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: Find the normal by determinant:
$\begin{vmatrix}\hat i & \hat j & \hat k\\ 3 & 2 & -1\\ 1 & -2 & -2\end{vmatrix} = (-6, 5, -8)$.

Step 2: Angle with the z axis:
Direction cosine of the normal along $z$ is $\frac{|-8|}{5\sqrt5}$.
This equals $\cos\theta$, so $\theta = \cos^{-1}\frac{8}{5\sqrt5}$. Note $\frac{8}{5\sqrt5} \approx 0.716$, so $\theta \approx 44^{\circ}$, close to but not equal to $\frac\pi4$.

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