Question

A vector $\vec{ v }$ in the first octant is inclined to the $x$ axis at $60^{\circ}$, to the $y$-axis at $45^{\circ}$ 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

Answer: (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 \quad(\gamma \rightarrow \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.
$\Rightarrow a+\sqrt{2} b+c=1$