Question

Let $\vec{\alpha} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ and $\vec{\beta} = \hat{i} + 2\hat{j} - 4\hat{k}$. Let $\vec{\beta_1}$ be parallel to $\vec{\alpha}$ and $\vec{\beta_2}$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta} = \vec{\beta_1} + \vec{\beta_2}$, then the value of $5\vec{\beta_2} \cdot (\hat{i} + \hat{j} + \hat{k})$ is

• A. 9
• B. 7
• C. 6
• D. 11

Answer 1

Answer: (B)

7

Answer 2

Let $\vec{w} = \hat{i} + \hat{j} + \hat{k}$. We are given $\vec{\beta} = \vec{\beta_1} + \vec{\beta_2}$, where $\vec{\beta_1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta_2}$ is perpendicular to $\vec{\alpha}$. This implies $\vec{\beta_1} = \text{proj}_{\vec{\alpha}} \vec{\beta}$ and $\vec{\beta_2} = \vec{\beta} - \text{proj}_{\vec{\alpha}} \vec{\beta}$.

We need to calculate $5\vec{\beta_2} \cdot \vec{w}$. $5\vec{\beta_2} \cdot \vec{w} = 5 \left( \vec{\beta} - \frac{\vec{\beta} \cdot \vec{\alpha}}{\|\vec{\alpha}\|^2} \vec{\alpha} \right) \cdot \vec{w}$ $5\vec{\beta_2} \cdot \vec{w} = 5 \left( \vec{\beta} \cdot \vec{w} - \frac{\vec{\beta} \cdot \vec{\alpha}}{\|\vec{\alpha}\|^2} (\vec{\alpha} \cdot \vec{w}) \right)$

Calculate the dot products and squared magnitude: $\vec{\beta} \cdot \vec{\alpha} = (1)(4) + (2)(3) + (-4)(5) = 4 + 6 - 20 = -10$ $\|\vec{\alpha}\|^2 = 4^2 + 3^2 + 5^2 = 16 + 9 + 25 = 50$ $\vec{\beta} \cdot \vec{w} = (1)(1) + (2)(1) + (-4)(1) = 1 + 2 - 4 = -1$ $\vec{\alpha} \cdot \vec{w} = (4)(1) + (3)(1) + (5)(1) = 4 + 3 + 5 = 12$

Substitute these values: $5\vec{\beta_2} \cdot \vec{w} = 5 \left( -1 - \frac{-10}{50} (12) \right)$ $5\vec{\beta_2} \cdot \vec{w} = 5 \left( -1 - \left(-\frac{1}{5}\right) (12) \right)$ $5\vec{\beta_2} \cdot \vec{w} = 5 \left( -1 + \frac{12}{5} \right)$ $5\vec{\beta_2} \cdot \vec{w} = 5 \left( \frac{-5 + 12}{5} \right)$ $5\vec{\beta_2} \cdot \vec{w} = 5 \left( \frac{7}{5} \right) = 7$