{|\mathbf{a}||\mathbf{b}|} & =\frac{4}{21}\,\,\,\, \text { [From Eq. (i)] }\\ \Rightarrow & & |\mathbf{a}||\mathbf{b}| & =21 \\ \Rightarrow & & \sqrt{8+p^2} \times 7 & =21 \Rightarrow 8+p^2=9 \end{aligned} $

$ \begin{aligned} \Rightarrow & p^2 =1 \\ \therefore & p = \pm 1 \end{aligned} $

Let $A, B, C, D, E, P$ have position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}$ and $\mathbf{p}$ respectively.

Given, $D$ and $E$ divide the line segment $B C$ and $A E$ in the ratio $2: 1$.

By section formula,

$ \begin{aligned} & \mathbf{d}=\frac{2 c+b}{2+1} \Rightarrow 3 d=2 c+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & e=\frac{c+2 a}{2+1} \Rightarrow 3 e=c+2 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eq. (i), $3 \mathbf{d}-\mathbf{b}=\mathbf{2} \mathbf{c}$

From Eq. (ii), $3 \mathbf{e}-2 \mathbf{a}=\mathbf{c}$

$\therefore \quad 6 e-4 a=2 c$

Equating both values of 2 c ,

$ \begin{aligned} & & 3 \mathrm{~d}-\mathrm{b} & =6 \mathrm{e}-4 \mathrm{a} \\ \Rightarrow & & 3 \mathrm{~d}+4 \mathrm{a} & =6 \mathrm{e}+\mathrm{b} \\ \Rightarrow & & \frac{3 \mathrm{~d}+4 \mathrm{a}}{3+4} & =\frac{6 \mathrm{e}+\mathrm{b}}{6+1} \end{aligned} $

LHS is the position vector of the point which divides segment $A D$ internally in the ratio $3: 4$. RHS is the position

vector of the point which divides segment $B E$ internally in the ratio $6: 1$. But $P$ is the point of intersection of $A D$ and $B E$.

$\therefore P$ divides $A D$ internally in the ratio 3 : 4 and $P$ divides $B E$ internally in the ratio $6: 1$.

Let position vector of

$ \begin{aligned} & A: \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & B: 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & C:-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ & D: a \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & A B: \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \\ & A C:-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}} \\ & A D:(a-1) \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

Given $\mathbf{A}, \mathbf{B}, \mathbf{C}$ and $\mathbf{D}$ are coplanar.

$\left[\begin{array}{lll}\mathbf{A B} & \mathbf{A C} & \mathbf{A D}\end{array}\right]=0$

$ \left|\begin{array}{ccc} 1 & 5 & -7 \\ -4 & 3 & -8 \\ a-1 & 0 & 1 \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow(a-1)(-40+21)+0+1(3+20)=0 \\ & \Rightarrow \quad-19(a-1)+23=0 \\ & \Rightarrow \quad-19 a+19+23=0 \Rightarrow a=\frac{42}{19} \end{aligned} $

Let $\mathbf{a}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$

$\mathbf{a}$ be a vector in the plane containing vectors

$\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

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

$ \begin{aligned} &\left|\begin{array}{ccc} a & b & c \\ 1 & 2 & 1 \\ 2 & -1 & 1 \end{array}\right|=0 \\ & \Rightarrow 3 a+b-5 c=0 \quad \ldots(\mathbf{i}) \\ & \Rightarrow \mathbf{a} \perp(\mathbf{i}+\mathbf{j}+3 \mathbf{k}) \Rightarrow \mathbf{a} \cdot(\mathbf{i}+\mathbf{j}+3 \mathbf{k})=0 \end{aligned} $

$ \begin{aligned} &a+b+3 c=0\\ &\text { Projection of } \mathbf{a} \text { on } \mathbf{b} \text { is } 3 \sqrt{6} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\text {. } \end{aligned} $

$ \begin{aligned} & \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}=3 \sqrt{6} \Rightarrow \frac{a+2 b+c}{\sqrt{1+4+1}}=3 \sqrt{6} \\ & a+2 b+c=18 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

on subtracting Eq. (ii) from Eq. (i),

$ 2 a-8 c=0 \Rightarrow a=4 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

on multiplying 2 in Eq. (ii) and subtracting

Eq. (iii), we get

$ a+5 c=-18 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

From Eqs. (iv) and (v), $c=-2$ and $a=-8$

On putting in Eq. (iii), $b=14$

$ \begin{gathered} \mathbf{a}=-8 \hat{\mathbf{i}}+14 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ |a|^2=8^2+14^2+2^2=64+196+4=264 \end{gathered} $

$ \begin{aligned} & \text { Let } \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \text { c }=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{d}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & l=\mathbf{b} \cdot \mathbf{c}=1-6-2=-7 \\ & m=\mathbf{c} \cdot \mathbf{a}=1+3-2=2 \\ & {[m \mathbf{b}+l \mathbf{a} \mathbf{b} \mathbf{d}]=[m \mathbf{b} \mathbf{b} \mathbf{d}]+[l \mathbf{l} \mathbf{a} \mathbf{b} \mathbf{d}]} \\ & \Rightarrow[\mathrm{labd}]=l[\mathrm{abd}] \\ & =-7\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{array}\right|=-7 \\ & {[1+3+5]=-63} \end{aligned} $

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

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

Comparing both sides,

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

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

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

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

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

Their resultant say $R$ is given by

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

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

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

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

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

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

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

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

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

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

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

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

Given,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

and also given

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

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

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

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

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

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

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

In $,$\triangle A D C$,

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

In $\triangle M D C$,

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

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

In $\triangle A D X$,

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

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

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

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

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

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

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

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

Now, $x=2 y$

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

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

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

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

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

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

Given that, $\left| {{{\widehat u + \widehat v} \over 2} + \widehat u \times \widehat v} \right| = 1$

$ \Rightarrow {\left| {{{\widehat u + \widehat v} \over 2} + \widehat u \times \widehat v} \right|^2} = 1$

$ \Rightarrow {{2 + 2\cos \theta } \over 4} + {\sin ^2}\theta = 1$ [$\because$ $\widehat u\,.\,(\widehat u \times \widehat v) = \widehat v(\widehat u \times \widehat v) = 0$]

$ \Rightarrow {\cos ^2}{\theta \over 2} = \cos 2\theta $

$ \Rightarrow \theta = n\pi \, \pm \,{\theta \over 2},\,n \in z$

$ \Rightarrow \theta = {{2\pi } \over 3}$

$ \Rightarrow \left| {\widehat u \times \widehat v} \right| = \sin {{2\pi } \over 3} = \sin {\pi \over 3} = \left| {{{\widehat u - \widehat v} \over 2}} \right|$

Suppose $\overrightarrow r = x\overrightarrow a + y\overrightarrow b + 2\overrightarrow c $

and $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = k$

$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightarrow b \times \{ (\overrightarrow r - \overrightarrow c ) \times \overrightarrow b \} + \overrightarrow c \times \{ (\overrightarrow r - \overrightarrow a ) \times \overrightarrow c \} = \overrightarrow 0 $

$ \Rightarrow {k^2}(\overrightarrow r - \overrightarrow b ) - {k^2}x\overrightarrow a + {k^2}(\overrightarrow r - \overrightarrow c ) - {k^2}y\overrightarrow b + {k^2}(\overrightarrow r - \overrightarrow a ) - {k^2}z\overrightarrow c = \overrightarrow 0 $

$ \Rightarrow 3\overrightarrow r -(\overrightarrow a + \overrightarrow b + \overrightarrow c ) - \overrightarrow r = \overrightarrow 0 $

$ \Rightarrow \overrightarrow r = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$

${\left| {3\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {2\overrightarrow a + 3\overrightarrow b } \right|^2}$

$\left( {3\overrightarrow a + \overrightarrow b } \right).\left( {3\overrightarrow a + \overrightarrow b } \right) = \left( {2\overrightarrow a + 3\overrightarrow b } \right).\left( {2\overrightarrow a + 3\overrightarrow b } \right)$

$9\overrightarrow a .\,\overrightarrow a + 6\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow b = 4\overrightarrow a \,.\,\overrightarrow a + 12\overrightarrow a \,.\,\overrightarrow b + 9\overrightarrow b \,.\,\overrightarrow b $

$5{\left| {\overrightarrow a } \right|^2} - 6\overrightarrow a \,.\,\overrightarrow b = 8{\left| {\overrightarrow b } \right|^2}$

$5{(8)^2} - 6.8\,.\,\left| {\overrightarrow b } \right|\cos 60^\circ = 8{\left| {\overrightarrow b } \right|^2}$ $\because$ $\left( \matrix{ {1 \over 8}\left| {\overrightarrow a } \right| = 1 \hfill \cr \Rightarrow \left| {\overrightarrow a } \right| = 8 \hfill \cr} \right)$

$40 - 3\left| {\overrightarrow b } \right| = {\left| {\overrightarrow b } \right|^2}$

$ \Rightarrow {\left| {\overrightarrow b } \right|^2} + 3\left| {\overrightarrow b } \right| - 40 = 0$

$\left| {\overrightarrow b } \right| = - 8$, $\left| {\overrightarrow b } \right| = 5$

(rejected)

$A(\widehat j)\,.\,B(10\widehat i)$

$H(h\widehat j + 10\widehat k)$

$G(10\widehat i + h\widehat j + 10\widehat k)$

$\overrightarrow {AG} = 10\widehat i + h\widehat j + 10\widehat k$

$\overrightarrow {BH} = - 10\widehat i + h\widehat j + 10\widehat k$

$\cos \theta = {{\overrightarrow {AG} \overrightarrow {BH} } \over {\left| {\overrightarrow {AG} } \right|\left| {\overrightarrow {BH} } \right|}}$

${1 \over 5} = {{{h^2}} \over {{h^2} + 200}}$

$4{h^2} = 200 \Rightarrow h = 5\sqrt 2 $

$\left| {\overrightarrow a } \right| = \sqrt 3 $; $\overrightarrow a .\overrightarrow c = 3$; $\overrightarrow a \times \overrightarrow b = - 2\widehat i + \widehat j + \widehat k$, $\overrightarrow a \times \overrightarrow c = \overrightarrow b $

Cross with $\overrightarrow a $,

$\overrightarrow a \times (\overrightarrow a \times \overrightarrow c ) = \overrightarrow a \times \overrightarrow b $

$ \Rightarrow (\overrightarrow a .\overrightarrow c )\overrightarrow a - {a^2}\overrightarrow c = \overrightarrow a \times \overrightarrow b $

$ \Rightarrow 3\overrightarrow a - 3\overrightarrow c = - 2\widehat i + \widehat j + \widehat k$

$ \Rightarrow 3\widehat i + 3\widehat j + 3\widehat k - 3\overrightarrow c = - 2\widehat i + \widehat j + \widehat k$

$ \Rightarrow \overrightarrow c = {{5\widehat i} \over 3} + {{2\widehat j} \over 3} + {{2\widehat k} \over 3}$

$\therefore$ $\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a \times \overrightarrow b ).\overrightarrow c = {{ - 10} \over 3} + {2 \over 3} + {2 \over 3} = - 2$

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

$ = 1.2\cos \theta \overrightarrow b - \overrightarrow c $

$ \Rightarrow \overrightarrow a = 2\cos \theta \overrightarrow b - \overrightarrow c $

${\left| {\overrightarrow a } \right|^2} = {(2\cos \theta )^2} + {2^2} - 2.2\cos \theta \overrightarrow b \,.\,\overrightarrow c $

$ \Rightarrow 2 = 4{\cos ^2}\theta + 4 - 4\cos \theta .2\cos \theta $

$ \Rightarrow - 2 = - 4{\cos ^2}\theta $

$ \Rightarrow {\cos ^2}\theta = {1 \over 2}$

$ \Rightarrow {\sec ^2}\theta = 2$

$ \Rightarrow {\tan ^2}\theta = 1$

$ \Rightarrow \theta = {\pi \over 4}$

$ \therefore $ $1 + \tan \theta = 2$

$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$

$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$

$\overrightarrow a + \overrightarrow b = 3\widehat j + 5\widehat k;\overrightarrow a.\overrightarrow b = - 1 + 2 + 6 = 7$

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

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

$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b - 0} \right)} \right) \times \overrightarrow b $

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

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

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

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

$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ i & j & k \cr 1 & 1 & 2 \cr { - 1} & 2 & 3 \cr } } \right| = - \widehat i - 5\widehat j + 3\widehat k$

$\therefore$ $7\left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$

$\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {7\left( { - \widehat i - 5\widehat j + 3\widehat k} \right)} \right)$

$7\left( {0\widehat i + 3\widehat j + 5\widehat k} \right) \times \left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 3 & 5 \cr { - 1} & { - 5} & 3 \cr } } \right|$

$ \Rightarrow 34\widehat i - (5)\widehat j + (3\widehat k)$

$ \Rightarrow 34\widehat i - 5\widehat j + 3\widehat k$

$\therefore$ $7\left( {0\widehat i + 3\widehat j + 5\widehat k} \right) \times \left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$

= $7(34\widehat i - 5\widehat j + 3\widehat k)$

Because vectors are coplanar

Hence, $\left| {\matrix{ a & a & c \cr 1 & 0 & 1 \cr c & c & b \cr } } \right| = 0$

$ \Rightarrow {c^2} = ab \Rightarrow c = \sqrt {ab} $

$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$

$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$

$\sin \theta = \pm \,{4 \over 5}$

$\therefore$ $\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $

$ = 10.\left( { \pm \,{3 \over 5}} \right) = \pm 6$

$\left| {\overrightarrow a .\,\overrightarrow b } \right| = 6$

If the vectors are co-planar,

$\left| {\matrix{ {a + b + 2} & {a + 2b + c} & { - b - c} \cr {b + 1} & {2b} & { - b} \cr {b + 2} & {2b} & {1 - b} \cr } } \right| = 0$

Now, ${R_3} \to {R_3} - {R_2},{R_1} \to {R_1} - {R_2}$

So, $\left| {\matrix{ {a + 1} & {a + c} & { - c} \cr {b + 1} & {2b} & { - b} \cr 1 & 0 & 1 \cr } } \right| = 0$

$ = (a + 1)2b - (a + c)(2b + 1) - c( - 2b)$

$ = 2ab + 2b - 2ab - a - 2bc - c + 2bc$

$ = 2b - a - c = 0$

$\overrightarrow a = \lambda \overrightarrow b + \mu \overrightarrow c = \widehat i(2\lambda + \mu ) + \widehat j(\lambda - \mu ) + \widehat k(\lambda + \mu )$

$\overrightarrow a \,.\,\overrightarrow d = 0 = 3(2\lambda + \mu ) + 2(\lambda - \mu ) + 6(\lambda + \mu )$

$ \Rightarrow 14\lambda + 7\mu = 0 \Rightarrow \mu = - 2\lambda $

$ \Rightarrow \overrightarrow a = (0)\widehat i - 3\lambda \widehat j + ( - \lambda )\widehat k$

$ \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {10} \left| \lambda \right| = \sqrt {10} \Rightarrow \left| \lambda \right| = 1$

$\lambda = 1$ or $ - 1$

$[\overrightarrow a \overrightarrow b \overrightarrow c ]$ = 0

$[\overrightarrow a \overrightarrow b \overrightarrow c ] + [\overrightarrow a \overrightarrow b \overrightarrow d ] + [\overrightarrow a \overrightarrow c \overrightarrow d ] = [\overrightarrow a \overrightarrow b + \overrightarrow c \overrightarrow d ]$

$ = \left| {\matrix{ 0 & { - 3\lambda } & \lambda \cr 3 & 0 & 2 \cr 3 & 2 & 6 \cr } } \right|$

$ = 3\lambda (12) + \lambda (6) = 42\lambda = - 42$

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

$ = \overrightarrow a ( - \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow b ) = - 2\left( {\overrightarrow a \times (\overrightarrow b \times \overrightarrow c )} \right)$

$ = - 2(\overrightarrow a \times \overrightarrow a ) = \overrightarrow 0 $

(2) Projection of $\overrightarrow a $ on $\overrightarrow b \times \overrightarrow c $

$ = {{\overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )} \over {\left| {\overrightarrow b \times \overrightarrow c } \right|}} = {{\overrightarrow a .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right| = 2$

(3) $\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] + \left[ {\overrightarrow c \overrightarrow a \overrightarrow b } \right] = 2\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = 2\overrightarrow a .(\overrightarrow b \times \overrightarrow c )$

$ = 2\overrightarrow a .\overrightarrow a = 2{\left| {\overrightarrow a } \right|^2} = 8$

(4) $\overrightarrow a \times \overrightarrow b = \overrightarrow c $ and $\overrightarrow b \times \overrightarrow c = \overrightarrow a $

$ \Rightarrow \overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are mutually $ \bot $ vectors.

$\therefore$ $\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow c } \right| \Rightarrow \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| \Rightarrow \left| {\overrightarrow b } \right| = {{\left| {\overrightarrow c } \right|} \over 2}$

Also, $\left| {\overrightarrow b \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right| \Rightarrow \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| = 2 \Rightarrow \left| {\overrightarrow c } \right| = 2$ & $\left| {\overrightarrow b } \right| = 1$

${\left| {3\overrightarrow a + \overrightarrow b - 2\overrightarrow c } \right|^2} = (3\overrightarrow a + \overrightarrow b - 2\overrightarrow c ).(3\overrightarrow a + \overrightarrow b - 2\overrightarrow c )$

$ = 9{\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2}$

$ = (9 \times 4) + 1 + (4 \times 4)$

$ = 36 + 1 + 16 = 53$

Projection of $\overrightarrow {BA} $

on ${\overrightarrow {BC} }$ is equal to

$ = \left| {\overrightarrow {BA} } \right|\cos \angle ABC$

$ = 7\left| {{{{7^2} + {3^2} - {5^2}} \over {2 \times 7 \times 3}}} \right| = {{11} \over 2}$

$\left| {\overrightarrow a } \right| = 3 = a;\overrightarrow a \,.\,\overrightarrow c = c$

Now, $\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $

$ \Rightarrow {c^2} + {a^2} - 2\overrightarrow c \,.\,\overrightarrow a = 8$

$ \Rightarrow {c^2} + 9 - 2(c) = 8$

$ \Rightarrow {c^2} - 2c + 1 = 0 \Rightarrow c = 1 = \left| {\overrightarrow c } \right|$

Also, $\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$

Given, $(\overrightarrow a \times \overrightarrow b ) = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin {\pi \over 6}$

$ = (3)(1)(1/2)$

$ = 3/2$

$\overrightarrow a $ is perpendicular to $\overrightarrow b $

$ \therefore $ $\overrightarrow a $ . $\overrightarrow b $ = 0

Given, | $\overrightarrow a $ $\times$ $\overrightarrow b $ | = | $\overrightarrow a $ |

and | $\overrightarrow a $ | = | $\overrightarrow b $ |

$ \therefore $ | $\overrightarrow a $ $\times$ $\overrightarrow b $ | = | $\overrightarrow a $ | = | $\overrightarrow b $ | = k(assume)

Now, angle between $\overrightarrow a $ and ($\overrightarrow a $ + $\overrightarrow b $ + ($\overrightarrow a $ $\times$ $\overrightarrow b $))

$\cos \theta = {{\overrightarrow a ((\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ))} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$

$ = {{|\overrightarrow a {|^2} + \,\overrightarrow a .\,\overrightarrow b + \overrightarrow a (\overrightarrow a \times \overrightarrow b )} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$

[Note $\overrightarrow a .(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow a \overrightarrow a \overrightarrow b ] = 0$]

$ = {{|\overrightarrow a {|^2} + 0 + 0} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$

Now, $|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2}$

$ = |\overrightarrow a {|^2} + |\overrightarrow a {|^2} + |\overrightarrow a \times \overrightarrow b {|^2} + $

$2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .(\overrightarrow a \times \overrightarrow b ) + 2\overrightarrow a .(\overrightarrow a \times \overrightarrow b )$

$ = {k^2} + {k^2} + {k^2}$

$ \therefore $ $|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2} = 3{k^2}$

$ \Rightarrow |\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )| = \sqrt 3 k$

$ \therefore $ $\cos \theta = {{{k^2}} \over {k(\sqrt 3 k)}} = {1 \over {\sqrt 3 }}$

$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$

$|\overrightarrow a | = 8,|\overrightarrow b | = 7,|\overrightarrow c | = 10$

Projection of $\overrightarrow {AB} $ on $\overrightarrow {AC} $

= $|\overrightarrow {AB} |cos\theta $

$ = 10\left( {{{|\overrightarrow {AB} {|^2} + |\overrightarrow {CA} {|^2} - |\overrightarrow {BC} {|^2}} \over {2.|\overrightarrow {CA} ||\overrightarrow {AB} {|}}}} \right)$

$ = 10\left( {{{{{10}^2} + {7^2} - {8^2}} \over {2(10)(7)}}} \right)$

$ = 10\left( {{{85} \over {140}}} \right)$

$ = {{85} \over {14}}$

${\left| {\overrightarrow a } \right|_{old}} = {\left| {\overrightarrow a } \right|_{new}}$

(3p)² + 1 = (p + 1)² + 10

$ \Rightarrow $ 9p² $-$ p² $-$ 2p $-$ 10 = 0

$ \Rightarrow $ 8p² $-$ 2p $-$ 10 = 0

$ \Rightarrow $ 4p² $-$ p $-$ 5 = 0

$ \Rightarrow $ 4p² $-$ 5p + 4p $-$ 5 = 0

$ \Rightarrow $ (4p $-$ 5) (p + 1) = 0

$ \Rightarrow $ p = ${5 \over 4}$, $-$ 1

$\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k\,$

$\overrightarrow {OP} \bot \overrightarrow {OQ} $

$\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$

$\overrightarrow {PQ} = \left( { - 1 - x} \right)\widehat i + \left( {2 - y} \right)\widehat j + \left( {3x + 1} \right)\widehat k$

$\left| {\overrightarrow {PQ} } \right| = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} $

$\sqrt {20} = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} $

20 = 1 + x² + 2x + 4 + y² $-$ 4y + 9x² + 1 + 6x

20 = 10x² + y² + 8x + 6 $-$ 4y

20 = 10x² + 4x² + 8x + 6 $-$ 8x

14 = 14x² $ \Rightarrow $ x² = 1

Also, $\overrightarrow {OP} .\,\overrightarrow {OQ} = 0$

$ - x + 2y - 3x = 0$

$4x = 2y$

y = 2x

$ \therefore $ y² = 4x² $ \Rightarrow $ y² = 4

x = 1 as x > 0 and y = 2

$ \therefore $ $\left| {\matrix{ x & y & { - 1} \cr { - 1} & 2 & {3x} \cr 3 & z & { - 7} \cr } } \right| = 0$

$ \Rightarrow $ $\left| {\matrix{ 1 & 2 & { - 1} \cr { - 1} & 2 & 3 \cr 3 & z & { - 7} \cr } } \right|$ = 0

$ \Rightarrow $ 1($-$14 $-$3z) $-$ 2(7 $-$ 9) $-$ 1($-$z $-$6) = 0

$ \Rightarrow $ $-$14 $-$3z + 4 + z + 6 = 0

$ \Rightarrow $ 2z = $-$4 $ \Rightarrow $ z = $-$2

$ \therefore $ x² + y² + z² = 9

$\overrightarrow a = (2, - 3,4)$, $\overrightarrow b = (7,1, - 6)$

$\overrightarrow r \times \overrightarrow a - \overrightarrow r \times \overrightarrow b = 0$

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

$\overrightarrow r = \lambda (\overrightarrow a - \overrightarrow b )$

$\overrightarrow r = \lambda ( - 5\widehat i - 4\widehat j + 10\widehat k)$

$\overrightarrow r \,.\,(2, - 3,1) = ?$

Given $\overrightarrow r \,.\,(1,2,1) = - 3$

$\lambda ( - 5 - 8 + 10) = - 3 \Rightarrow \lambda = 1$

$ \therefore $ $( - 5, - 4,10)\,.\,(2, - 3,1)$

= - 10 + 12 + 10 = 12

Given $\overrightarrow r $ $\times$ $\overrightarrow a $ = $\overrightarrow b $ $\times$ $\overrightarrow r $

$ \Rightarrow $ $\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $

$\overrightarrow r \times (\overrightarrow a + \overrightarrow b ) = 0$

$\overrightarrow r ||(\overrightarrow a + \overrightarrow b )$

$\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b )$

$(\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k)$

$ \because $ $\overrightarrow r \,.\,(2\widehat i + 5\widehat j - \alpha \widehat k) = - 1$

$\lambda \left[ {3\widehat i - \widehat j + 2\widehat k} \right]\,.\,\left[ {2\widehat i + 5\widehat j - \alpha \widehat k} \right] = - 1$

$ \Rightarrow \lambda (6 - 5 - 2\alpha ) = - 1$

$\lambda (1 - 2\alpha ) = - 1$ .... (1)

$\overrightarrow r \,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$

$\lambda (3\widehat i - \widehat j + 2\widehat k)\,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$

$ \Rightarrow \lambda [3\alpha - 2 + 2] = 3 \Rightarrow \lambda \alpha = 1$ .... (2)

From (1) & (2)

$\lambda \left[ {1 - {2 \over \lambda }} \right] = - 1$

$\lambda - 2 = - 1 \Rightarrow \lambda = 1\,\alpha = 1$

$\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k$

$\alpha + |\overrightarrow r {|^2} = 1 + 14 = 15$

($\alpha$, $\beta$) $ \equiv $ (2 cos 75$^\circ$, 2 sin 75$^\circ$)

Area = ${1 \over 2}$ (2 cos 75$^\circ$) (2 sin 75$^\circ$)

= sin(150$^\circ$) = ${1 \over 2}$ square unit

$\overrightarrow {{a_2}} = \lambda \overrightarrow {{a_1}} $

$\widehat i + y\widehat j + z\widehat k = \lambda (x\widehat i - \widehat j + \widehat k)$

$1 = \lambda x,y = - \lambda ,z = \lambda $

$x\widehat i + y\widehat j + z\widehat k = {1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k$

Unit vector $ = {{{1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k} \over {\sqrt {{1 \over {{\lambda ^2}}} + {\lambda ^2} + {\lambda ^2}} }}$

$ = {{\widehat i - {\lambda ^2}\widehat j + {\lambda ^2}\widehat k} \over {\sqrt {1 + 2{\lambda ^4}} }}$

Let ${\lambda ^2} = 1$, possible unit vector $ = {{\widehat i - \widehat j + \widehat k} \over {\sqrt 3 }}$

$\overrightarrow a \,.\,\overrightarrow b = 0$

$\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ) = (\overrightarrow a \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow a \,.\,\overrightarrow a )\overrightarrow b = - |\overrightarrow a {|^2}\overrightarrow b $

Now, $\overrightarrow a \times (\overrightarrow a \times ( - |\overrightarrow a {|^2}\overrightarrow b ))$

$ = - |\overrightarrow a {|^2}(\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ))$

$ = - |\overrightarrow a {|^2}( - |\overrightarrow a {|^2}\overrightarrow b ) = |\overrightarrow a {|^4}\overrightarrow b $