Question

The angle between the straight line \( \vec{r} = (\hat{i}+2\hat{j}+\hat{k}) + s(\hat{i}-\hat{j}+\hat{k}) \) and the plane \( \vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k}) = 4 \) is

• A. \( \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \)
• B. \( \sin^{-1}\left(\frac{\sqrt{2}}{6}\right) \)
• C. \( \sin^{-1}\left(\frac{\sqrt{2}}{3}\right) \)
• D. \( \sin^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
• E. \( \sin^{-1}\left(\frac{2}{3}\right) \)

Hint:

Angle between line and plane is complementary to angle between line and normal.

Answer 1

The Correct Option is A

Concept:
• Angle between line and plane: \[ \sin\theta = \frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}||\vec{n}|} \]
• \( \vec{d} \) = direction vector of line
• \( \vec{n} \) = normal vector to plane

Step 1: Extract direction vector of line.
From: \[ \vec{r} = (\hat{i}+2\hat{j}+\hat{k}) + s(\hat{i}-\hat{j}+\hat{k}) \] Direction vector: \[ \vec{d} = (1,-1,1) \]

Step 2: Extract normal vector of plane.
From: \[ \vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k}) = 4 \] Normal vector: \[ \vec{n} = (2,-1,1) \]

Step 3: Geometric interpretation.

• Line direction vector makes angle with plane
• Equivalent to angle between line and projection on plane
• Hence sine formula is used

Step 4: Compute dot product.
\[ \vec{d}\cdot\vec{n} = 1\cdot2 + (-1)(-1) + 1\cdot1 \] \[ = 2 + 1 + 1 = 4 \]

Step 5: Compute magnitudes.
\[ |\vec{d}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \] \[ |\vec{n}| = \sqrt{4 + 1 + 1} = \sqrt{6} \]

Step 6: Apply formula.
\[ \sin\theta = \frac{4}{\sqrt{3}\cdot\sqrt{6}} \] \[ = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} \] \[ = \frac{2\sqrt{2}}{3} \]

Step 7: Final Answer.
\[ \boxed{\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)} \]