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MHT CET · Maths · Three Dimensional Geometry

The direction cosines of a line which lies in ZoX plane and makes an angle of \(30^{\circ}\)
with Z-axis are

  1. A \(0, \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\)
  2. B \(\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\)
  3. C \(0, \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\)
  4. D \(\frac{\sqrt{3}}{2}, 0, \pm \frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\)

Step-by-step Solution

Detailed explanation

Direction cosines are the cosines of the angles which a line makes with the positive coordinate axes in anticlockwise direction. They are represented by \(\langle l, m, n\rangle\) where \(\mathrm{l}, \mathrm{m}, \mathrm{n}\) correspond to the \(\mathrm{x}\)-axis, \(\mathrm{y}\)-axis, and \(\mathrm{z}\)-axis respectively. One important property of direction cosines is that the sum of their squares is unity.
i.e., \(l^{2}+m^{2}+n^{2}=1\)
If the line lies in the zox plane, it implies that the \(y\)-axis is normal to the line as the \(y\)-axis is normal to the ZOX plane. Which means the angle the line makes with the \(y\)-axis is \(90^{\circ}\) or \(\frac{\pi}{2}\) radians. Thus, \(m=\cos \left(\frac{\pi}{2}\right)=0\)

It is given that the line makes an angle of \(30^{\circ}\) or \(\frac{\pi}{6}\) radians with the positive z-axis. Thus, \(n=\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\)
Now we use the property \(l^{2}+m^{2}+n^{2}=1\)
Substituting the values of \(m\) and \(n\), we get the following values of । \(l=\pm \frac{1}{2}\)
We get two values of I because the only information given is that the line makes an angle of \(\frac{\pi}{6}\) with the z-axis, so it can make an angle of either \(\frac{\pi}{3}\) or \(\frac{2 \pi}{3}\) with the \(x\)-axis, which correspond to the two values of I.
Therefore, the direction cosines are
\(\langle l, m, n\rangle=\left\langle\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\right\rangle\)