Question

If $\vec{A} = \hat{i} + \hat{j} + 3\hat{k}$, $\vec{B} = -\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{C} = 2\hat{i} - 2\hat{j} - 8\hat{k}$, then the angle between the vectors $\vec{P} = \vec{A} + \vec{B} + \vec{C}$ and $\vec{Q} = (\vec{A} \times \vec{B})$ is (in degree)}

• A. $0^\circ$
• B. $45^\circ$
• C. $90^\circ$
• D. $60^\circ$

Hint:

You don't need to calculate the actual cross product! If $\vec{P}$ lies in the plane of $\vec{A}$ and $\vec{B}$, it is automatically $90^\circ$ to $\vec{A} \times \vec{B}$.

Answer 1

The Correct Option is C

Step 1: Concept
The cross product $\vec{A} \times \vec{B}$ is always perpendicular to both $\vec{A}$ and $\vec{B}$, and by extension, any linear combination of them ($m\vec{A} + n\vec{B}$).

Step 2: Meaning
Check if $\vec{C}$ is a multiple of $\vec{A}$ or $\vec{B}$. $\vec{C} = -2(-\hat{i} + \hat{j} + 4\hat{k}) = -2\vec{B}$.

Step 3: Analysis
$\vec{P} = \vec{A} + \vec{B} + (-2\vec{B}) = \vec{A} - \vec{B}$.
Since $\vec{Q} = \vec{A} \times \vec{B}$, $\vec{Q}$ is perpendicular to $\vec{A}$ and $\vec{Q}$ is perpendicular to $\vec{B}$.
Therefore, $\vec{Q}$ is perpendicular to any vector in the plane of $\vec{A}$ and $\vec{B}$, such as $\vec{P} = \vec{A} - \vec{B}$.

Step 4: Conclusion
The angle between $\vec{P}$ and $\vec{Q}$ is $90^\circ$. Final Answer: (C)