Question

The direction ratios of the normal to the plane through $(1,0,0)$ and $(0,1,0)$ 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

Answer: (B)

$1,1,\sqrt{2}$

Let the equation of the plane through $(1,0,0)$ and $(0,1,0)$ be

$ax+by+cz+d=0$

where $\overrightarrow{n}=(a,b,c)$ is unit normal vector to the plane.

$\Rightarrow \sqrt{a^2+b^2+c^2}=1$

$\Rightarrow a+d=0$ and $b+d=0$Subtracting these two we get $a-b=0$ ....$\left(1\right)$

Also since this plane makes angle $\frac{\pi }{4}$ with $x+y=3$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{(a,b,c).(1,1,0)}{1.\sqrt{2}}$

$\Rightarrow a+b=1$ ........$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get

$2a=1$ or $a=\frac{1}{2}$

From $\left(1\right)$ we have $a=b=\frac{1}{2}$

$\Rightarrow c^2=1-\frac{1}{4}-\frac{1}{4}=\frac{1}{2}$

or $c=\pm \frac{1}{\sqrt{2}}$

$\therefore$ Direction ratios of normal is $(a,b,c)=(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}})$

$=(1,1,\sqrt{2})$ by multiplying by $2$