Three Dimensional Geometry Question 10
Question 10 - 25 January - Shift 2
The foot of perpendicular of the point $(2,0,5)$ on the line $\frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}$ is $(\alpha, \beta, \gamma)$. Then.Which of the following is NOT correct?
(1) $\frac{\alpha \beta}{\gamma}=\frac{4}{15}$
(2) $\frac{\alpha}{\beta}=-8$
(3) $\frac{\beta}{\gamma}=-5$
(4) $\frac{\gamma}{\alpha}=\frac{5}{8}$
Show Answer
Answer: (3)
Solution:
Formula: A point and a line, Direction Cosines And Direction Ratio
$L: \frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text{ (let) }$
Let foot of perpendicular is
$P(2 \lambda-1,5 \lambda+1,-\lambda-1)$
$\overrightarrow{{}PA}=(3-2 \lambda) \hat{i}-(5 \lambda+1) \hat{j}+(6+\lambda) \hat{k}$
Direction ratio of line $\Rightarrow \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$
Now, $\Rightarrow \overrightarrow{{}PA} \cdot \overrightarrow{{}b}=0$
$\Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0$
$\Rightarrow \lambda=\frac{-1}{6}$
$P(2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv P(\alpha, \beta, \gamma)$
$\Rightarrow \alpha=2(-\frac{1}{6})-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3}$
$\Rightarrow \beta=5(-\frac{1}{6})+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6}$
$\Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6}$
Therefore, option (3) is here correct option according to question.
