Question

Let $\vec{a} = \alpha\hat{i} + 3\hat{j} - \hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} + \beta\hat{k}$ and $\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}$ where $\alpha, \beta \in \mathbb{R}$, be three vectors. If the projection of $\vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\vec{b} \times \vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$, then the value of $(\alpha + \beta)$ is ______.

• A. 5
• B. 3
• C. 4
• D. 6

Hint:

When dealing with cross product equalities, you only actually need to evaluate one single component (like the $\hat{i}$ term) to find your unknown! Using the others merely serves as a self-check.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The projection of $\vec{a}$ on $\vec{c}$ is given by $\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}$. The cross product $\vec{b} \times \vec{c}$ is calculated using the determinant method.

Step 2: Formula Application:
1. $|\vec{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = 3$. 2. Projection $= \frac{\alpha(1) + 3(2) - 1(-2)}{3} = \frac{\alpha + 8}{3}$.

Step 3: Explanation:
Given projection $= 10/3 \implies \alpha + 8 = 10 \implies \alpha = 2$. Now, $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & \beta \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2-2\beta) - \hat{j}(-6-\beta) + \hat{k}(6+1)$. Comparing the $\hat{i}$ component: $2 - 2\beta = -6 \implies 2\beta = 8 \implies \beta = 4$. Thus, $\alpha + \beta = 2 + 4 = 6$.

Step 4: Final Answer:
The value of $(\alpha + \beta)$ is 6.