Question

A vector \(\overset{̄}{r}\) of magnitude \(3\sqrt{2}\) units which makes angles of \(\frac{π}{4}\) and \(\frac{π}{2}\) respectively with Y and Z axes is

• A. \(\overset{̄}{r} = \pm 3\hat{i}+3\hat{j}\)
• B. \(\overset{̄}{r} = \hat{i}+\hat{j}\)
• C. \(\overset{̄}{r} = \pm 2\hat{i}+3\hat{j}\)
• D. \(\overset{̄}{r} = \pm 5\hat{i}+\hat{j}\)

Hint:

Use direction cosines: l squared plus m squared plus n squared equals 1, then multiply by the magnitude.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A vector of magnitude \(r\) with direction cosines \(l, m, n\) is \(\bar{r} = r(l\hat{i} + m\hat{j} + n\hat{k})\), where \(l^2 + m^2 + n^2 = 1\).

Step 2: Find m and n.
The angle with the Y axis is \(\pi/4\), so \(m = \cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}\). The angle with the Z axis is \(\pi/2\), so \(n = 0\).

Step 3: Find l.
\(l^2 + \dfrac{1}{2} + 0 = 1\), so \(l = \pm\dfrac{1}{\sqrt{2}}\).

Step 4: Build the vector.
\[ \bar{r} = 3\sqrt{2}\left(\pm\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right) = \pm 3\hat{i} + 3\hat{j} \]

Step 5: Check.
Its magnitude is \(\sqrt{9 + 9} = 3\sqrt{2}\). Options (B), (C) and (D) have a different magnitude.

Final Answer:
\(\bar{r} = \pm 3\hat{i} + 3\hat{j}\), option (A). \[ \boxed{\bar{r} = \pm 3\hat{i} + 3\hat{j}} \]