First slide

Planes in 3D

Question

If the plane $3 (x - 2) + (y - 2) + 6 (z + 3) = 0$ contains the line $\frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z + 3}{1}$ whose inclination with X-axis is 60⁰, then it satisfies the equation

Moderate

A

$6 a^{2} + 36 a + 37 = 0$
B

$36 a^{2} + 37 = 0$
C

$36 a^{2} + 37 a + 36 = 0$
D

$a + 3 = 0$

Solution

The d.r’s of line are $(a, b, 1)$
$∴$ The d.c.’s of line are $(\frac{a}{\sqrt{a^{2} + b^{2} + 1}}, \frac{b}{\sqrt{a^{2} + b^{2} + 1}}, \frac{1}{\sqrt{a^{2} + b^{2} + 1}})$
$∴ \cos 60^{0} = \frac{a}{\sqrt{a^{2} + b^{2} + 1}}$ $\Rightarrow \frac{1}{2} = \frac{a}{\sqrt{a^{2} + b^{2} + 1}}$
i.e. $2 a = \sqrt{a^{2} + b^{2} + 1}$ ,
i.e., $3 a^{2} = b^{2} + 1$ ............(1)
As the plane contains the given line, we have $3 (a) + 1 (b) + 6 (1) = 0$ ................(2)
Eliminating b from (1) & (2),
we get $3 a^{2} = {\{- (6 + 3 a)\}}^{2} + 1$
i.e. $6 a^{2} + 36 a + 37 = 0$ ,

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