Question

If the direction cosines of a vector of magnitude 3 are \( \frac{2}{3}, \frac{-a}{3}, \frac{2}{3} \), \( a > 0 \), then the vector is:

• A. \( 2\hat{i} + \hat{j} + 2\hat{k} \)
• B. \( 2\hat{i} - \hat{j} + 2\hat{k} \)
• C. \( \hat{i} - 2\hat{j} + 2\hat{k} \)
• D. \( \hat{i} + 2\hat{j} + 2\hat{k} \)
• E. \( \hat{i} + 2\hat{j} - 2\hat{k} \)

Hint:

Direction cosines are essentially the components of the unit vector in the same direction. Multiplying them by the magnitude gives the original vector.

Answer 1

The Correct Option is B

Concept: Direction cosines \( (l, m, n) \) satisfy the identity \( l^2 + m^2 + n^2 = 1 \). Furthermore, a vector \( \vec{r} \) with magnitude \( |\vec{r}| \) and direction cosines \( l, m, n \) is given by \( \vec{r} = |\vec{r}|(l\hat{i} + m\hat{j} + n\hat{k}) \).

Step 1: Find the value of \( a \) using the direction cosine identity.
Given \( l = \frac{2}{3}, m = \frac{-a}{3}, n = \frac{2}{3} \). \[ \left(\frac{2}{3}\right)^2 + \left(\frac{-a}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = 1 \] \[ \frac{4}{9} + \frac{a^2}{9} + \frac{4}{9} = 1 \] \[ \frac{8 + a^2}{9} = 1 \quad \Rightarrow \quad 8 + a^2 = 9 \] \[ a^2 = 1 \quad \Rightarrow \quad a = \pm 1 \] Since \( a > 0 \), we have \( a = 1 \).

Step 2: Determine the direction cosines.
Substitute \( a = 1 \) back into the components: \[ l = \frac{2}{3}, m = -\frac{1}{3}, n = \frac{2}{3} \]

Step 3: Construct the vector.
Given magnitude \( |\vec{r}| = 3 \): \[ \vec{r} = 3 \left( \frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \right) \] \[ \vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} \]