Three Dimensional Geometry Question 4
Question 4 - 24 January - Shift 2
If the foot of the perpendicular drawn from $(1,9,7)$ to the line passing through the point $(3,2,1)$ and parallel to the planes $x+2 y+z=0$ and $3 y-z=3$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to
(1) -1
(2) 3
(3) 1
(4) 5
Show Answer
Answer: (4)
Solution:
Formula: Equation of Straight Line, Direction Cosines And Direction Ratios
Direction ratio of line $= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 0 & 3 & -1\end{vmatrix} $
$=\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3)$
$=-5 \hat{i}+\hat{j}+3 \hat{k}$
$M(-5 \lambda+3, \lambda+2,3 \lambda+1)$
$\overrightarrow{{}PM} \perp(-5 \hat{i}+\hat{j}+3 \hat{k})$
$-5(-5 \lambda+2)+(\lambda-7)+3(3 \lambda-6)=0$
$\Rightarrow 25 \lambda+\lambda+9 \lambda-10-7-18=0$
$\Rightarrow \lambda=1$
Point $\mathbf{M}=(-2,3,4)=(\alpha, \beta, \gamma)$
$\alpha+\beta+\gamma=5$
