Three Dimensional Geometry Question 22
Question 22 - 30 January - Shift 2
A vector $\overrightarrow{{}v}$ 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 $\overrightarrow{{}v}$, then
(1) $\sqrt{2} a+b+c=1$
(2) $a+b+\sqrt{2} c=1$
(3) $a+\sqrt{2} b+c=1$
(4) $\sqrt{2} a-b+c=1$
Show Answer
Answer: (3)
Solution:
Formula: Equation of a plane, Direction ratio and direction cosines
$\hat{v}=\cos 60^{\circ} \hat{i}+\cos 45^{\circ} \hat{j}+\cos \gamma \hat{k}$
$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \gamma=1 \quad(\gamma \to$ 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.
$\Rightarrow a+\sqrt{2} b+c=1$
