First slide

Planes in 3D

Question

The direction ratios of a normal to the plane through (1, 0, 0) and (0, 1, 0), which makes an angle of $\frac{π}{4}$ with the plane x + y = 3, are

Moderate

A

$< 1, \sqrt{2}, 1 >$
B

$< 1, 1, \sqrt{2} >$
C

$< 1, 1, 2 >$
D

$< \sqrt{2}, 1, 1 >$

Solution

Any plane through ( 1, 0, 0) is $a (x - 1) + by + cz = 0 - - - - (i)$
It passes through (0, 1, 0). Therefore, $a (0 - 1) + b (1) + c (0) = 0 or - a + b = 0 - - - - (ii)$
(i) makes an angle of $\frac{π}{4}$ with x + y=3,therefore
$\begin{array}{l} \cos \frac{π}{4} = \frac{a (1) + b (1) + c (0)}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{1 + 1 + 0}} \\ or \frac{1}{\sqrt{2}} = \frac{a + b}{\sqrt{2} \sqrt{a^{2} + b^{2} + c^{2}}} \\ or a + b = \sqrt{a^{2} + b^{2} + c^{2}} \end{array}$
Squaring, we get
$\begin{array}{l} a^{2} + b^{2} + 2 ab = a^{2} + b^{2} + c^{2} \\ or 2 ab = c^{2} or 2 a^{2} = c^{2} \\ or c = \sqrt{2} a \end{array}$
$\begin{aligned} Hence a : b : c = a : a : \sqrt{2} a \\ = 1 : 1 : \sqrt{2} \end{aligned}$

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