Question

Let $\vec{a} = \alpha\,\hat{i} - 3\hat{j} - 2\hat{k}$ and $\vec{c} = \hat{i} - 2\hat{j} + 2\hat{k}$. If the projection of $\vec{a}$ on $\vec{c}$ is 2, then the value of $\alpha$ is

• A. 2
• B. 4
• C. 3
• D. 5
• E. 6

Hint:

Projection of $\vec{a}$ on $\vec{c}$ is a scalar: $\text{proj} = \dfrac{\vec{a}\cdot\vec{c}}{|\vec{c}|}$. Do not confuse this with the vector projection, which is $\dfrac{\vec{a}\cdot\vec{c}}{|\vec{c}|^2}\vec{c}$.

Answer 1

The Correct Option is B

Step 1: Concept:
• Projection of vector \(\vec{a}\) on \(\vec{c}\) is: \[ \text{Projection} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} \]
• Given that projection = 2, we use this to find \(\alpha\).

Step 2: Calculation:
• Compute dot product: \[ \vec{a} \cdot \vec{c} = \alpha(1) + (-3)(-2) + (-2)(2) \] \[ = \alpha + 6 - 4 = \alpha + 2 \]
• Find magnitude of \(\vec{c}\): \[ |\vec{c}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3 \]
• Substitute into projection formula: \[ \frac{\alpha + 2}{3} = 2 \]

Step 3: Final Answer:
• Solve: \[ \alpha + 2 = 6 \Rightarrow \alpha = 4 \]
• \[ \boxed{\alpha = 4} \]