Question

If $|\vec{a}| = 3$, $|\vec{b}| = 2$, then $(2\vec{a} + 3\vec{b})\cdot(2\vec{a} - 3\vec{b})$ is equal to

• A. 6
• B. $-12$
• C. 12
• D. $-6$
• E. 0

Hint:

$(A+B)\cdot(A-B) = |A|^2 - |B|^2$ holds for vectors just as $(a+b)(a-b)=a^2-b^2$ does for scalars. The cross-terms cancel because the dot product is commutative.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Use the identity $(\vec{p}+\vec{q})\cdot(\vec{p}-\vec{q}) = |\vec{p}|^2 - |\vec{q}|^2$ where $\vec{p} = 2\vec{a}$ and $\vec{q} = 3\vec{b}$.

Step 2: Detailed Explanation:
\[ (2\vec{a}+3\vec{b})\cdot(2\vec{a}-3\vec{b}) = |2\vec{a}|^2 - |3\vec{b}|^2 = 4|\vec{a}|^2 - 9|\vec{b}|^2 \] \[ = 4(9) - 9(4) = 36 - 36 = 0 \]

Step 3: Final Answer:
$(2\vec{a}+3\vec{b})\cdot(2\vec{a}-3\vec{b}) = 0$.