Question

The direction ratios of a 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 $A(1,0,0)$ & $B(0,1,0)$ be two points on the plane and direction ratio of its normal be $ai+bj+ck$, then $(ai+bj+ck).\overrightarrow{BA}=0$
$\Rightarrow (ai+bj+ck).(i-j)=0$
$\Rightarrow a=b$ ....(1)
Since, $ai+bj+ck$ makes an angle $\frac{\pi }{4}$ with $x+y-3=0$
Therefore, $\cos \frac{\pi }{4}=\frac{(ai+bj+ck).(i+j)}{\sqrt{(a^2+b^2+c^2)(1^2+1^2)}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{a+b}{\sqrt{2(a^2+b^2+c^2)}}=\frac{2a}{\sqrt{2(2a^2+c^2)}}$ ....[ from (1) ]
$\Rightarrow 2a^2+c^2=4a^2$
$\Rightarrow c=\pm \sqrt{2}a$
Therefore, direction ratios are $(a,b,c)=(a,a,\pm \sqrt{2}a)=(1,1,\pm \sqrt{2})$
Ans: B