Download App

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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 11. three dimension geometry4 Marks
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.

$\Rightarrow \quad \lambda=-1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The sum of the series ${1.3^2} + {2.5^2} + {3.7^2} + ..........$ upto $20$ terms is
→
If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to :
→
If ${S_n} = \frac{{n(n + 1)\left( {n + 2} \right)}}{6}$ then $\sum\limits_{n = 1}^\infty  {\frac{1}{{{t_n}}}}  = $
→
Let $f(\theta)=\sin \left(\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right)$, where $-\frac{\pi}{4}<\theta<\frac{\pi}{4}$. Then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is
→
Let $f(x)$ be defined for all $x > 0$ and be continuous. Let $f(x)$ satisfy $f\left( {\frac{x}{y}} \right) = f(x) - f(y)$ for all $x, y$ and $f(e) = 1,$ then
→
The period of the function $f\left( x \right)$ = $\frac{x}{3} - \left[ {\frac{x}{3} - 5} \right] + \frac{x}{4} - \left[ {\frac{x}{4} - 5} \right] + \frac{x}{5} - \left[ {\frac{x}{5} - 5} \right]$ is (where $[.]$ is $G.I.F.$ )
→
The mean and variance of a binomial distribution are $6$ and $4$. The parameter $n$ is
→
$\beta \gamma$ be the foot of perpendicular from the point $(1,2,3)$ on the line $\frac{x+3}{5}=\frac{y 1}{2}=\frac{z+4}{3}$ .then $19(\alpha+\beta+\gamma)$ is equal to :
→
$\lim _{x \rightarrow \frac{x}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$ is equal to $:$
→
Let $(a, b)$ be the point of intersection of the curve $x^{2}=2 y$ and the straight line $y-2 x-6=0$ in the second quadrant. Then the integral $I=\int_{a}^{b} \frac{9 x^{2}}{1+5^{x}} d x$ is equal to :
→