Question

Let $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$. Then $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) =$

• A. 6
• B. 7
• C. 8
• D. 9
• E. 10

Hint:

The difference of squares identity works for vectors in a dot product just like it does for real numbers.

Answer 1

The Correct Option is C

Step 1: Concept
Using the identity: $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$.

Step 2: Calculation
$|\vec{a}|^2 = 3^2 + 2^2 + 2^2 = 9 + 4 + 4 = 17$. $|\vec{b}|^2 = 1^2 + 2^2 + (-2)^2 = 1 + 4 + 4 = 9$.

Step 3: Calculation
$17 - 9 = 8$. Final Answer: (C)