Vector Algebra

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} $