Question

A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\prime}$, to the $y$-axis at $45$ and to the $z$-axis at an acute angle If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then

• A. $\sqrt{2} a-b+c=1$
• B. $a+\sqrt{2} b+c=1$
• C. $\sqrt{2} a+b+c=1$
• D. $a+b+\sqrt{2} c=1$

Answer 1

The Correct Option is B

$\hat{v}=\cos 60^{\circ }\hat{i}+\cos 45^{\circ }\hat{j}+\cos \gamma \hat{k}$
$\Rightarrow \frac{1}{4}+\frac{1}{2}+{\cos }^2\gamma =1(\gamma \to \text{Acute})$
$\Rightarrow \cos \gamma =\frac{1}{2}$
$\Rightarrow \gamma =60^{\circ }$
Equation of plane is
$\frac{1}{2}(x-\sqrt{2})+\frac{1}{\sqrt{2}}(y+1)+\frac{1}{2}(z-1)=0$
$\Rightarrow x+\sqrt{2}y+z=1$
(a, b, c) lies on it.
$\Rightarrow a+\sqrt{2}b+c=1$
So, the corrrect option is (B) : $a+\sqrt{2} b+c=1$