Question

If \( \vec{u} = \hat{i} + \hat{j}, \vec{v} = \hat{i} - \hat{j}, \vec{w} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \hat{n} \) is a unit vector such that \( \vec{u} \cdot \hat{n} = 0 \) and \( \vec{v} \cdot \hat{n} = 0 \), then \( |\vec{w} \cdot \hat{n}| \) is

• A. 3
• B. 0
• C. 1
• D. 2

Hint:

If $\vecu=\hati+\hatj, \vecv=\hati-\hatj, \vecw=\hati+2\hatj+3\hatk$ and $\hatn$ is a unit vector such that $\vecu·\hatn=0$ and $\vecv·\hatn=0$, then $|\vecw·\hatn|$ is

Answer 1

The Correct Option is A

Step 1: Concept
A vector perpendicular to both $\vec{u}$ and $\vec{v}$ must be parallel to their cross product.
Step 2: Analysis
$\vec{u}\times\vec{v} = (\hat{i}+\hat{j})\times(\hat{i}-\hat{j}) = -2\hat{k}$. The unit vector $\hat{n}$ is therefore $\pm\hat{k}$.
Step 3: Evaluation
Calculate the dot product: $\vec{w}\cdot\hat{n} = (\hat{i}+2\hat{j}+3\hat{k})\cdot(\pm\hat{k})$.
Step 4: Conclusion
The absolute value $|\vec{w}\cdot\hat{n}| = | \pm 3 | = 3$.
Final Answer: (a)