Question

The acute angle \(θ\) between the \(xy\)-plane and the plane passing through the point \((1,2,4)\) and parallel to the vectors with direction ratios \(3,2,-1\) and \(1,-2,-2\) is...

• A. \(cos^{-1}(\frac{8}{5\sqrt{5}})\)
• B. \(cos^{-1}(\frac{5}{8\sqrt{5}})\)
• C. \(sin^{-1}(\frac{8}{5\sqrt{5}})\)
• D. \(\frac{π}{4}\)

Hint:

The normal of the plane is the cross product of the two direction vectors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A plane parallel to two vectors has normal along their cross product. The angle between two planes equals the angle between their normals. The xy-plane has normal \(\hat k\).

Step 2: Key Formula or Approach:
\(\bar n = (3, 2, -1)\times(1, -2, -2)\).

Step 3: Detailed Explanation:
\(\bar n = \hat i[2(-2) - (-1)(-2)] - \hat j[3(-2) - (-1)(1)] + \hat k[3(-2) - 2(1)]\)
\(= \hat i(-4 - 2) - \hat j(-6 + 1) + \hat k(-6 - 2) = -6\hat i + 5\hat j - 8\hat k\).
\(|\bar n| = \sqrt{36 + 25 + 64} = \sqrt{125} = 5\sqrt5\).
\[ \cos\theta = \frac{|\bar n\cdot\hat k|}{|\bar n|} = \frac{8}{5\sqrt5} \]
The acute angle is \(\theta = \cos^{-1}\frac{8}{5\sqrt5}\). The point \((1, 2, 4)\) only fixes the position of the plane, not its tilt.

Final Answer:
\(\theta = \cos^{-1}\left(\frac{8}{5\sqrt5}\right)\), option (A). \[ \boxed{\cos^{-1}\left(\frac{8}{5\sqrt{5}}\right)} \]