Question

A unit vector parallel to the straight line $\vec{r} = -(5 + 4s)\hat{i} + (7 - 2s)\hat{j} + (3 + 4s)\hat{k}$, where $s$ is the parameter of the line, is

• A. $\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}$
• B. $\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{6}$
• C. $\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{5}$
• D. $\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{4}$
• E. $\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{7}$

Hint:

Collect all coefficients of the parameter $s$ to form your direction vector. Ignore the static constant vector when looking for parallel vectors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A vector equation of a line is typically written as $\vec{r} = \vec{a} + s\vec{d}$. The vector $\vec{d}$ is the direction vector parallel to the line. To find a unit vector parallel to the line, we normalize this direction vector.

Step 2: Detailed Explanation:
The given equation is:
\[ \vec{r} = -(5 + 4s)\hat{i} + (7 - 2s)\hat{j} + (3 + 4s)\hat{k} \]
Rearrange to separate the constant vector and the parameter $s$:
\[ \vec{r} = (-5\hat{i} + 7\hat{j} + 3\hat{k}) + s(-4\hat{i} - 2\hat{j} + 4\hat{k}) \]
The direction vector parallel to the line is $\vec{d} = -4\hat{i} - 2\hat{j} + 4\hat{k}$.
Find the magnitude of $\vec{d}$:
\[ |\vec{d}| = \sqrt{(-4)^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \]
The unit vector parallel to the line is:
\[ \hat{u} = \frac{\vec{d}}{|\vec{d}|} = \frac{-4\hat{i} - 2\hat{j} + 4\hat{k}}{6} \]
Divide each component by 2:
\[ \hat{u} = \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} \]

Step 3: Final Answer:
The unit vector is $\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}$.