Download App

11. three dimension geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths11. three dimension geometry1 Mark
MCQ
If two lines $L_1$ and $L_2$ in space, are defined by ${L_1} = \{ x = \sqrt \lambda  y + \left( {\sqrt \lambda   - 1} \right),z = \left( {\sqrt \lambda   - 1} \right)y + \sqrt \lambda  \} $ and ${L_2} = \{ x = \sqrt \mu  y + \left( {1 - \sqrt \mu  } \right),z = \left( {1 - \sqrt \mu  } \right)y + \sqrt \mu  \} $ then $L_1$ is perpendicular to $L_2$, for all non-negative reals $\lambda $ and $ \mu $, such that
  • A
    $\sqrt \lambda   + \sqrt \mu   = 1$
  • B
    $\lambda  \ne \mu $
  • C
    $\lambda  + \mu  = 0$
  • ✓
    $\lambda  = \mu $
✓

Answer

Correct option: D.
$\lambda  = \mu $
d
For $L_{1}$

$x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}$          .....$(i)$

$z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$         ......$(ii)$

From $(i)$ and $(ii)$

$\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$         ....$(A)$

The equation $(\mathrm{A})$ is the equation of line $\mathrm{L}_{1}$

Similarly equation ofline $\mathrm{L}_{2}$ is

$x-\frac{(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y-0}{1}=\frac{z-\sqrt{\mu}}{1-\sqrt{\mu}}$     ....$(B)$

Since $\mathrm{L}_{1} \perp \mathrm{L}_{2},$ therefore

$\sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0$

$\Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu}$

$\Rightarrow \lambda=\mu$

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

More "M.C.Q (1 Marks)" from 11. three dimension geometry→11. three dimension geometry — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $y=y(x)$ is the solution curve of the differential equation $\left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0$, $x>2, y(4)=\frac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals: . . . . . . . 
→
The differential equation of the family of curves represented by the equation ${x^2} + {y^2} = {a^2}$ is
→
The derivative of function $\cos (\sin x)$ is :
→
Let $A$ be a $2 \times 2$ symmetric matrix such that $A\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right]$ and the determinant of $A$ be 1.

If $A^{-1}=\alpha A+\beta I$, where $I$ is an identity matrix of order $2 \times 2$, then $\alpha+\beta$ equals....................

→
A function $y = f (x)$ satisfies $(x + 1) .\,\,$$ f ‘ (x) - 2 (x2 + x) f (x) = \frac{{{e^{{x^2}}}}}{{(x + 1)}}\,\,$,  If $f (0) = 5 ,$ then $f (x)$ is
→
Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{ i }+\frac{1}{2} \hat{ j }+2 \hat{ k }$. If $\overrightarrow{ a } \times(2 \hat{ i }+\hat{ k })=2 \hat{ i }-13 \hat{ j }-4 \hat{ k }$, then the projection of the vector $\vec{a}$ on the vector $2 \hat{ i }+2 \hat{ j }+\hat{ k }$ is
→
If $\cos\alpha,\cos\beta,\cos\gamma$ are the direction cosines of a vector $\vec{\text{a}}$ then $\cos2\alpha+\cos2\beta+\cos2\gamma$ is equal to:
  1. 2
  2. 3
  3. -1
  4. 0
→
Let $\mathrm{A}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0\end{array}\right) .$ Then $\mathrm{A}^{2025}-\mathrm{A}^{2020}$ is equal to :
→
Consider system of equations  $ x + y -az = 1$  ;  $2x + ay + z = 1$   ; $ax + y -z = 2$
→
If $\sin \left(\frac{y}{x}\right)=\log _0|x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3}$, then $\alpha^2$ is equal to
→