Question

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}$ and $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$, then $\alpha + \beta$ is equal to

• A. 2
• B. -1
• C. 0
• D. 1

Hint:

The condition $|\vec{u} + \vec{v}| = |\vec{u} - \vec{v}|$ is a classic vector identity that always means $\vec{u}$ is perpendicular to $\vec{v}$ ($\vec{u} \cdot \vec{v} = 0$). Memorizing this saves time from doing the algebraic expansion every time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given condition $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$ signifies that the diagonals of a parallelogram formed by vectors $\vec{a}$ and $\vec{b}$ are equal in length. Geometrically, this means the parallelogram is a rectangle, which implies that vectors $\vec{a}$ and $\vec{b}$ must be perpendicular.

Step 2: Key Formula or Approach:
1. Recognize that $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \iff \vec{a} \cdot \vec{b} = 0$. Alternatively, square both sides: $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \implies |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) \implies 4(\vec{a} \cdot \vec{b}) = 0 \implies \vec{a} \cdot \vec{b} = 0$. 2. Calculate the dot product of the given vectors $\vec{a}$ and $\vec{b}$ and set it to zero. 3. Solve the resulting equation to find the value of $\alpha + \beta$.

Step 3: Detailed Explanation:
Given vectors: $\vec{a} = 2\hat{i} + 2\hat{j} - 1\hat{k}$ $\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}$ As established, the condition $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$ implies that the vectors are orthogonal: $\vec{a} \cdot \vec{b} = 0$ Now, compute the dot product: $(2)(\alpha) + (2)(\beta) + (-1)(2) = 0$ $2\alpha + 2\beta - 2 = 0$ Divide the entire equation by 2: $\alpha + \beta - 1 = 0$ $\alpha + \beta = 1$ The question asks for the value of $\alpha + \beta$, which we have directly found to be 1.

Step 4: Final Answer:
The value of $\alpha + \beta$ is 1.