Question

The angle between a normal to the plane \( 2x - y + 2z - 1 = 0 \) and the \( z \)-axis is:

• A. \( \cos^{-1}\left(\frac{1}{3}\right) \)
• B. \( \sin^{-1}\left(\frac{2}{3}\right) \)
• C. \( \cos^{-1}\left(\frac{2}{3}\right) \)
• D. \( \sin^{-1}\left(\frac{1}{3}\right) \)
• E. \( \sin^{-1}\left(\frac{3}{5}\right) \)

Hint:

The angle between a vector and any coordinate axis is determined simply by the direction cosine of that vector for that specific axis. Here, the direction cosine for the $z$-axis is $C / |\vec{n}|$.

Answer 1

The Correct Option is C

Concept: The normal vector \( \vec{n} \) to a plane \( Ax + By + Cz + D = 0 \) is given by \( \vec{n} = A\hat{i} + B\hat{j} + C\hat{k} \). The unit vector along the \( z \)-axis is \( \hat{k} \). The angle \( \theta \) between two vectors \( \vec{u} \) and \( \vec{v} \) is found using the dot product formula: \[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \]

Step 1: Identify the normal vector and the \( z \)-axis vector.
Normal vector to the plane \( 2x - y + 2z - 1 = 0 \) is \( \vec{n} = 2\hat{i} - \hat{j} + 2\hat{k} \). Vector along the \( z \)-axis is \( \hat{k} \) (or \( 0\hat{i} + 0\hat{j} + 1\hat{k} \)).

Step 2: Calculate the magnitudes and dot product.
Magnitude of \( \vec{n} \): \[ |\vec{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] Magnitude of \( \hat{k} \): \[ |\hat{k}| = 1 \] Dot product \( \vec{n} \cdot \hat{k} \): \[ (2)(0) + (-1)(0) + (2)(1) = 2 \]

Step 3: Find the angle.
\[ \cos \theta = \frac{2}{3 \times 1} = \frac{2}{3} \] \[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]