Question

Let $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + p\hat{k}$ be two vectors. If $(\vec{a}, \vec{b}) = 60^\circ$, then $p =$

• A. $\frac{\sqrt{7}}{3\sqrt{2}}$
• B. $\frac{3\sqrt{5}}{\sqrt{7}}$
• C. $\frac{\sqrt{3}}{\sqrt{7}}$
• D. $\frac{\sqrt{5}}{\sqrt{7}}$

Hint:

When solving an equation involving square roots by squaring both sides, always check for extraneous solutions. In this case, the equation $3\sqrt{5+p^2} = 4p$ itself implies that $4p$ must be non-negative (since the square root is non-negative). This condition $p \ge 0$ allows you to discard the negative solution obtained after squaring.

Answer 1

The Correct Option is B

Step 1: Use Dot Product Formula: \[ \cos \theta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}| |\bar{b}|} \] Given \( \theta = 60^\circ \), so \( \cos 60^\circ = \frac{1}{2} \).
Step 2: Calculate Magnitudes and Dot Product: \[ \bar{a} \cdot \bar{b} = (1)(2) + (2)(-1) + (2)(p) = 2 - 2 + 2p = 2p \] \[ |\bar{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] \[ |\bar{b}| = \sqrt{2^2 + (-1)^2 + p^2} = \sqrt{4 + 1 + p^2} = \sqrt{5 + p^2} \]
Step 3: Form Equation and Solve: \[ \frac{1}{2} = \frac{2p}{3\sqrt{5+p^2}} \] Cross-multiply: \[ 3\sqrt{5+p^2} = 4p \] Squaring both sides (assuming \( p>0 \) since cosine is positive): \[ 9(5 + p^2) = 16p^2 \] \[ 45 + 9p^2 = 16p^2 \] \[ 7p^2 = 45 \] \[ p^2 = \frac{45}{7} \] \[ p = \sqrt{\frac{45}{7}} = \frac{3\sqrt{5}}{\sqrt{7}} \]