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3D Geometry question

2007 · Shift 0 · Q44

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Mathematics
3D Geometry
2007 · Shift 0 · Q44

3D Geometry question

2007 · Shift 0 · Q44

JEE MainMathematics3D GeometryMCQ+4 / −1

If a line makes an angle of

\pi /4

with the positive directions of each of

x

-axis and

y

-axis, then the angle that the line makes with the positive direction of the

z

-axis is :

A
${\pi \over 4}$
B
${\pi \over 2}$
C
${\pi \over 6}$
D
${\pi \over 3}$

View written solutionFree

Correct answer: B

Let the angles made by the line with the positive directions of the $x$ -, $y$ -, and $z$ -axes be $\alpha, \beta, \gamma$ respectively.
For a line in 3D, the direction cosines satisfy

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.

Given:

\alpha = \frac{\pi}{4}, \qquad \beta = \frac{\pi}{4}.

So,

\cos^2\alpha = \cos^2\frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2},

and similarly,

\cos^2\beta = \frac{1}{2}.

Substitute into the direction cosine relation:

\frac{1}{2} + \frac{1}{2} + \cos^2\gamma = 1.

This gives

1 + \cos^2\gamma = 1

so,

\cos^2\gamma = 0.

Hence,

\cos\gamma = 0.

Therefore,

\gamma = \frac{\pi}{2}.

Since the angle with the positive direction of the $z$ -axis lies in $[0,\pi]$ , this is valid.

Checking options:

A: $\frac{\pi}{4}$ — incorrect
B: $\frac{\pi}{2}$ — correct
C: $\frac{\pi}{6}$ — incorrect
D: $\frac{\pi}{3}$ — incorrect

Therefore, the correct answer is B.

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