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: Understanding the Concept
The angle between two vectors can be found using the dot product formula. We are given the angle and the vectors (with one unknown component) and need to solve for the unknown.
Step 2: Key Formula or Approach
The cosine of the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by: \[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \] We will calculate each part of this formula ($\vec{a} \cdot \vec{b}$, $|\vec{a}|$, $|\vec{b}|$) in terms of $p$, substitute the given angle $\theta=60^\circ$, and then solve the resulting equation for $p$.
Step 3: Detailed Explanation
Given vectors and angle: $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ $\vec{b} = 2\hat{i} - \hat{j} + p\hat{k}$ $\theta = 60^\circ$, so $\cos 60^\circ = \frac{1}{2}$. 1. Calculate the dot product $\vec{a \cdot \vec{b}$:} \[ \vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (2)(p) = 2 - 2 + 2p = 2p \] 2. Calculate the magnitudes $|\vec{a|$ and $|\vec{b}|$:} \[ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3 \] \[ |\vec{b}| = \sqrt{2^2 + (-1)^2 + p^2} = \sqrt{4+1+p^2} = \sqrt{5+p^2} \] 3. Substitute into the dot product formula and solve for p: \[ \cos 60^\circ = \frac{2p}{3\sqrt{5+p^2}} \] \[ \frac{1}{2} = \frac{2p}{3\sqrt{5+p^2}} \] Cross-multiply: \[ 3\sqrt{5+p^2} = 4p \] For this equality to hold, since the left side is positive, the right side must also be positive, which implies $p>0$. Square both sides of the equation: \[ (3\sqrt{5+p^2})^2 = (4p)^2 \] \[ 9(5+p^2) = 16p^2 \] \[ 45 + 9p^2 = 16p^2 \] \[ 45 = 16p^2 - 9p^2 \] \[ 45 = 7p^2 \] \[ p^2 = \frac{45}{7} \] \[ p = \pm \sqrt{\frac{45}{7}} = \pm \frac{\sqrt{9 \times 5}}{\sqrt{7}} = \pm \frac{3\sqrt{5}}{\sqrt{7}} \] Since we established that $p$ must be positive, we take the positive root. \[ p = \frac{3\sqrt{5}}{\sqrt{7}} \] Step 4: Final Answer
The value of $p$ is $\frac{3\sqrt{5}}{\sqrt{7}}$.