Question

If $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are perpendicular and $\vec{b} = 3\hat{i} - 4\hat{j} + 2\hat{k}$, then $|\vec{a}|$ is equal to:

• A. $\sqrt{41}$
• B. $\sqrt{39}$
• C. $\sqrt{19}$
• D. $\sqrt{29}$
• E. $\sqrt{31}$

Hint:

In a parallelogram with sides $\vec{a}$ and $\vec{b}$, the diagonals are $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$. If the diagonals are perpendicular, the parallelogram is a rhombus, meaning all sides are equal ($|\vec{a}| = |\vec{b}|$).

Answer 1

The Correct Option is D

Concept: Two vectors are perpendicular if their dot product is zero. The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is $\sqrt{x^2 + y^2 + z^2}$.

Step 1: Use the perpendicular condition.
$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0$. Expanding the dot product: \[ \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 0 \] Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, this simplifies to: \[ |\vec{a}|^2 - |\vec{b}|^2 = 0 \quad \Rightarrow \quad |\vec{a}|^2 = |\vec{b}|^2 \] This means $|\vec{a}| = |\vec{b}|$.

Step 2: Calculate the magnitude of $\vec{b}$.
Given $\vec{b} = 3\hat{i} - 4\hat{j} + 2\hat{k}$: \[ |\vec{b}| = \sqrt{3^2 + (-4)^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29} \]

Step 3: Conclusion.
Since $|\vec{a}| = |\vec{b}|$, then $|\vec{a}| = \sqrt{29}$.