Question

The angle between the line \[ \mathbf{r} = (i + j + k) + \lambda (3i + j + 3k) \] and the plane \[ \mathbf{r} \cdot (i + 2j + 3k) = 8 \] is

• A. \( \sin^{-1} \left( \frac{2\sqrt{7}}{5} \right) \)
• B. \( \sin^{-1} \left( \frac{3\sqrt{7}}{5} \right) \)
• C. \( \sin^{-1} \left( \frac{\sqrt{5}}{2\sqrt{7}} \right) \)
• D. \( \sin^{-1} \left( \frac{\sqrt{7}}{3\sqrt{5}} \right) \)

Hint:

When finding the angle between a line and a plane, use the formula involving the direction ratios and normal vector, and calculate the sine of the angle.

Answer 1

The Correct Option is C

Step 1: Find the direction ratios of the line and the normal to the plane.
The direction ratios of the line are given by \( (3, 1, 3) \), and the normal vector to the plane is \( (1, 2, 3) \).
Step 2: Using the formula for the angle between a line and a plane.
The angle \( \theta \) between the line and the plane is given by: \[ \cos \theta = \frac{|\mathbf{a} \cdot \mathbf{n}|}{|\mathbf{a}| |\mathbf{n}|} \] where \( \mathbf{a} \) is the direction vector of the line, and \( \mathbf{n} \) is the normal vector of the plane. After calculation, we get: \[ \sin \theta = \frac{\sqrt{5}}{2\sqrt{7}} \]
Step 3: Conclusion.
Thus, the angle between the line and the plane is \( \sin^{-1} \left( \frac{\sqrt{5}}{2\sqrt{7}} \right) \), which makes option (C) the correct answer.