Question

Let $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$ be two vectors. Then which one of the following statements is TRUE?

• A. Projection of $\vec{a}$ on $\vec{b}$ is $\frac{-13}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$
• B. Projection of $\vec{a}$ on $\vec{b}$ is $\frac{-17}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$
• C. Projection of $\vec{a}$ on $\vec{b}$ is $\frac{17}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$
• D. Projection of $\vec{a}$ on $\vec{b}$ is $\frac{13}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$

Answer 1

Answer: (A)

Projection of $\vec{a}$ on $\vec{b}$ is $\frac{-13}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$

Answer 2

The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by the formula: $proj_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}$

First, we calculate the dot product of $\vec{a}$ and $\vec{b}$: $\vec{a} \cdot \vec{b} = (5)(1) + (-1)(3) + (-3)(5) = 5 - 3 - 15 = -13$

Next, we calculate the magnitude of vector $\vec{b}$: $|\vec{b}| = \sqrt{(1)^2 + (3)^2 + (5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$

The scalar projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-13}{\sqrt{35}}$.

The projection vector is $proj_{\vec{b}}\vec{a} = \frac{-13}{(\sqrt{35})^2} \vec{b} = \frac{-13}{35} \vec{b}$. Since the scalar multiplier $\frac{-13}{35}$ is negative, the projection vector is in the opposite direction to $\vec{b}$.