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 Question:
We are given two independent constraints. The first (scalar projection) will allow us to solve for $\alpha$. The second (vector cross product) will allow us to solve for $\beta$. Finally, we sum them.

Step 2: Detailed Explanation:
Part 1: Finding $\alpha$ using projection.
The scalar projection of vector $\vec{a}$ onto vector $\vec{c}$ is given by the formula:
$\text{Proj}_{\vec{c}}\vec{a} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}$
Calculate the dot product $\vec{a} \cdot \vec{c}$:
$\vec{a} \cdot \vec{c} = (\alpha)(1) + (3)(2) + (-1)(-2) = \alpha + 6 + 2 = \alpha + 8$
Calculate the magnitude of vector $\vec{c}$:
$|\vec{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
Substitute these into the projection formula and set it equal to the given value $\frac{10}{3}$:
$\frac{\alpha + 8}{3} = \frac{10}{3}$
$\alpha + 8 = 10 \implies \alpha = 2$
Part 2: Finding $\beta$ using the cross product.
Calculate the cross product $\vec{b} \times \vec{c}$ using a determinant:
$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & \beta \\ 1 & 2 & -2 \end{vmatrix}$
Expand the determinant:
$= \hat{i}[(-1)(-2) - (\beta)(2)] - \hat{j}[(3)(-2) - (\beta)(1)] + \hat{k}[(3)(2) - (-1)(1)]$
$= \hat{i}[2 - 2\beta] - \hat{j}[-6 - \beta] + \hat{k}[6 + 1]$
$= (2 - 2\beta)\hat{i} + (\beta + 6)\hat{j} + 7\hat{k}$
We are given that $\vec{b} \times \vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$.
Equate the corresponding $\hat{i}$ components:
$2 - 2\beta = -6$
$-2\beta = -8 \implies \beta = 4$
(We can verify this with the $\hat{j}$ component: $\beta + 6 = 4 + 6 = 10$, which matches perfectly).
Part 3: Final Calculation
The question asks for the sum $(\alpha + \beta)$:
$\alpha + \beta = 2 + 4 = 6$.

Step 3: Final Answer:
The value of $(\alpha + \beta)$ is 6, matching option (d).