[shahkb4]

Solution

\[ \mathrm{L}_{1} \xrightarrow{\mathrm{drs}} 4,1,1, \mathrm{~L}_{2} \xrightarrow{\mathrm{drs}} 1,1,0 \]

required $\mathrm{L}, \mathrm{L} \perp \mathrm{L}_{1} \& \mathrm{~L} \perp \mathrm{~L}_{2}$

\[ \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{array}\right|=-\hat{i}+\hat{j}+3 \hat{j} \]

line $\mathrm{L} \xrightarrow{\mathrm{drs}}-1,1,3$
line $\mathrm{L} \frac{\mathrm{x}-1}{-1}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-1}{3}=\lambda \mathrm{Q}(-\lambda+1, \lambda+1,3 \lambda+1)$
Q lies on yz plane

\[ \begin{aligned} & \mathrm{Q}_{\mathrm{x}}=0 \Rightarrow-\lambda+1=0 \Rightarrow \lambda=1 \\ & \mathrm{Q}(0,2,4) \mathrm{S}(1,0,-1) \\ & \overrightarrow{\mathrm{PQ}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{PS}}=-\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \end{aligned} \]

area of parallelogram $=|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PS}}|=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 1 & 3 \\ 0 & 1 & -2\end{array}\right|$

\[ =|\hat{\imath}-2 \hat{\jmath}+\hat{k}|=\sqrt{6} \text { Answer(6) } \]

[shahkb4]

Solution

\[ \begin{aligned} & \mathrm{L}_{1}: \frac{\mathrm{x}}{2}=\frac{\mathrm{y}+\mathrm{a}}{1}=\frac{\mathrm{z}}{1}=\lambda \\ & \mathrm{A}(2 \lambda, \lambda-\mathrm{a}, \lambda) \\ & \mathrm{L}_{2}: \frac{\mathrm{x}-2 \mathrm{~b}}{2}=\frac{\mathrm{y}-\mathrm{a}}{2}=\frac{\mathrm{z}+2 \mathrm{~b}}{-5}=\mu \\ & \mathrm{A}(2 \mu+2 \mathrm{~b}, \mu+\mathrm{a},-5 \mu-2 \mathrm{~b}) \\ & \Rightarrow \lambda=\mu+\mathrm{b}, \lambda=\mu+2 \mathrm{a}, \mu+2 \mathrm{a}=-5 \mu-2 \mathrm{~b} \end{aligned} \]

From first two, $\mathrm{b}=2 \mathrm{a}$
From third, $6 \mu=-2 b-2 a=-4 a-2 a$

\[ \lambda=a, \mu=-a \]

We have Point $\mathrm{A}(2 \mathrm{a}, 0, \mathrm{a})$ and $\mathrm{P}(\mathrm{a}, 2, \mathrm{a})$

\[ \begin{aligned} & A P \xrightarrow{\mathrm{drs}} a,-2,0 ; L_{1} \xrightarrow{\mathrm{drs}} 2,1,1 \\ & A P \perp L_{1} \\ & 2 a-2+0=0 \Rightarrow a=1 \\ & b=2 a=2 \therefore a+b=3 \text { Answer(3) } \end{aligned} \]

[shahkb4]

Solution

\[ \begin{aligned} & \mathrm{R}(2 \lambda+1,3 \lambda+2,4 \lambda+3) \text { on } \mathrm{L}_{1} \\ & \mathrm{R}(5 \mu+4,2 \mu+1, \mu) \text { on } \mathrm{L}_{2} \\ & 2 \lambda+1=5 \mu+4 \mid 3 \lambda+2=2 \mu+1 \\ & =\lambda=\frac{5 \mu}{2}+\frac{3}{2} \left\lvert\, \Rightarrow \frac{15 \mu}{2}+\frac{9}{2}+1=2 \mu\right. \\ & \mathrm{P}(2 \lambda+1,3 \lambda+2,4 \lambda+3) \text { on } \mathrm{L}_{1} \\ & \mathrm{Q}(5 \mu+4,2 \mu+1, \mu) \text { on } \mathrm{L}_{2} \\ & |\overrightarrow{\mathrm{PR}}|^{2}=29 \\ & \Rightarrow(2 \lambda+2)^{2}+(3 \lambda+3)^{2}+(4 \lambda+4)^{2}=29 \\ & \Rightarrow 29 \lambda^{2}+4+9+16+8 \lambda+18 \lambda+32 \lambda=29 \\ & \Rightarrow 29 \lambda^{2}+58 \lambda=0 \Rightarrow 29 \lambda(\lambda+2)=0 \\ & \Rightarrow \lambda=0, \mathrm{P}(1,2,3) \text { (accepted) } \end{aligned} \]

or $\lambda=-2, \mathrm{P}(-3,-4,-5)$ (rejected)
as P lies in First octant

\[ |\overrightarrow{\mathrm{PQ}}|^{2}=(5 \mu+3)^{2}+(2 \mu-1)^{2}+(\mu-3)^{2}=\frac{47}{3} \]

\[ \begin{aligned} & \Rightarrow 25 \mu^{2}+9+30 \mu+4 \mu^{2}+1-4 \mu+\mu^{2}+9-6 \mu=\frac{47}{3} \\ & \Rightarrow 30 \mu^{2}+20 \mu+\frac{10}{3}=0 \Rightarrow(3 \mu+1)^{2}=0 \Rightarrow \mu=-\frac{1}{3} \\ & \mathrm{Q}\left(\frac{7}{3}, \frac{1}{3},-\frac{1}{3}\right) \mathrm{R}(-1,-1,-1) \\ & \begin{aligned} 27(\mathrm{QR})^{2} & =27\left(\left(\frac{10}{3}\right)^{2}+\left(\frac{4}{3}\right)^{2}+\left(+\frac{2}{3}\right)^{2}\right) \\ & =3(120)=360 \text { Answer }(2) \end{aligned} \end{aligned} \]

[shahkb4]

Solution

First Method:
line $\mathrm{L}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$

\[ \begin{aligned} & \mathrm{D}(\lambda, 2 \lambda+1, \lambda \lambda+2) \\ & \mathrm{AD} \xrightarrow{\mathrm{drs}} \lambda-1,2 \lambda+1-6,3 \lambda+2-3 \\ & \text { or } \lambda-1,2 \lambda-5,3 \lambda-1 \end{aligned} \]

\[ \begin{aligned} & \mathrm{BC} \xrightarrow{\mathrm{drs}} 1,2,3 \\ & \mathrm{AD} \perp \mathrm{BC} \Rightarrow(\lambda-1) 1+(2 \lambda-5) 2+3(3 \lambda-1)=0 \Rightarrow 14 \lambda=14 \Rightarrow(\lambda=1) \\ & \mathrm{D}(1,3,5) \mathrm{A}(1,6,3) \\ & \mathrm{AD}=\sqrt{0+3^{2}+2^{2}}=\sqrt{13} \\ & \Delta \mathrm{ABC}=\frac{1}{2}(\mathrm{BC}) \mathrm{AD}=\frac{1}{2}(10) \sqrt{13}=5 \sqrt{13} \end{aligned} \]

Second Method:

B(4,9, $\alpha$ ) lies on line BC

\[ \frac{4}{1}=\frac{9-1}{2}=\frac{\alpha-2}{3} \Rightarrow \alpha=14 \]

B(4,9,14); A(1,6,3)

\[ \begin{aligned} & \overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+11 \hat{\mathrm{k}} ; \overrightarrow{\mathrm{BC}}=10 \frac{(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})}{\sqrt{1+4+9}} \\ & \cos \theta=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{BC}}|}=\frac{10}{\sqrt{14}} \frac{(3+6+33)}{10 \sqrt{9+9+121}} \\ & \cos \theta=\frac{3 \sqrt{14}}{\sqrt{139}} ; \sin \theta=\sqrt{1-\cos ^{2} \theta}=\frac{\sqrt{13}}{\sqrt{139}} \\ & \Delta \mathrm{ABC}=\frac{1}{2}(\mathrm{AB})(\mathrm{BC}) \sin \theta=\frac{1}{2}|3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}| 10\left(\frac{\sqrt{13}}{\sqrt{139}}\right) \\ & =5 \sqrt{13} \sqrt{9+9+121}\left(\frac{1}{\sqrt{139}}\right)=5 \sqrt{13} \end{aligned} \]

Third Method:

\[ \begin{aligned} \overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}} & +3 \hat{\mathrm{j}}+11 \hat{\mathrm{k}} ; \overrightarrow{\mathrm{BC}}=\frac{10}{\sqrt{14}}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ \Delta \mathrm{ABC} & =\frac{1}{2}|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|=\frac{1}{2} \frac{10}{\sqrt{14}}\left|\begin{array}{ccc} \hat{1} & \hat{j} & \hat{k} \\ 3 & 3 & 11 \\ 1 & 2 & 3 \end{array}\right| \\ & =\frac{5}{\sqrt{14}}|-13 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}|=\frac{5}{\sqrt{14}} \sqrt{169+4+9}=\frac{5 \sqrt{182}}{\sqrt{14}}=5 \sqrt{13} \text { Answer( } 1 \text { ) } \end{aligned} \]

[shahkb4]

Solution

\[ \begin{aligned} & 4 \ell+\mathrm{m}-\mathrm{n}=0 \Rightarrow \mathrm{n}=4 \ell+\mathrm{m} \\ & \mathrm{n}(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0 \\ & \Rightarrow(4 \ell+\mathrm{m})(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0 \\ & \Rightarrow 8 \ell \mathrm{~m}+40 \ell^{2}+2 \mathrm{~m}^{2}+10 \ell \mathrm{~m}+3 \ell \mathrm{~m}=0 \\ & \Rightarrow 40 \ell^{2}+16 \ell \mathrm{~m}+5 \ell \mathrm{~m}+2 \mathrm{~m}^{2}=0 \\ & \Rightarrow 8 \ell(5 \ell+2 \mathrm{~m})+\mathrm{m}(5 \ell+2 \mathrm{~m})=0 \\ & \Rightarrow(5 \ell+2 \mathrm{~m})(8 \ell+\mathrm{m})=0 \end{aligned} \]

\[ \begin{array}{l|l} 5 \mathrm{l}+2 \mathrm{~m}=0 & 8 \ell+\mathrm{m}=0 \\ \Rightarrow \mathrm{~m}=-\frac{5 \ell}{2} & \Rightarrow \mathrm{~m}=-8 \ell \\ \mathrm{n}=4 \ell+\mathrm{m} & \mathrm{n}=4 \ell+\mathrm{m}=4 \ell-8 \ell \\ \mathrm{n}=4 \ell-\frac{5 \ell}{2}=\frac{3 \ell}{2} & \Rightarrow \mathrm{n}=-4 \ell \\ \ell, \mathrm{~m}, \mathrm{~m} \rightarrow \ell,-\frac{5 \ell}{2}, \frac{3 \ell}{2} & \ell, \mathrm{~m}, \mathrm{n} \rightarrow \ell,-8 \ell,-4 \ell \end{array} \]

or line $\mathrm{L}_{1} \mathrm{drs} 2,-5,3 \mid$ line $\mathrm{L}_{2} \mathrm{drs}-1,8,4$

\[ \cos \theta=\left|\frac{2(-1)-5(8)+3(4)}{\sqrt{4+25+9} \sqrt{1+64+16}}\right|=\frac{30}{\sqrt{38} \times 9}=\frac{10}{3 \sqrt{38}} \operatorname{Answer}(4) \]

[shahkb4]

Solution

line $\mathrm{L} \equiv \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}+3}{6}=\lambda=\mathrm{t}$

\[ \begin{aligned} & \mathrm{A}(2 \lambda-1,3 \lambda-1,6 \lambda-3) \\ & \mathrm{L}_{1}: \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-9}{0}=\mu \\ & \mathrm{A}(2 \mu-1,3 \mu-1,9) \\ & 2 \lambda-1=2 \mu-1 \Rightarrow \lambda=\mu \\ & 6 \lambda-3=9 \Rightarrow 6 \lambda=12 \Rightarrow \lambda=2 \\ & \mathrm{~A}(3,5,9) \end{aligned} \]

Point $\mathrm{P}(2 \mathrm{t}-1,3 \mathrm{t}-1,6 \mathrm{t}-3)$

\[ \left.\begin{array}{l} A P=7 \\ \Rightarrow(2 t-4)^{2}+(3 t-6)^{2}+(6 t-12)^{2}=49 \\ \Rightarrow(t-1)(t-3)=0 \Rightarrow t=1,3 \\ t=1 \quad P_{1}(1,2,3) \\ t=3 \quad P_{2}(5,8,15) \end{array}\right] . \sum_{a, b, c \in S}(a+b+c)=1+2+3+5+8+15=34 \text { Answer }(1) \text {. } \]

[shahkb4]

Solution

\[ \begin{aligned} & \text { line } \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} \\ & P Q \xrightarrow[\text { drs }]{ } 5-1, b-2, c-a \\ & \text { or } 4, b-2, c-a \\ & \text { line } \xrightarrow{\text { drs }} 3,2,-2 \\ & P Q \perp \text { line } \Rightarrow 4(3)+2(b-2)-2(c-a)=0 \\ & \Rightarrow b-c+a+4=0 \end{aligned} \]

$\mathrm{M}=$ Mid Point of PQi.e. $\mathrm{P}(1,2, \mathrm{a}) \mathrm{Q}(5, \mathrm{~b}, \mathrm{c})$, $\mathrm{M}\left(3, \frac{\mathrm{~b}+2}{2}, \frac{\mathrm{a}+\mathrm{c}}{2}\right)$ lies on given line

\[ \begin{aligned} & \frac{3-6}{3}=\frac{\frac{b+2}{2}-7}{2}=\frac{\frac{a+c}{2}-7}{-2} \\ & \Rightarrow-1=\frac{b-12}{4}=\frac{a+c-14}{-4} \end{aligned} \]

from (1): $8-\mathrm{c}+\mathrm{a}+4=0$

\[ \Rightarrow \mathrm{c}=\mathrm{a}+12 \]

From (2), $\mathrm{a}+\mathrm{a}+12=18 \Rightarrow \mathrm{a}=3$

\[ \begin{aligned} & c=3+12=15 \\ & a^{2}+b^{2}+c^{2}=9+64+225=298 \text { Answer }(3) \end{aligned} \]

[shahkb4]

Solution

let point B on given line

\[ \begin{aligned} & \mathrm{B}(2 \lambda-1,3 \lambda+3,1-\lambda) \\ & \mathrm{AB} \xrightarrow{\mathrm{drs}} 2 \lambda-1-5,3 \lambda+3-4,1-\lambda-2 \\ & \text { or } 2 \lambda-6,3 \lambda-1,-\lambda-1 \end{aligned} \]

line $\xrightarrow{\mathrm{drs}} 2,3,-1$
$\mathrm{AB} \perp$ line

\[ \begin{aligned} & 2(2 \lambda-6)+3(3 \lambda-1)-1(-\lambda-1)=0 \\ & \Rightarrow 4 \lambda-12+9 \lambda-3+\lambda+1=0 \Rightarrow 14 \lambda=14 \Rightarrow \lambda=1 \\ & \mathrm{~B}(1,6,0) \equiv(\alpha, \beta, \gamma) \\ & \alpha=1, \beta=6, \gamma=0 \end{aligned} \]

required length of Projection

\[ =\frac{|(\alpha \hat{\imath}+\beta \hat{\jmath}+\gamma \hat{k}) \cdot(6 \hat{\imath}+2 \hat{j}+3 \hat{k})|}{\sqrt{36+4+9}}=\left|\frac{6 \alpha+2 \beta+3 \gamma}{7}\right|=\frac{6+12+0}{7}=\frac{18}{7} \operatorname{Answer}(3) \]

[shahkb4]

Solution

\[ \begin{aligned} & \alpha=\beta=\gamma \\ & \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\ & \Rightarrow 3 \cos ^{2} \alpha=1 \Rightarrow \cos \alpha=\frac{1}{\sqrt{3}} \end{aligned} \]

direction ratios $\rightarrow 1,1,1$

\[ \begin{aligned} & \text { line } \frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}=\lambda \\ & B(\lambda-3, \lambda+5, \lambda+2) \\ & A B \xrightarrow{d r s} \lambda-3+2, \lambda+5-r, \lambda+2-1 \text { or } \lambda-1, \lambda+5-r, \lambda+1 \\ & \text { line } \xrightarrow{\mathrm{drs}} 1,1,1 \end{aligned} \]

line $\perp \mathrm{AB} \Rightarrow(\lambda-1) 1+(\lambda+5-\mathrm{r}) 1+(\lambda+1) 1=0$

\[ \begin{aligned} & \Rightarrow \lambda=\frac{r-5}{3} \\ & \mathrm{~B}(\lambda-3, \lambda+5, \lambda+2) \equiv\left(\frac{r-14}{3}, \frac{r+10}{3}, \frac{r+1}{3}\right) \\ & \mathrm{A}(-2, r, 1) \\ & \mathrm{AB}^{2}=\frac{14}{3} \Rightarrow\left(\frac{r-14}{3}+2\right)^{2}+\left(\frac{r+10}{3}-r\right)^{2}+\left(\frac{r+1}{3}-1\right)^{2}=\frac{14}{3} \\ & \Rightarrow\left(\frac{r-8}{3}\right)^{2}+\left(\frac{10-2 r}{3}\right)^{2}+\left(\frac{r-2}{3}\right)^{2}=\frac{14}{3} \\ & \Rightarrow r^{2}+64-16 r+100+4 r^{2}-40 r+r^{2}-4 r+4=14 \times 3=42 \\ & \Rightarrow 6 r^{2}-60 r+12 r=0 \Rightarrow r^{2}-10 r+21=0 \\ & \Rightarrow(r-7)(r-3)=0 \Rightarrow r=7,3 \text { Sum }=10 \text { Answer }(4) \end{aligned} \]

[shahkb4]

Solution

line $\mathrm{L}_{1} \frac{\mathrm{x}-2}{-3}=\frac{\mathrm{y}-6}{2}=\frac{\mathrm{z}-7}{4}=\lambda_{1}$
line $\mathrm{L}_{2} \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-5}{3}=\lambda_{2}$
Point $\mathrm{C}\left(-3 \lambda_{1}+2,2 \lambda_{1}+6,4 \lambda_{1}+7\right)$ on line $\mathrm{L}_{1}$
Point $\mathrm{D}\left(2 \lambda_{2}+4, \lambda_{2}+3,3 \lambda_{2}+5\right)$ on line $\mathrm{L}_{2}$

\[ \begin{aligned} & \mathrm{CD} \xrightarrow{\mathrm{drs}} 2 \lambda_{2}+3 \lambda_{1}+2, \lambda_{2}-2 \lambda_{1}-3,3 \lambda_{2}-4 \lambda_{1}-2 \\ & \text { or }-3,5,16 \\ & \frac{2 \lambda_{2}+3 \lambda_{1}+2}{-3}=\frac{\lambda_{2}-2 \lambda_{1}-3}{5}=\frac{3 \lambda_{2}-4 \lambda_{1}-2}{16} \end{aligned} \]

From first \& middle

\[ \begin{aligned} & 10 \lambda_{2}+15 \lambda_{1}+10=-3 \lambda_{2}+6 \lambda_{1}+9 \\ & \Rightarrow 13 \lambda_{2}+9 \lambda_{1}+1=0 \end{aligned} \]

From middle \& third

\[ \begin{aligned} & 16 \lambda_{2}-32 \lambda_{1}-48=15 \lambda_{2}-20 \lambda_{1}-10 \\ & \Rightarrow \lambda_{2}=12 \lambda_{1}+38 \\ & 13\left(12 \lambda_{1}+38\right)+9 \lambda_{1}+1=0 \\ & \Rightarrow 165 \lambda_{1}+495=0 \Rightarrow \lambda_{1}=-3 \\ & \Rightarrow \lambda_{2}=-36+38 \Rightarrow \lambda_{2}=2 \text { Ans. } \\ & \lambda_{1}=-3, \lambda_{2}=2 \end{aligned} \]

we get $\mathrm{C}(11,0,-5) \mathrm{D}(8,5,11)$

\[ \mathrm{CD}^{2}=9+25+256=290 \text { Answer }(2) \]

[shahkb4]

Solution

\[ \begin{aligned} & \text { line } \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z}{1}=\lambda \\ & P(2 \lambda+1,-3 \lambda-1, \lambda) \cdot A(1,-1,0) \\ & A P=\sqrt{4 \lambda^{2}+9 \lambda^{2}+\lambda^{2}}=4 \sqrt{14} \\ & \Rightarrow 14 \lambda^{2}=16 \times 14 \Rightarrow \lambda= \pm 4 \\ & \lambda=4 P(9,-13,4) ; O P=\sqrt{81+169+16}=\sqrt{266} \text { greatest } \\ & \lambda=-4 P(-7,11,-4) ; O P=\sqrt{49+121+16}=\sqrt{186} \text { least } \\ & P(-7,11,-4) \equiv(\alpha, \beta, \gamma) ; \alpha=-7, \beta=11, \gamma=-4 \\ & \text { Shortest distance bw skew lines }=\frac{\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|}{\| \hat{i}} \hat{j} \frac{\hat{k}}{a_{1}} b_{1} c_{1}\left\|a_{2} b_{2} c_{2}\right\| \\ & \qquad\left|\begin{array}{ccc} -2 & 1 & -7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{array}\right| \\ & \qquad=\frac{|-2(-1)+5-7(-3)|}{|-\hat{1}+5 \hat{j}-3 \hat{k}|}=\frac{28}{\sqrt{35}}=4 \sqrt{\frac{7}{5}} \text { Answer(2) } \\ & \left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 1 \end{array}\right| \end{aligned} \]

[shahkb4]

Solution

As P is equidistant from lines AB and AC . It means P lies on angle bisector $\angle \mathrm{BAC}$ $\cos 2 \theta=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{AC}}|}=\frac{(3 \hat{\imath}+\hat{\jmath}-\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+3 \hat{k})}{\sqrt{9+1+1} \sqrt{1+1+9}}$

\[ \begin{aligned} & \Rightarrow \cos 2 \theta=\frac{3-1-3}{11}=-\frac{1}{11} \\ & \Rightarrow 1-2 \sin ^{2} \theta=-\frac{1}{11} \Rightarrow \sin \theta=\sqrt{\frac{6}{11}} \end{aligned} \]

area of $\Delta \mathrm{ABP}=\frac{1}{2}(\mathrm{AB})(\mathrm{AP}) \sin \theta$

\[ \begin{aligned} & =\frac{1}{2} \sqrt{9+1+1}\left(\frac{\sqrt{5}}{2}\right) \sqrt{\frac{6}{11}} \\ & =\frac{1}{2} \times \frac{\sqrt{5}}{2} \sqrt{6}=\frac{\sqrt{30}}{4} \text { Answer }(3) \end{aligned} \]

[shahkb4]

Solution

Point S which is equidistant from lines PQ and PR , lies on angle bisector of $\angle \mathrm{QPR}$

\[ \begin{aligned} & |\overrightarrow{\mathrm{PR}}|=\sqrt{a^{2}+b^{2}+16}=9 \\ & \Rightarrow a^{2}+b^{2}+16=81 \\ & \Rightarrow a^{2}+b^{2}=65 \ldots(1) \\ & |\overrightarrow{\mathrm{PQ}}|=\sqrt{4+1+4}=3 \end{aligned} \]

S divides QR in 1:3 ratio

\[ \begin{aligned} & S\left(\frac{a-6}{4}, \frac{b-3}{4}, \frac{1}{2}\right) \\ & P S=\frac{(a-6) \hat{i}}{4}+\frac{(b-3) \hat{j}}{4}+\frac{\hat{k}}{2} \\ & \overrightarrow{P S}=\frac{\hat{\imath}-7 \hat{\jmath}+2 \hat{k}}{4}=\frac{(a-6) \hat{\imath}+(b-3) \hat{\jmath}+2 \hat{k}}{4} \end{aligned} \]

We get $\mathrm{a}=7, \mathrm{~b}=-4$

\[ 3 a-4 b=3(7)-4(-4)=21+16=37 \text { Answer }(37) \]

[shahkb4]

Solution

Any vector $\vec{v}$ in plane of $\vec{a} \& \vec{b}$

\[ \begin{aligned} & \vec{v}=x \vec{a}+y \vec{b}=x(2 \hat{i}-\hat{j}-\hat{k})+y(\hat{i}+3 \hat{j}-\hat{k}) \\ & \Rightarrow \vec{v}=(2 x+y) \hat{i}+(3 y-x) \hat{j}-(x+y) \hat{k} \end{aligned} \]

Projection length $\overrightarrow{\mathrm{v}}$ on $\overrightarrow{\mathrm{c}}=\frac{|\overrightarrow{\mathrm{v}} \cdot \overrightarrow{\mathrm{c}}|}{|\overrightarrow{\mathrm{c}}|}=\frac{1}{\sqrt{14}}$

\[ \begin{aligned} & \Rightarrow \frac{|2(2 x+y)+1(3 y-x)-7(x+y)|}{\sqrt{4+1+9}}=\frac{1}{\sqrt{14}} \\ & \Rightarrow|4 x+2 y+3 y-x-3 x-3 y|=1 \\ & \Rightarrow|2 y|=1 \Rightarrow y= \pm \frac{1}{2} \\ & |\vec{V}|=\sqrt{(2 x+y)^{2}+(3 y-x)^{2}+(x+y)^{2}} \\ & =\sqrt{4 x^{2}+y^{2}+4 x y+9 y^{2}+x^{2}-6 x y+x^{2}+y^{2}+2 x y} \\ & =\sqrt{6 x^{2}+11 y^{2}}=\sqrt{6 x^{2}+\frac{11}{4}}=\frac{\sqrt{24 x^{2}+11}}{2} \end{aligned} \]

(1) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=\frac{\sqrt{21}}{2} \Rightarrow 24 \mathrm{x}^{2}+11=21$

\[ \Rightarrow \mathrm{x}^{2}=\frac{10}{24} \Rightarrow \mathrm{x}= \pm \sqrt{\frac{5}{12}} \]

(2) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=13 \Rightarrow 24 \mathrm{x}^{2}+11=169 \times 4$

\[ \Rightarrow x \approx \pm 5.26 \]

(3) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=\frac{\sqrt{35}}{2} \Rightarrow \mathrm{x}=1$
(4) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=7 \Rightarrow 24 \mathrm{x}^{2}+11=4 \times 49$

\[ \Rightarrow \mathrm{x} \approx \pm 2.27 \text { Answer(1,2,3,4) } \]

[shahkb4]

Solution

\[ \begin{aligned} & 2 \vec{a} \times \vec{c}+3 \vec{b} \times \vec{c}=0 \Rightarrow(2 \vec{a}+3 \vec{b}) \times \vec{c}=0 \\ & \Rightarrow \vec{c} \| 2 \vec{a}+3 \vec{b} \Rightarrow \vec{c}=\lambda(2 \vec{a}+3 \vec{b}) \\ & \Rightarrow \vec{c}=\lambda(2(2 \hat{i}-5 \hat{j}+5 \hat{k})+3(\hat{i}-\hat{j}+3 \hat{k})) \\ & \Rightarrow \vec{c}=\lambda(7 \hat{i}-13 \hat{j}+19 \hat{k}) \\ & (\vec{a}-\vec{b}) \cdot \vec{c}=(\hat{i}-4 \hat{j}+2 \hat{k}) \cdot \lambda(7 \hat{i}-13 \hat{j}+19 \hat{k})=-97 \\ & \Rightarrow \lambda(7+52+38)=-97 \Rightarrow \lambda=-1 \\ & \vec{c}=-7 \hat{i}+13 \hat{j}-19 \hat{k} \\ & \vec{c} \times \hat{k}=-7 \hat{i} \times \hat{k}+13 \hat{j} \times \hat{k}-19 \hat{k} \times \hat{k} \\ & \Rightarrow \vec{c} \times \hat{k}=7 \hat{j}+13 \hat{i}-0 \\ & |\vec{c} \times \hat{k}|^{2}=49+169=218 \text { Answer }(3) \end{aligned} \]

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