[shahkb4]

shahkb4

3 June, 2026 at 8:15 pm #1080

Solution

$\mathrm{L}_{1} \xrightarrow{\text { drs }} 4,1,1, \mathrm{~L}_{2} \xrightarrow{\text { drs }} 1,1,0$
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 $\mathrm{L} \xrightarrow{\text { drs }}-1,1,3$
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
$\mathrm{Q}_{\mathrm{x}}=0 \Rightarrow-\lambda+1=0 \Rightarrow \lambda=1$
$\mathrm{Q}(0,2,4) \mathrm{S}(1,0,-1)$
$\overrightarrow{\mathrm{PQ}}=-\hat{i}+\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{PS}}=-\hat{j}-2 \hat{k}$

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

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