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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 62 of 160

Marks: +1, -0

The angle between the planes

2x - y + z = 6

x + y + 2z = 3

is ______

$\frac{\pi}{3}$
$\cos^{-1}\left(\frac{1}{6}\right)$
$\frac{\pi}{4}$
$\frac{\pi}{6}$

Solution:

Given, planes are

2x - y + z = 6

and

x + y + 2z = 3

General equation of plane is

ax + by + cz = d

Normal vector of plane is

a\hat{i} + b\hat{j} + c\hat{k}

.

For plane

2x - y + z = 6

n_1 = 2\hat{i} - \hat{j} + \hat{k}

For plane

x + y + 2z = 3

n_2 = \hat{i} + \hat{j} + 2\hat{k}

Angle between

n_1

and

n_2

is given as,

\cos \theta = \frac{\left| n_1 \cdot n_2 \right|}{\left| n_1 \right| \left| n_2 \right|}

= \frac{(2-1+2)}{\sqrt{4+1+1} \sqrt{1+1+4}}

= \frac{3}{6} = \frac{1}{2}

\cos \theta = \cos \frac{\pi}{3}

∴ Angle between given planes is

\frac{\pi}{3}

.

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