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STD 12 - 11. three dimension geometry — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 11. three dimension geometry1 Mark
MCQ
From a point $P(\lambda, \lambda, \lambda)$, perpendiculars $P Q$ and $P R$ are drawn respectively on the lines $y=x, z=1$ and $y$ $=-x, z=-1$. If $P$ is such that $\angle Q P R$ is a right angle, then the possible value(s) of $\lambda$ is(are)
  • A
    $\sqrt{2}$
  • B
    $1$
  • ✓
    $-1$
  • D
    $-\sqrt{2}$
✓

Answer

Correct option: C.
$-1$
c
Line is

$\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}=\alpha $$\quad..............(1)$

$Q(\alpha, \alpha, 1)$

Direction ratio of $PQ$ are

$\lambda-\alpha, \lambda-\alpha, \lambda-1$

Since $PQ$ is perpendicular to $(1)$

$\therefore \quad \lambda-\alpha+\lambda-\alpha+0=0 $

$ \lambda=\alpha$

$\therefore \quad$ Direction ratio of $P Q$ are

$0,0, \lambda-1$

Another line is

$\frac{x-0}{-1}= \frac{y-0}{1}=\frac{z+1}{0}=\beta $

$\therefore \quad R(-\beta, \beta,-1) $

$\therefore \quad \text { Direction ratio of PR are } $

$ \lambda+\beta, \lambda-\beta, \lambda+1$

Since $P Q$ is perpendicular to $(ii)$

$\therefore -\lambda-\beta+\lambda-\beta=0 $

$ \beta=0 $

$\therefore \quad R(0,0,-1) $

$\text { and } \text { Direction ratio of } P Q \text { are } \lambda, \lambda, \lambda+1 $

$\text { Since } P Q \perp PR $

$\therefore 0+0+\lambda^2-1=0 \Rightarrow \lambda= \pm 1 \Rightarrow B, C$

For $\lambda=1$ the point is on the line so it will be rejected.

$\Rightarrow \quad \lambda=-1$

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