Question

Vector of magnitude 3 making equal angles with \( x \) and \( y \) axes and perpendicular to \( z \) axis is

• A. \( \hat{i} + 2\sqrt{2}\hat{j} \)
• B. \( 3\hat{k} \)
• C. \( \frac{3\sqrt{2}}{2}\hat{i} + \frac{3\sqrt{2}}{2}\hat{j} \)
• D. \( \sqrt{3}\hat{i} + \sqrt{3}\hat{j} + \sqrt{3}\hat{k} \)

Hint:

"Perpendicular to \( z \)-axis" immediately tells you there is no \( \hat{k} \) component.
For any vector in 2D space making equal angles with axes, the direction cosines are \( (1/\sqrt{2}, 1/\sqrt{2}) \).

Answer 1

The Correct Option is C

Concept:
• A vector perpendicular to the \( z \)-axis lies in the \( xy \)-plane, meaning its \( z \)-component is zero.
• Direction cosines \( (L, M, N) \) satisfy \( L^2 + M^2 + N^2 = 1 \).
• A vector of magnitude \( r \) is given by \( \vec{v} = r(L\hat{i} + M\hat{j} + N\hat{k}) \).

Step 1: Determine the direction cosines
Let the vector make angles \( \alpha, \beta, \gamma \) with the axes.
Given \( \alpha = \beta \) (equal angles with \( x \) and \( y \)).
Since it is perpendicular to the \( z \)-axis, \( \gamma = 90^\circ \).
Then \( L = \cos \alpha \), \( M = \cos \alpha \), and \( N = \cos 90^\circ = 0 \).

Step 2: Solve for the direction cosines
Using the identity \( L^2 + M^2 + N^2 = 1 \):
\[ \cos^2 \alpha + \cos^2 \alpha + 0^2 = 1 \]
\[ 2\cos^2 \alpha = 1 \implies \cos^2 \alpha = \frac{1}{2} \]
\[ \cos \alpha = \pm \frac{1}{\sqrt{2}} \]

Step 3: Construct the vector
Magnitude \( r = 3 \).
\[ \vec{v} = 3(L\hat{i} + M\hat{j} + N\hat{k}) \]
Assuming positive components for the matching option:
\[ \vec{v} = 3\left( \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k} \right) \]
\[ \vec{v} = \frac{3}{\sqrt{2}}\hat{i} + \frac{3}{\sqrt{2}}\hat{j} \]

Step 4: Rationalize the coefficients
\[ \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]
So, \( \vec{v} = \frac{3\sqrt{2}}{2}\hat{i} + \frac{3\sqrt{2}}{2}\hat{j} \).
This matches option (C).