Question

The angle between the line \[ \frac{x - 1}{2} = \frac{y + 3}{1} = \frac{z + 7}{2} \] and the plane \[ \mathbf{r} \cdot (6\hat{i} - 2\hat{j} - 3\hat{k}) = 5 \] is

• A. \( \sin^{-1} \left( \frac{4}{21} \right) \)
• B. \( \cos^{-1} \left( \frac{4}{21} \right) \)
• C. \( \sin^{-1} \left( \frac{5}{7} \right) \)
• D. \( \cos^{-1} \left( \frac{5}{7} \right) \)

Hint:

To find the angle between a line and a plane, use the direction ratios of the line and the normal vector to the plane, and apply the formula for the sine of the angle between them.

Answer 1

The Correct Option is A

Step 1: Find the direction ratios of the line and the normal to the plane.
The direction ratios of the line are \( (2, 1, 2) \), and the normal vector to the plane is \( \mathbf{n} = (6, -2, -3) \), as given by the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) in the plane equation.
Step 2: Use the formula for the angle between a line and a plane.
The formula for the angle \( \theta \) between a line with direction ratios \( a_1, b_1, c_1 \) and a plane with normal vector \( a_2, b_2, c_2 \) is: \[ \sin \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}. \] Substituting the values for the direction ratios of the line and the normal vector of the plane, we get: \[ \sin \theta = \frac{|2(6) + 1(-2) + 2(-3)|}{\sqrt{2^2 + 1^2 + 2^2} \cdot \sqrt{6^2 + (-2)^2 + (-3)^2}}. \] Simplifying, we get: \[ \sin \theta = \frac{|12 - 2 - 6|}{\sqrt{9} \cdot \sqrt{49}} = \frac{4}{21}. \] Thus, the angle is \( \sin^{-1} \left( \frac{4}{21} \right) \), making option (A) the correct answer.