Question

Let $\vec{\alpha} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}$, and $\vec{\gamma} = 2\hat{i} + \hat{j} + 6\hat{k}$.If $\vec{\alpha}$ and $\vec{\beta}$ are both perpendicular to a vector $\vec{\delta}$ and $\vec{\delta} \cdot \vec{\gamma} = 10$, then the magnitude of $\vec{\delta}$ is: }

• A. $2\sqrt{3}$
• B. $\sqrt{3}$
• C. $1/\sqrt{3}$
• D. $\sqrt{3}/2$

Hint:

A vector perpendicular to two given vectors is always a scalar multiple of their cross product.

Answer 1

The Correct Option is A

Step 1: Concept
Since $\vec{\delta}$ is perpendicular to both $\vec{\alpha}$ and $\vec{\beta}$, it must be parallel to their cross product $\vec{\alpha} \times \vec{\beta}.
Step 2: Analysis
First, calculate the cross product: $$\vec{\alpha} \times \vec{\beta} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & -1
2 & -1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3+2) + \hat{k}(-1-4) = 5\hat{i} - 5\hat{j} - 5\hat{k}$$ Thus, $\vec{\delta} = \lambda(5\hat{i} - 5\hat{j} - 5\hat{k}) = 5\lambda(\hat{i} - \hat{j} - \hat{k})$. Using the condition $\vec{\delta} \cdot \vec{\gamma} = 10$: $$5\lambda(\hat{i} - \hat{j} - \hat{k}) \cdot (2\hat{i} + \hat{j} + 6\hat{k}) = 10$$ $$5\lambda(2 - 1 - 6) = 10 \implies 5\lambda(-5) = 10 \implies \lambda = -2/5$$
Step 3: Conclusion
Substituting $\lambda$ back: $\vec{\delta} = 5(-2/5)(\hat{i} - \hat{j} - \hat{k}) = -2\hat{i} + 2\hat{j} + 2\hat{k}$. The magnitude is $|\vec{\delta}| = \sqrt{(-2)^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$. Final Answer:(A)