Question

Let \(O\) be the origin and \(\vec r\) be the position vector of a point \(P\). If \(\overline{OP}\) makes angles \(\frac{\pi}{3}\) and \(\frac{\pi}{6}\) with \(\overline{i}\) and \(\overline{j}\) respectively, then a vector along \(\overline{OP}\) with magnitude 2 is

• A. \(\vec i+\sqrt3\,\vec j\)
• B. \(\vec j+\sqrt3\,\vec k\)
• C. \(\sqrt{3}\vec i+\vec j\)
• D. \(\sqrt3\,\vec j+\vec k\) \bigskip

Hint:

Multiply unit vector by required magnitude to get actual vector.

Answer 1

The Correct Option is A

Concept: Direction cosines satisfy: \[ \cos\alpha,\cos\beta,\cos\gamma \] and unit vector is formed using them.

Step 1: Find direction cosines.
\[ \cos\alpha=\cos\frac{\pi}{3}=\frac12,\quad \cos\beta=\cos\frac{\pi}{6}=\frac{\sqrt3}{2} \]

Step 2: Find third direction cosine.
\[ l^2+m^2+n^2=1 \] \[ n=\sqrt{1-\frac14-\frac34}=0 \] So direction vector: \[ \frac12\vec i+\frac{\sqrt3}{2}\vec j \]

Step 3: Multiply magnitude 2.
\[ \vec r =2\left(\frac12\vec i+\frac{\sqrt3}{2}\vec j\right) \] \[ =\vec i+\sqrt3\,\vec j \] \[ \boxed{(A)} \]