Question

If $|\vec{a}| = \sqrt{26}, \; |\vec{b}| = \sqrt{3}$ and $\vec{a} \times \vec{b} = 5\hat{i} + \hat{j} - 4\hat{k}$, then $\vec{a} \cdot \vec{b} =$:

• A. $\pm 12$
• B. $\pm 4$
• C. $\pm 10$
• D. $\pm 8$
• E. $\pm 6$

Hint:

Use identity linking dot and cross product to avoid finding vectors explicitly.

Answer 1

Answer: (E)

Concept:
• $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$

Step 1: Find $|\vec{a} \times \vec{b}|^2$
\[ = 5^2 + 1^2 + (-4)^2 = 25 + 1 + 16 = 42 \]

Step 2: Apply identity
\[ 42 + (\vec{a} \cdot \vec{b})^2 = 26 \times 3 = 78 \]

Step 3: Solve
\[ (\vec{a} \cdot \vec{b})^2 = 78 - 42 = 36 \] \[ \vec{a} \cdot \vec{b} = \pm 6 \] Final Conclusion:
\[ = \pm 6 \]