Let a line L passing through the point P(1,1,1) be perpendicular to the lines (x-4)/(4)=(y…

Question

Let a line L passing through the point $P(1,1,1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line L intersect the $yz-$ plane at the point Q . Another line parallel to L and passing through the point $S(1,0,-1)$ intersects the $yz$-plane at the point R . Then the square of the area of the parallelogram PQRS is equal to $____$ .

Answer: 6

Accepted Answer

$\begin{gathered} {L}_{1}:\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}={\lambda }_{1} \ {L}_{2}:\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}={\lambda }_{2} \end{gathered}$

given that $L$ is passing through $P(1,1,1)$

and perpendicular to ${L}_{1}$ and ${L}_{2}$.

dr's of $L=$ dr's of ${L}_{1}	imes$ dr's of ${L}_{2}$

$\begin{gathered} =|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 4 \quad  1 \quad  1 \ 1 \quad  1 \quad  0\end{array}| \ = \quad  \hat{i}(0-1)-\hat{j}(0-1)+\hat{k}(4-1) \ = \quad  -\hat{i}+\hat{j}+3\hat{k} \end{gathered}$

Equation of line $L:\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-1}{3}=\lambda $

It is given that $L$ intersect $yz$ plane at $Q$ for $Q$ on $y-z$ plane $x=0$

$\lambda =\frac{0-1}{-1}=1$

$\frac{y-1}{1}=\lambda \Rightarrow y-1=1\Rightarrow y=2$

$\frac{z-1}{3}=\lambda \Rightarrow \frac{z-1}{3}=1\Rightarrow z=4$

Point $Q$ is $(0,2,4)$

Another line parallel to $L$ let ${L}^{'}$ passing through $S(1,0,-1)$.

The area of parallelogram $PQRS$

$\begin{gathered} = |\overset{	o \leftarrow }{PQ}	imes \overset{	o \leftarrow }{PS}| \ \overset{	o \leftarrow }{PQ}=Q-P=(0,2,4)-(1,1,1)=(-1,1,3) \ \overset{	o \leftarrow }{PS}=S-P=(1,0,-1)-(1,1,1)=(0,-1,-2) \end{gathered}$

$=\hat{i}(-2+3)-\hat{j}(2-0)+\hat{k}(1-0)$

$=\hat{i}-2\hat{j}+\hat{k}$

$=|\overset{	o \leftarrow }{PQ}	imes \overset{	o \leftarrow }{PS}|=\sqrt{{1}^{2}+(-2{)}^{2}+(1{)}^{2}}=\sqrt{6}$

Square of area $=6$

Solutions · 1