Question

The direction ratios of a normal to the plane through $(1,0,0),(0,1,0)$ which makes an angle of $\frac{\pi }{4}$ with the plane $x+y=3$ are

• A. $1,\sqrt{2},1$
• B. $1,1,\sqrt{2}$
• C. $1,1,2$
• D. $\sqrt{2},1,1$

Answer 1

The correct option is B $1,1,\sqrt{2}$

Let the equation of the plane be $Ax+By+Cz=D.$

Angle between two planes is given by cos$\theta =\frac{A_1{A}_2+B_1{B}_2+C_1{C}_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}$

$\therefore \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}=\frac{A\times 1+B\times 1}{\sqrt{A^2+B^2+C^2}\sqrt{1^2+1^2}}$

$\Rightarrow A^2+B^2+C^2=(A+B{)}^2$

$\Rightarrow A^2+B^2+C^2=A^2+B^2+2AB$

$\Rightarrow C^2=2AB$ ...... $\left(I\right)$

$\because$ The plane passes through $(1,0,0)$ and $(0,1,0)$.

$\Rightarrow A=D$ and $B=D$

From $\left(I\right)$, we have,

$C^2=2A^2$

$\Rightarrow C=\sqrt{2}A$

$\therefore$ equation of plane reduces to $Ax+Ay+\sqrt{2}Az=A$

$x+y+\sqrt{2}z=1$

$\therefore$ direction ratios of normal are ($1,1,\sqrt{2}$).

$\therefore$ ans. is option B.