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: Angles with axes
Angles with X and Y are $45^{\circ}$ and $60^{\circ}$, so the x and y components are $12\cos 45^{\circ}=6\sqrt{2}$ and $12\cos 60^{\circ}=6$.

Step 2: Use the length
$|OP|^{2}=144\Rightarrow72+36+z^{2}=144\Rightarrow z^{2}=36\Rightarrow z=\pm 6$.

Step 3: Match
Since cos 60 is positive, the j-component is +6 and the k-component is either sign, which matches the third option.

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