Question

The acute angle between the vector \(2\hat{i}+\hat{j}-3\hat{k}\) and the plane containing the vectors \(2\hat{i}+3\hat{j}-\hat{k}\) and \(\hat{i}-\hat{j}+2\hat{k}\) is

• A. \(sin^{-1}(\frac{4}{\sqrt{42}})\)
• B. \(cos^{-1}(\frac{4}{\sqrt{42}})\)
• C. \(sin^{-1}(\frac{3}{\sqrt{42}})\)
• D. \(cos^{-1}(\frac{3}{\sqrt{42}})\)

Hint:

Find the plane normal as a cross product and use sin(theta) = |v . n| / (|v||n|).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The normal to the plane containing \(\vec p = 2\hat i + 3\hat j - \hat k\) and \(\vec q = \hat i - \hat j + 2\hat k\) is \(\vec n = \vec p\times\vec q\). The angle \(\theta\) between a vector and a plane satisfies \(\sin\theta = \dfrac{|\vec v\cdot\vec n|}{|\vec v||\vec n|}\).

Step 2: Normal vector:
\[ \vec n = \begin{vmatrix}\hat i & \hat j & \hat k \\ 2 & 3 & -1 \\ 1 & -1 & 2\end{vmatrix} = (6 - 1)\hat i - (4 + 1)\hat j + (-2 - 3)\hat k = 5\hat i - 5\hat j - 5\hat k \]
Divide by 5: \(\vec n \parallel (1, -1, -1)\).

Step 3: Dot product:
\(\vec v = (2, 1, -3)\). \(\vec v\cdot\vec n = 2 - 1 + 3 = 4\). \(|\vec v| = \sqrt{14}\), \(|\vec n| = \sqrt3\).

Step 4: Angle:
\[ \sin\theta = \frac{4}{\sqrt{14}\sqrt3} = \frac{4}{\sqrt{42}} \Rightarrow \theta = \sin^{-1}\frac{4}{\sqrt{42}} \]
Option (B) and (D) use cosine, which would give the angle with the normal, not with the plane. Option (A).

Final Answer:
The sine of the angle is 4 over root 42. \[ \boxed{\text{(A) }\sin^{-1}\dfrac{4}{\sqrt{42}}} \]