Three Dimensional Geometry Question 27
Question 27 - 31 January - Shift 1
Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$. then $\alpha^{2}$ is equal to
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Answer: 180
Solution:
Formula: A point and a line, Direction Cosines And Direction Ratio, Area of triangle (6.7)
Any point on $L((2 \lambda+1),(-\lambda-1),(\lambda+3))$
$2(2 \lambda+1)+(-\lambda-1)+3(\lambda+3)=16$
$6 \lambda+10=16 \Rightarrow \lambda=1$
$\therefore P=(3,-2,4)$
$DR$ of $QR=\langle 2 \lambda,-\lambda, \lambda+6\rangle$
$DR$ of $L=\langle 2,-1,1\rangle$
$4 \lambda+\lambda+\lambda+6=0 \quad 6 \lambda+6=0 \Rightarrow \lambda=-1$
$Q=(-1,0,2)$
$ \overrightarrow{{}QR}=2 \hat{i}-\hat{j}-5 \hat{k} \quad \overrightarrow{{}QP}=4 \hat{i}-2 \hat{j}+2 \hat{k} $
$ \begin{aligned} & \overrightarrow{{}QR} \times \overrightarrow{{}QP}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -5 \\ 4 & -2 & 2 \end{vmatrix} =-12 \hat{i}-24 \hat{j} \\ & \alpha=\frac{1}{2} \times \sqrt{144+576} \Rightarrow \alpha^{2}=\frac{720}{4}=180 \end{aligned} $
