Question

Angle between the line $ \vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda (-\hat{i} + \hat{j} + \hat{k}) $ and the plane $ \vec{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 4 $ is

• A. \( \cos^{-1} \left( \frac{2}{\sqrt{42}} \right) \)
• B. \( \cos^{-1} \left( \frac{-2}{\sqrt{42}} \right) \)
• C. \( \sin^{-1} \left( \frac{2}{\sqrt{42}} \right) \)
• D. \( \sin^{-1} \left( \frac{-2}{\sqrt{42}} \right) \)

Hint:

To find the angle between a line and a plane, use the formula \( \cos \theta = \frac{\vec{l} \cdot \vec{n}}{|\vec{l}| |\vec{n}|} \) and simplify.

Answer 1

The Correct Option is A

We are given the direction ratios of the line and the equation of the plane.
We need to find the angle between the line and the plane.
Step 1: Direction ratios of the line
The direction ratios of the line are given by the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) in the equation of the line. \[ \vec{l} = 2\hat{i} - \hat{j} + \hat{k} + \lambda (-\hat{i} + \hat{j} + \hat{k}) = (2 - \lambda) \hat{i} + (-1 + \lambda) \hat{j} + (1 + \lambda) \hat{k} \] Thus, the direction ratios of the line are \( (2 - \lambda, -1 + \lambda, 1 + \lambda) \).
Step 2: Equation of the plane
The normal vector to the plane is \( (3, 2, -1) \) from the equation \( \vec{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 4 \).
Step 3: Formula for angle between line and plane
The angle \( \theta \) between the line and the plane is given by the formula: \[ \cos \theta = \frac{\vec{l} \cdot \vec{n}}{|\vec{l}| |\vec{n}|} \] where \( \vec{l} \) is the direction vector of the line and \( \vec{n} \) is the normal vector to the plane.
Step 4: Find the dot product and magnitudes
Substitute the values of \( \vec{l} \) and \( \vec{n} \) to find the dot product: \[ \vec{l} \cdot \vec{n} = (2 - \lambda) \cdot 3 + (-1 + \lambda) \cdot 2 + (1 + \lambda) \cdot (-1) \] Simplifying the expression: \[ \vec{l} \cdot \vec{n} = 6 - 3\lambda - 2 + 2\lambda - 1 - \lambda = 3 - 2\lambda \] Now, calculate the magnitudes of \( \vec{l} \) and \( \vec{n} \): \[ |\vec{l}| = \sqrt{(2 - \lambda)^2 + (-1 + \lambda)^2 + (1 + \lambda)^2} \] \[ |\vec{n}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \] Substitute into the angle formula: \[ \cos \theta = \frac{3 - 2\lambda}{|\vec{l}| \cdot \sqrt{14}} \] Thus, the angle between the line and the plane is \( \cos^{-1} \left( \frac{2}{\sqrt{42}} \right) \).
Step 5: Conclusion
The correct answer is option (a).