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:
The equation of the line is given in parametric form. We can separate it into a point vector and a direction vector. A unit vector parallel to the line is simply the normalized direction vector.
Step 2: Key Formula or Approach:
Rewrite \(\vec{r}\) as \(\vec{a} + s\vec{d}\), where \(\vec{d}\) is the direction vector.
The unit vector is \(\hat{d} = \frac{\vec{d}}{|\vec{d}|}\).
Step 3: Detailed Explanation:
Given equation:
\[ \vec{r} = -(5 + 4s)\hat{i} + (7 - 2s)\hat{j} + (3 + 4s)\hat{k} \] Separate the constant terms and the terms with 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 of the line is \(\vec{d} = -4\hat{i} - 2\hat{j} + 4\hat{k}\).
We can factor out a positive \(2\) to match the numerator format of the options:
\[ \vec{d} = 2(-2\hat{i} - \hat{j} + 2\hat{k}) \] Let's find the unit vector parallel to \(\vec{d}_{simp} = -2\hat{i} - \hat{j} + 2\hat{k}\):
Magnitude \(|\vec{d}_{simp}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\).
Therefore, the unit vector is:
\[ \hat{u} = \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} \] Step 4: Final Answer:
The unit vector is \(\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}\).