Vector Algebra

$ \text { ∴ Vector normal to plane } \mathbf{n}=\mathbf{b} \times \mathbf{c} $

$ \begin{aligned} & \therefore \quad \mathbf{n}=\frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|} \\ & \text { Now } \mathbf{A r} \cdot \mathbf{n}=0 \\ & (\mathbf{r}-\mathbf{a}) \cdot \frac{(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|}=0 \\ & \Rightarrow \quad \mathbf{r} \cdot \frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|}=\mathbf{a} \cdot \frac{(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|} \\ & \quad \mathbf{r} \cdot \frac{(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|}=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{|\mathbf{b} \times \mathbf{c}|} \end{aligned} $

Given that, $|\mathbf{x}|=|\mathbf{y}|=|\mathbf{z}|=\sqrt{2}$ and $\theta=60^{\circ}$

Thus, $\mathbf{x} \cdot \mathbf{y}=|\mathbf{x}||\mathbf{y}| \cos \theta$

$ \begin{aligned} & =(\sqrt{2})(\sqrt{2}) \cos 60^{\circ} \\ & =2 \times \frac{1}{2}=1 \end{aligned} $

Similarly $\mathbf{y} \cdot \mathbf{z}=\mathbf{z} \cdot \mathbf{x}=1$

and $\mathbf{x} \cdot \mathbf{x}=|\mathbf{x}||\mathbf{x}|$

$ \begin{aligned} \cos 0^{\circ} & =(\sqrt{2})(\sqrt{2}) \times 1=2 \\ \mathbf{y} \cdot \mathbf{y} & =2, \mathbf{z} \cdot \mathbf{z}=2 \end{aligned} $

Now, $\mathbf{a}=\mathbf{x} \times(\mathbf{y} \times \mathbf{z})=(\mathbf{x} \cdot \mathbf{z}) \mathbf{y}-(\mathbf{x} \cdot \mathbf{y}) \mathbf{z}$

$ =1 \cdot \mathbf{y}-1 \cdot \mathbf{z} $

$ \begin{equation*} a=y-z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Also, $\quad \mathbf{b}=\mathbf{y} \times(\mathbf{z} \times \mathbf{x})=(\mathbf{y} \cdot \mathbf{z}) \mathbf{z}-(\mathbf{y} \cdot \mathbf{z}) \mathbf{x}$

$ =1 \cdot \mathbf{z}-1 \cdot \mathbf{x} $

$ \begin{equation*} \mathbf{b}=\mathbf{z}-\mathbf{x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $

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

$ \begin{align*} & \mathbf{a}+\mathbf{b}=(\mathbf{y}-\mathbf{z})+(\mathbf{z}-\mathbf{x}) \\ & \mathbf{y}-\mathbf{x}=\mathbf{a}+\mathbf{b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{align*} $

Again $\quad \mathbf{c}=\mathbf{x} \times \mathbf{y}$

Taking cross product with $\mathbf{x}$ on both sides

$ \begin{align*} & \mathbf{x} \times \mathbf{c}=\mathbf{x} \times(\mathbf{x} \times \mathbf{y})=(\mathbf{x} \cdot \mathbf{y}) \mathbf{x}-(\mathbf{x} \cdot \mathbf{x}) \mathbf{y} \\ & \mathbf{x} \times \mathbf{c}=(1) \mathbf{x}-(2) \mathbf{y} \\ & \Rightarrow \mathbf{x} \times \mathbf{c}=\mathbf{x}-2 \mathbf{y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{align*} $

Finally, $\mathbf{c}=\mathbf{x} \times \mathbf{y}$

Taking cross product with $\mathbf{y}$ on both sides

$ \begin{align*} & \mathbf{y} \times \mathbf{c}=\mathbf{y} \times(\mathbf{x} \times \mathbf{y}) \\ & \mathbf{y} \times \mathbf{c}=(\mathbf{y} \cdot \mathbf{y}) \mathbf{x}-(\mathbf{y} \cdot \mathbf{x}) \mathbf{y}=(2) \mathbf{x}-(1) \mathbf{y} \\ & \mathbf{y} \times \mathbf{c}=2 \mathbf{x}-\mathbf{y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{align*} $

Subtracting Eq. (v) from Eq. (iv), we get

$ \begin{aligned} \mathbf{x} \times \mathbf{c}-\mathbf{y} \times \mathbf{c} & =(+\mathbf{x}-2 \mathbf{y})-(2 \mathbf{x}-\mathbf{y}) \\ (\mathbf{x}-\mathbf{y}) \times \mathbf{c} & =-\mathbf{x}-\mathbf{y} \end{aligned} $

or $\quad \mathbf{x}+\mathbf{y}=(\mathbf{y}-\mathbf{x}) \times \mathbf{c}$

From Eq. (iii), $\mathbf{y}-\mathbf{x}=\mathbf{a}+\mathbf{b}$

$ \begin{equation*} \therefore \quad \mathbf{x}+\mathbf{y}=(\mathbf{a}+\mathbf{b}) \times \mathbf{c} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{equation*} $

Subtracting Eq. (iii) from Eq. (vi), we get

$ \begin{aligned} & (\mathbf{x}+\mathbf{y})-(\mathbf{y}-\mathbf{x})=(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b}) \\ & 2 \mathbf{x}=(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b}) \\ & \text { or } \quad \mathbf{x}=\frac{1}{2}[(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b})] \end{aligned} $

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

$ |\mathbf{a}|=\sqrt{4+4+1}=3 $

Now given $|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$

Squaring on both sides,

$ \begin{aligned} &|\mathbf{c}-\mathbf{a}|^2=8 \Rightarrow|\mathbf{c}|^2+|\mathbf{a}|^2-2 \mathbf{a} \cdot \mathbf{c}=8 \\ &|\mathbf{c}|^2+9-2|\mathbf{c}|=8 \\ & {[\because|\mathbf{a}|=3 \text { and } \mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|] } \\ & \therefore|\mathbf{c}|^2-2|\mathbf{c}|+1=0 \\ &(|\mathbf{c}|-1)^2=0 \Rightarrow|\mathbf{c}|-1=0 \Rightarrow|\mathbf{c}|=1 \end{aligned} $

Now, we find $(\mathbf{a} \times \mathbf{b})=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -2 & 1 \\ 0 & -1 & 1\end{array}\right|$

$ \begin{aligned} & (\mathbf{a} \times \mathbf{b})=\hat{\mathbf{i}}(-2+1)-\hat{\mathbf{j}}(2-0)+\hat{\mathbf{k}}(-2) \\ & (\mathbf{a} \times \mathbf{b})=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

$ \therefore \quad|\mathbf{a} \times \mathbf{b}|=\sqrt{1+4+4}=3 $

Now, given angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$.

$ \begin{aligned} \therefore|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}| & =|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \cdot \sin \frac{\pi}{3} \\ & =3 \times 1 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2} \end{aligned} $

Given,

$ \begin{aligned} l(3 \mathbf{a}+2 \mathbf{b}+\mathbf{c}) & +m(2 \mathbf{a}+2 \mathbf{b}+3 \mathbf{c}) +n(\mathbf{a}+2 \mathbf{b}+5 \mathbf{c})=0 \end{aligned} $

Where $l, m, n$ are scalar

$ \begin{aligned} \therefore \quad \mathbf{a}(3 l+2 m+n) & +2 \mathbf{b}(l+m+n) +\mathbf{c}(l+3 m+5 n)=0 \end{aligned} $

$\mathbf{a , b , c}$ are independent vector

$ \begin{align*} \therefore \quad 3 l+2 m+n & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ l+m+n & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ l+3 m+5 n & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{align*} $

From Eqs. (ii) and (iii), $m+2 n=0$ Putting the value of $m$ in Eq. (i), we get

$ \begin{aligned} & & 3 l+2(-2 n)+n & =0 \\ \Rightarrow & & 3 l-3 n & =0 \\ \Rightarrow & & 1 & =n \\ \therefore& & l=n, m+2 n & =0 \end{aligned} $

We have,

$ \mathbf{r}=t \mathbf{a}+(1-t) \mathbf{b} $

where $\mathbf{a}$ and $\mathbf{b}$ are two non collinear vector,

$ \mathbf{r}-\mathbf{b}=t(\mathbf{a}-\mathbf{b}) $

Clearly, $\mathbf{r}$ is a point on the third side of triangle where $\mathbf{a}, \mathbf{b}$ are two sides of triangle when $t \in[0,1]$

Given, $|\mathbf{b}|=2|\mathbf{a}|$ and $|\mathbf{c}|=3|\mathbf{a}|$

Angle between each pair of vector is $\pi / 3$.

$ \begin{aligned} \mid \mathbf{a}+\mathbf{b}+ & +\left.\mathbf{c}\right|^2=|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2 +2|\mathbf{a}||\mathbf{b}| \cos \pi / 3+2|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3} +2|\mathbf{a}||\mathbf{c}| \cos \frac{\pi}{3} \end{aligned} $

$ \begin{aligned} 25=|\mathbf{a}|^2 & +4|\mathbf{a}|^2+9|\mathbf{a}|^2+4|\mathbf{a}|^2 \times \frac{1}{2} +2 \times 6|\mathbf{a}|^2 \times \frac{1}{2}+6|\mathbf{a}|^2 \times \frac{1}{2} \end{aligned} $

$ \begin{aligned} & & 25 & =25|\mathbf{a}|^2 \\ \Rightarrow & & |\mathbf{a}| & =1 \\ \therefore& & |\mathbf{b}| & =2 \Rightarrow|\mathbf{c}|=3 \\ \Rightarrow & & |\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}| & =1+2+3=6 \end{aligned} $

Given, $|\mathbf{b}|=\frac{1}{2}|\mathbf{a}|$ and $|\mathbf{c}|=\frac{\sqrt{3}}{2}|\mathbf{a}|$

and $\quad \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{a} \cdot \mathbf{c}=0$

$ \begin{aligned} |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 & =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2 \\ & =4|\mathbf{b}|^2+|\mathbf{b}|^2+3|\mathbf{b}|^2=8|\mathbf{b}|^2 \\ \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}| & =2 \sqrt{2}|\mathbf{b}| \end{aligned} $

Angle between $(\mathbf{a}+\mathbf{b}+\mathbf{c})$ and $\mathbf{b}$ is

$ \begin{aligned} \cos \theta & =\frac{(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot \mathbf{b}}{|\mathbf{a}+\mathbf{b}+\mathbf{c}||\mathbf{b}|} \\ \Rightarrow \quad \cos \theta & =\frac{|\mathbf{b}|^2}{2 \sqrt{2}|\mathbf{b}|^2}=\frac{1}{2 \sqrt{2}} \\ \theta & =\cos ^{-1}\left(\frac{1}{2 \sqrt{2}}\right) \end{aligned} $

Let point $P(\mathbf{r})$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$ and $A(\hat{\mathbf{i}})$ and $B(\hat{\mathbf{j}})$

$ \begin{aligned} \mathbf{A P} & =(x-1) \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow \quad \mathbf{A B} & =-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ \text { Area of } \triangle A B P & =\frac{1}{2}|\mathbf{A P} \times \mathbf{A B}| \\ \mathbf{A P} \times \mathbf{A B} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x-1 & y & z \\ -1 & 1 & 0 \end{array}\right| \\ & =-z \hat{\mathbf{i}}+z \hat{\mathbf{j}}+(x+y-1) \hat{\mathbf{k}} \\ |\mathbf{A P} \times \mathbf{A B}| & =\sqrt{2 z^2+(x+y-1)^2} \\ \therefore \quad 1 & =\frac{1}{2} \sqrt{2 z^2+(x+y-1)^2} \\ & {[\because \text { Area of } \triangle A B P=1] } \\ & =(x+y-1)^2+2 z^2=4 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\mathbf{O A}=12 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}-18 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \mathbf{O B}=-3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}} \\ & \text { and } \quad \mathbf{O C}=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-24 \hat{\mathbf{k}} \end{aligned} $

$ \begin{gathered}\Rightarrow \mathbf{A B}=-15 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+9 \hat{\mathbf{k}} \\ |\mathbf{A B}|=\sqrt{(15)^2+(6)^2+(9)^2} \\ =\sqrt{225+36+81}=\sqrt{342} \\ \mathbf{B C}=6 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-15 \hat{\mathbf{k}} \end{gathered} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad \mathbf{B C}=\sqrt{36+81+225}=\sqrt{342} \\ & \text { and } \quad \mathbf{A C}=-9 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ & \therefore \quad|\mathbf{A C}|=\sqrt{81+225+36}=\sqrt{342} \end{aligned}\\ &\text { ∴ Position vector of Incentre }\\ &=\sqrt{342}(12 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}) \end{aligned} $

$ \begin{aligned} & +\sqrt{342}(-3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}) \\ & +\sqrt{342}(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-24 \hat{\mathbf{k}}) \\ & \hline \sqrt{342}+\sqrt{342}+\sqrt{342} \end{aligned} $

$ \begin{aligned} & =\frac{12 \hat{\mathbf{i}}-15 \hat{\mathbf{j}}-51 \hat{\mathbf{k}}}{3} \\ & =4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-17 \hat{\mathbf{k}} \end{aligned} $

(c) We have, Equation of plane

$ \begin{aligned} & \mathbf{r}=\mathbf{b}+\mathbf{c}+x(\mathbf{a}-\mathbf{b})+y(\mathbf{c}+\mathbf{a}) \\ & =(1-x) \mathbf{b}+(1+y) \mathbf{c}+(x+y) \mathbf{a} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

And equation of line

$ \mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{c}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

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

$ \begin{aligned} & \mathbf{b}+(1+y) \mathbf{c} \\ & \quad \begin{aligned}\Rightarrow\,\,\, x+y & =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \\ 1-x & =t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ 1+y & =-t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Solving Eqs. (iii), (iv) and (v), we get }\\ &\begin{aligned} & t=\frac{-1}{2}, x=\frac{3}{2} \text { and } y=\frac{-1}{2} \\ \therefore & \mathbf{r}=\mathbf{a}-\frac{1}{2}(\mathbf{b}-\mathbf{c}) \\ \Rightarrow & \mathbf{r}=\mathbf{a} \frac{1}{2} \mathbf{b}+\frac{1}{2} \mathbf{c} \end{aligned} \end{aligned} $

$ \begin{aligned} & =l \mathbf{a}+m \mathbf{b}+n \mathbf{c} \\ & \therefore \quad l=1, m \frac{-1}{2} \text { and } n=\frac{1}{2} \\ & \text { Now, } 3 l+4 m+2 n \\ & \quad=3(1)+4\left(-\frac{1}{2}\right)+2\left(\frac{1}{2}\right) \\ & =3-2+1=2 \end{aligned} $

$ \begin{aligned} &\text { The vertices of triangle are }\\ &2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \mathbf{A B} & =3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ \mathbf{B C} & =-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \mathbf{A C} & =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ |\mathbf{A B}| & =\sqrt{9+1+4}=\sqrt{14} \\ |\mathbf{B C}| & =\sqrt{4+9+1}=\sqrt{14} \\ |\mathbf{A C}| & =\sqrt{1+4+9}=\sqrt{14} \end{aligned} $

$\because \triangle A B C$ is an equilateral triangle

∴ Orthocentre of triangle is coincide with centroid

∴ Orthocentre of $\triangle A B C$ is

$ \left(\frac{\left.\begin{array}{r} 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}+5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+3 \hat{\mathbf{i}} \\ +5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{array}\right)}{3}\right) $

We have,

$ \begin{array}{ll} & \mathbf{A B}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}} \\ & \mathbf{A C}=s \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ \Rightarrow & \mathbf{C A}=-s \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ \text { and } & \mathbf{C B}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{array} $

Area of $\triangle A B C$ is $5 \sqrt{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \Rightarrow \quad \frac{1}{2}|\mathbf{C A} \times \mathbf{C B}|=5 \sqrt{6} $

Now, $\mathbf{C A} \times \mathbf{C B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -s & -3 & -4 \\ 3 & 1 & -2\end{array}\right|$

$ \begin{gathered} =\hat{\mathbf{i}}(6+4)-\hat{\mathbf{j}}(2 s+12)+\hat{\mathbf{k}}(-s+9) \\ =10 \hat{\mathbf{i}}-\hat{\mathbf{j}}(2 s+12)+\hat{\mathbf{k}}(-s+9) \\ \therefore \frac{1}{2} \sqrt{100+(2 s+12)^2+(-s+9)^2} \\ =5 \sqrt{6} \end{gathered} $

$ \begin{array}{llrl} \Rightarrow & & \sqrt{100+\left(4 s^2+144+48 s\right)+\left(s^2+81-18 s\right)} \\ & & =10 \sqrt{6} \\ \Rightarrow & & 325+5 s^2+30 s=600 \\ \Rightarrow & & 5 s^2+30 s-275=0 \\ \Rightarrow & & s^2+6 s-55=0 \\ \Rightarrow & & s^2+11 s-5 s-55=0 \\ \Rightarrow & & s(s+11)-5(s+11)=0 \\ \Rightarrow & & (s+11)(s-5)=0 \\ \Rightarrow & & s-5=0 \text { or } s+11 & \neq 0 \\ \Rightarrow & & s=5 \end{array} $

$ \begin{array}{ll} \text { Now, } & \mathbf{A B}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}} \\ & \mathbf{B C}=-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \text { and } & \mathbf{C A}=-s \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \end{array} $

$ \begin{aligned} &\text { By applying triangle law. }\\ &\begin{aligned} & \qquad \begin{array}{l} \mathbf{A B}+ \mathbf{B C}+\mathbf{C A}=\mathbf{O} \\ \Rightarrow(p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}})+(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ \quad+(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=\mathbf{O} \\ \Rightarrow(p-8) \hat{\mathbf{i}}+(q-4) \hat{\mathbf{j}}+(\mathbf{r}-2) \hat{\mathbf{k}}=\mathbf{O} \\ \Rightarrow \quad \quad p=8, q=4 \\ \text { and } \quad r=2 \end{array} \end{aligned} \end{aligned} $

Given $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors

$ \begin{aligned} & \therefore \quad|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1 \\ & \text { Now }, \mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c}) \\ & \Rightarrow \quad(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c}) \\ & \Rightarrow \quad(|\mathbf{a}||\mathbf{c}| \cos \beta) \mathbf{b}-(|\mathbf{a}||\mathbf{b}| \cos \alpha) \mathbf{c} =\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c}) \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \cos \beta \mathbf{b}-\cos \alpha \mathbf{c}=\frac{1}{\sqrt{2}} \mathbf{b}+\frac{1}{\sqrt{2}} \mathbf{c} \\ \Rightarrow & \cos \beta=\frac{1}{\sqrt{2}} \text { and } \cos \alpha=\frac{-1}{\sqrt{2}} \\ \Rightarrow & \cos \beta=\cos \frac{\pi}{4} \\ \text { and } & \cos \alpha=\cos \frac{3 \pi}{4} \\ \Rightarrow & \alpha=\frac{\pi}{4} \text { and } \alpha=\frac{3 \pi}{4} \\ \therefore & \alpha-\beta=\frac{3 \pi}{4}-\frac{\pi}{4}=\frac{2 \pi}{4}=\frac{\pi}{2} \end{array} $

$ \begin{aligned} &\text {We have, }\\ &\begin{aligned} & \mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b} \\ & \mathbf{O P}=x_1 \mathbf{a}+y_1 \mathbf{b}, \mathbf{O Q}=x_2 \mathbf{a}+y_2 \mathbf{b} \\ & \text { where, } x_1=\frac{-3}{4}, x_2=\frac{1}{3}, y_1=\frac{7}{4}, y_2=\frac{5}{3} \\ & \therefore \quad \mathbf{O P}=\frac{-3}{4} \mathbf{a}+\frac{7}{4} \mathbf{b} \\ & \quad \mathbf{O Q}=\frac{1}{3} \mathbf{a}+\frac{5}{3} \mathbf{b} \\ & \quad \mathbf{O P}=\frac{7 \mathbf{b}-3 \mathbf{a}}{7-3} \end{aligned} \end{aligned} $

$\therefore \mathbf{O P}$ divides $\mathbf{A B}$ externally in ratio $7: 3$.

Also, $\mathbf{A}^{\prime} \mathbf{O}=\mathbf{O A}=\mathbf{a}$

$ \begin{array}{ll} \therefore & \mathbf{O A}^{\prime}=-\mathbf{a} \\ & \mathbf{B}^{\prime} \mathbf{O}=\mathbf{O B}=\mathbf{b} \\ & \mathbf{O B}^{\prime}=-\mathbf{b} \end{array} $

Clearly $p$ lies inside the $\triangle A^{\prime} O B$ and $Q$ lies outside the $\triangle A O B$.

We have, $\mathbf{O P}=\frac{\mathbf{c}+3 \mathbf{d}}{4}$ and $\mathbf{O Q}=\frac{\mathbf{a}+\mathbf{c}}{2}$

It is given that

$ \begin{array}{ll} & \mathbf{A B}+\mathbf{A D}+\mathbf{B C}-2 \mathbf{D C}=\lambda \mathbf{P Q} \\ \Rightarrow \quad & (\mathbf{b}-\mathbf{a})+(\mathbf{d}-\mathbf{a})+(\mathbf{c}-\mathbf{b})-2(\mathbf{c}-\mathbf{d}) \\ & =\lambda\left[\frac{\mathbf{a}+\mathbf{c}}{2}-\frac{\mathbf{c}-3 \mathbf{d})}{4}\right] \\ \Rightarrow \quad & -2 \mathbf{a}-\mathbf{c}+3 \mathbf{d}=\frac{\lambda}{4}[2 \mathbf{a}+\mathbf{c}-3 \mathbf{d}] \\ \therefore \quad & \frac{2 \lambda}{4}=-2 \\ \Rightarrow \quad & \lambda=-4 \end{array} $

We have,

$ \begin{aligned} \mathbf{p} & =2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \therefore \quad \mathbf{p} \cdot \mathbf{q} & =(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =2-3-1=-2 \\ |\mathbf{p}| & =\sqrt{(2)^2+(-3)^2+(1)^2} \\ & =\sqrt{4+9+1}=\sqrt{14} \\ |\mathbf{q}| & =\sqrt{(1)^2+(1)^2+(-1)^2} \\ & =\sqrt{1+1+1}=\sqrt{3} \end{aligned} $

Now, $\mathbf{a}=$ orthogonal projections of $\mathbf{p}$ on $\mathbf{q}$

$ =\frac{\mathbf{p} \cdot \mathbf{q}}{|\mathbf{q}|^2} \mathbf{q}=\frac{-2}{3}(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) $

$\mathbf{b}=$ orthogonal projections of $\mathbf{q}$ on $\mathbf{p}$

$ =\frac{\mathbf{q} \cdot \mathbf{p}}{|\mathbf{p}|^2} \mathbf{p}=\frac{-2}{14}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) $

$ \begin{aligned} & \text { Now, } \mathbf{a} \times \mathbf{b}=\frac{4}{42}(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =\frac{4}{42}\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & -1 \\ 2 & -3 & 1 \end{array}\right| \\ & =\frac{4}{42}(-2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}) \\ & \text { and } \mathbf{a} \cdot \mathbf{b}=\frac{4}{42}(\hat{\mathbf{i}}+\hat{\mathbf{i}}-\hat{\mathbf{k}})(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =\frac{4}{42}(2-3-1)=\frac{-8}{42} \\ & \begin{aligned} \therefore \quad \frac{\mathbf{a} \times \mathbf{b}}{\mathbf{a} \cdot \mathbf{b}} & =\frac{\frac{4(-2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})}{42}}{\frac{-8}{42}} \\ = & \frac{2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{2} \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} & \mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \\ & \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

Since, $\mathbf{a} \perp \mathbf{x}$ and $\mathbf{b} \perp \mathbf{x}$

$ \begin{aligned} & \therefore \quad \mathbf{x}=\lambda(\mathbf{a} \times \mathbf{b})=\lambda\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -3 & 4 \\ 7 & 2 & -3 \end{array}\right| \\ & =\lambda(\hat{\mathbf{i}}+34 \hat{\mathbf{j}}+25 \hat{\mathbf{k}}) \end{aligned} $

Now, we have $\mathbf{x} \cdot \mathbf{c}=60$

$ \begin{array}{lc} \Rightarrow & (\lambda \hat{\mathbf{i}}+34 \lambda \hat{\mathbf{j}}+25 \lambda \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=60 \\ \Rightarrow & \lambda+34 \lambda+25 \lambda=60 \\ \Rightarrow & 60 \lambda=60 \Rightarrow \lambda=1 \\ \therefore & \mathbf{x}=\hat{\mathbf{i}}+34 \hat{\mathbf{j}}+25 \hat{\mathbf{k}} \end{array} $

We have given,

| a | = 1, | b | = 4, a . b = 2

Now, c = (2a $\times$ b) $-$ 3b

Multiplying by b both sides, we get

$b\,.\,c = (2a \times b)\,.\,b - 3|b{|^2}$

$ \Rightarrow b\,.\,c = 0 - 3{(4)^2} = - 48$

and $|c{|^2}\, = \,|(2a \times b)\,.\,3b{|^2}$

$|c{|^2}\, = \,{(2a \times b)^2} + 9|b{|^2} = 4{(a \times b)^2} + 9{(4)^2}$

$ \Rightarrow |c{|^2}\, = 4(|a{|^2}\,.\,|b{|^2} - {(a\,.\,b)^2}) + 144$

$ = 4(16 - 4) + 144 = 48 + 144 = 192$

$\therefore$ $b\,.\,c = |b||c|cos\theta $

$ \Rightarrow cos\theta = {{b\,.\,c} \over {|b||c|}} = {{ - 48} \over {4 \times \sqrt {192} }} = - {{\sqrt 3 } \over 2}$

$\theta = \pi - {\pi \over 6} = {{5\pi } \over 6}$

We have,

Vector r is coplanar with vector a and b

$\therefore$ $r = \lambda (b \times (a \times b))$

$ \Rightarrow r = \lambda ((b\,.\,b)a - (b\,.\,a)b)$

$ \Rightarrow r = \lambda (|b{|^2}a - (a\,.\,b)b)$

$ \Rightarrow r = \lambda [5a + b]$ [$\because$ $|b| = \sqrt {{2^2} + {1^2}} = \sqrt 5 $]

and $a\,.\,b = ( - \widehat i + \widehat j + \widehat k)\,.\,(2\widehat i + \widehat k) = - 1$

Given, $r\,.\,a = 7$

$\therefore$ $\lambda (5a + b)\,.\,a = 7$

$ \Rightarrow \lambda [5|a{|^2} + a\,.\,b] = 7$

$ \Rightarrow \lambda [15 - 1] = 7$ [$\because$ $|a{|^2} = 1 + 1 + 1 = 3$]

$ \Rightarrow \lambda = {1 \over 2}$

$\therefore$ $r = {1 \over 2}(5a + b)$

$r = {1 \over 2}( - 5\widehat i + 5\widehat j + 5\widehat k + 2\widehat i + \widehat k)$

$r = {{ - 3} \over 2}\widehat i + {5 \over 2}\widehat j + {6 \over 2}\widehat k$

z component of $r = {6 \over 2} = 3$

Now, $\overrightarrow {OX} = {{\overrightarrow {QR} } \over {QR}}$

and $\overrightarrow {OY} = {{\overrightarrow {RP} } \over {RP}}$

Therefore, $(\overrightarrow {OX} \times \overrightarrow {OY} ) = {{\overrightarrow {QR} } \over {QR}} \times {{\overrightarrow {RP} } \over {RP}} = {{\overrightarrow {QR} \times \overrightarrow {RP} } \over {PQ}}$

$ = {{PQ\sin R} \over {PQ}} = \sin R = \sin (\pi - (P + Q) = \sin (P + Q))$

We have $\widehat u = {u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k$

That is, $\left| {\widehat u} \right| = 1 = \sqrt {u_1^2 + u_2^2 + u_3^2} $

$ \Rightarrow u_1^2 + u_2^2 + u_3^2 = 1$

Also, it is given that

$\widehat \omega = {1 \over {\sqrt 6 }}(\widehat i + \widehat j + 2\widehat k)$

That is, $\left| {\widehat \omega } \right| = 1$

Now, $\left| {\widehat u \times \overrightarrow v } \right| = 1$

That is, $\left| {\widehat u} \right|\left| {\overrightarrow v } \right|\sin \theta = 1 \Rightarrow \left| {\overrightarrow v } \right| = {1 \over {\sin \theta }}$

which shows that there are infinitely many possible values exist for $\overrightarrow v $ (here $\theta$ is angle between the vectors $\overrightarrow v $ and $\widehat u$).

Hence, option (B) is correct.

Now, $\widehat \omega \,.\,(\widehat u \times \overrightarrow v ) = 1$

$\left| {\widehat \omega \,.\,(\widehat u \times \overrightarrow v )} \right| = 1$

That is, $\left| {\widehat \omega } \right|\left| {\widehat u \times \overrightarrow v } \right|\cos \alpha = 1$

where $\alpha$ is the angle between $\widehat \omega $ and $\widehat u \times \overrightarrow v $.

Therefore, (1)(1)cos$\alpha$ = 1

$\Rightarrow$ a = 0

which means that $\widehat \omega $ and $\widehat u \times \overrightarrow v $ are parallel vector or $\widehat \omega $ is perpendicular vector to $\widehat u$ and $\overrightarrow v $.

$\widehat u\,.\,\widehat \omega = 0$

$({u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k)\,.\,\left( {{{(\widehat i + \widehat j + 2\widehat k)} \over {\sqrt 6 }}} \right) = 0$

${u_1} + {u_2} + 2{u_3} = 0$

If $\widehat u$ lies in xy-plane then u₃ = 0. Therefore,

${u_1} + {u_2} = 0 \Rightarrow {u_1} = - {u_2} \Rightarrow \left| {{u_1}} \right| = \left| {{u_2}} \right|$

Hence, option (C) is correct.