Question

The angle between the line \(\overset{̄}{r} = (\hat{i}+2\hat{j}+\hat{k})+λ(\hat{i}+\hat{j}+\hat{k})\) and the plane \(\overset{̄}{r}\cdot (2\hat{i}-\hat{j}+\hat{k}) = 8\) is \(\ldots\)

• A. \(θ = sin^{-1}(\frac{\sqrt{2}}{3})\)
• B. \(θ = cos^{-1}(\frac{\sqrt{2}}{3})\)
• C. \(θ = cos^{-1}(\frac{2}{\sqrt{3}})\)
• D. \(θ = sin^{-1}(\frac{3}{\sqrt{2}})\)

Hint:

The angle with the plane satisfies sin(theta) = |d . n| / (|d||n|).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. So the formula uses sine and not cosine.

Step 2: Key Formula or Approach:
\[ \sin\theta = \frac{|\bar d\cdot\bar n|}{|\bar d|\,|\bar n|} \]
where \(\bar d\) is the direction of the line and \(\bar n\) is the normal to the plane.

Step 3: Detailed Explanation:
\(\bar d = \hat i + \hat j + \hat k\) and \(\bar n = 2\hat i - \hat j + \hat k\).
\[ \bar d\cdot\bar n = 2 - 1 + 1 = 2, \quad |\bar d| = \sqrt3, \quad |\bar n| = \sqrt6 \]
\[ \sin\theta = \frac{2}{\sqrt3\cdot\sqrt6} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt2} = \frac{\sqrt2}{3} \]
\[ \theta = \sin^{-1}\left(\frac{\sqrt2}{3}\right) \]
Option (B) uses cosine and so gives the angle with the normal. Option (D) has a ratio larger than 1, which cannot be a sine. Option (C) has a value larger than 1, so it is not a cosine either.

Final Answer:
\(\theta = \sin^{-1}\left(\dfrac{\sqrt2}{3}\right)\), option (A). \[ \boxed{\sin^{-1}\left(\frac{\sqrt2}{3}\right) \text{ (A)}} \]