Question

The position vector of a point \(P\), such that \(OP\) inclined to the X axis at \(45^{\circ }\) and to the Y axis at \(60^{\circ }\) and \(|OP|=12\) units is:

• A. \(\vec{r}=6\sqrt{2}\hat{i}-6\hat{j}+6\hat{k}\)
• B. \(\vec{r}=6\sqrt{2}\hat{i}\pm 6\hat{j}\pm 6\hat{k}\)
• C. \(\vec{r}=6\sqrt{2}\hat{i}+6\hat{j}\pm 6\hat{k}\)
• D. \(\vec{r}=6\sqrt{2}\hat{i}-6\hat{j}-6\hat{k}\)

Hint:

Use l^2 + m^2 + n^2 = 1 to find the third direction cosine, then multiply the direction cosines by the length.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The direction cosines are \(l=\cos 45^{\circ}\), \(m=\cos 60^{\circ}\) and n, with \(l^{2}+m^{2}+n^{2}=1\).

Step 2: Find n
\(l=\frac{1}{\sqrt{2}}\), \(m=\frac{1}{2}\).
\(\frac{1}{2}+\frac{1}{4}+n^{2}=1\Rightarrow n^{2}=\frac{1}{4}\Rightarrow n=\pm\frac{1}{2}\).

Step 3: Position vector
\(\bar{r}=12(l\hat{i}+m\hat{j}+n\hat{k})=12\left(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{2}\hat{j}\pm\frac{1}{2}\hat{k}\right)\).

Step 4: Simplify
\(\bar{r}=6\sqrt{2}\hat{i}+6\hat{j}\pm 6\hat{k}\).

Final Answer:
The position vector is 6 root 2 i + 6 j plus or minus 6 k. \[ \boxed{\text{(C) }\vec{r}=6\sqrt{2}\hat{i}+6\hat{j}\pm 6\hat{k}} \]