If a line makes $90^\circ$, $60^\circ$ and $30^\circ$ with the $x$, $y$ and $z$-axes respectively in the positive direction, then find its direction cosines.

Question

The angle between the lines, whose direction cosines $ l, m, n $ satisfy the equations: \[ l + m + n = 0 \quad \text{and} \quad 2l^2 + 2m^2 - n^2 = 0, \] is:

Answer 1

Step 1: Geometric reasoning for the x-axis angle:
A line making a $90^\circ$ angle with the x-axis is perpendicular to it, so the line has no component along the x-direction.
This means its direction cosine along x is zero, so $l = 0$, without directly plugging into a cosine formula.

Step 2: Using the given angles with y and z axes:
The line makes $60^\circ$ with the y-axis and $30^\circ$ with the z-axis, so by definition of direction cosine as the cosine of the angle with an axis:
\[ m = \cos 60^\circ = \frac{1}{2}, \qquad n = \cos 30^\circ = \frac{\sqrt3}{2} \]

Step 3: Confirm consistency:
Since direction cosines of any line always satisfy $l^2 + m^2 + n^2 = 1$, substitute the found values:
\[ 0 + \frac{1}{4} + \frac{3}{4} = 1 \]
This confirms the geometric reasoning agrees with the trigonometric values.

Final Answer:
The geometric shortcut for the right angle and direct cosine values for the others give the same result. \[ \boxed{l = 0,\ m = \frac{1}{2},\ n = \frac{\sqrt3}{2}} \]

If a line makes \(90^\circ\), \(60^\circ\) and \(30^\circ\) with the \(x\), \(y\) and \(z\)-axes respectively in the positive direction, then find its direction cosines.

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Solution and Explanation

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