Question

The acute angle between the line \[ \vec{r} = (i + 2j + k) + \lambda (i + j + k) \] and the plane \[ \vec{r} \cdot (2i - j + k) = 5 \] is

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

Hint:

For angle between a line and a plane, always use sine and the dot product of the line’s direction vector with the plane’s normal vector.

Answer 1

The Correct Option is A

Step 1: Identify direction ratios and normal vector.
The direction ratios of the given line are obtained from \( \lambda (i + j + k) \), hence \[ \vec{d} = \langle 1, 1, 1 \rangle \] The normal vector of the plane is \[ \vec{n} = \langle 2, -1, 1 \rangle \]
Step 2: Use the formula for angle between a line and a plane.
If \( \theta \) is the angle between the line and the plane, then \[ \sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}|\,|\vec{n}|} \]
Step 3: Compute dot product and magnitudes.
\[ \vec{d} \cdot \vec{n} = (1)(2) + (1)(-1) + (1)(1) = 2 \] \[ |\vec{d}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}, \quad |\vec{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6} \]
Step 4: Final calculation.
\[ \sin \theta = \frac{2}{\sqrt{3}\sqrt{6}} = \frac{\sqrt{2}}{3} \] \[ \theta = \sin^{-1}\!\left(\dfrac{\sqrt{2}}{3}\right) \]