Question

Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \). If \( \vec{b} \) is a vector such that \( \vec{a} \cdot \vec{b} = |\vec{b}|^2 \) and \( |\vec{a} - \vec{b}| = \sqrt{7} \), then \( |\vec{b}| = \)

• A. \( \sqrt{7} \)
• B. \( \sqrt{3} \)
• C. \( 7 \)
• D. \( 3 \)
• E. \( 7\sqrt{3} \)

Hint:

Whenever magnitude and dot product are both given, expand \( |\vec{a}\pm\vec{b}|^2 \) first. It simplifies the problem quickly.

Answer 1

The Correct Option is A

Concept: Use the identity \[ |\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 -2(\vec{a}\cdot\vec{b}) \]

Step 1: Find \( |\vec{a}|^2 \).
Given, \[ \vec{a}=\hat{i}-2\hat{j}+3\hat{k} \] \[ |\vec{a}|^2 = 1^2+(-2)^2+3^2 \] \[ =1+4+9=14 \] Thus, \[ |\vec{a}|^2=14 \]

Step 2: Use the given conditions.
Given, \[ |\vec{a}-\vec{b}|=\sqrt{7} \] So, \[ |\vec{a}-\vec{b}|^2=7 \] Also, \[ \vec{a}\cdot\vec{b}=|\vec{b}|^2 \] Substitute into the identity: \[ 7=14+|\vec{b}|^2-2|\vec{b}|^2 \] \[ 7=14-|\vec{b}|^2 \]

Step 3: Calculate \( |\vec{b}| \).
\[ |\vec{b}|^2=14-7=7 \] \[ |\vec{b}|=\sqrt{7} \] \[ \boxed{\sqrt{7}} \]