Question

A unit vector parallel to the straight line \( \frac{x - 2}{3} = \frac{3 + y}{-1} = \frac{z - 2}{-4} \) is:

• A. \( \frac{1}{\sqrt{26}}(3\hat{i} - \hat{j} + 4\hat{k}) \)
• B. \( \frac{1}{\sqrt{26}}(\hat{i} + 3\hat{j} - \hat{k}) \)
• C. \( \frac{1}{\sqrt{26}}(3\hat{i} - \hat{j} - 4\hat{k}) \)
• D. \( \frac{1}{\sqrt{26}}(3\hat{i} + \hat{j} + 4\hat{k}) \)
• E. \( \frac{1}{\sqrt{26}}(\hat{i} - 3\hat{j} + 4\hat{k}) \)

Hint:

A unit vector and its negative are both parallel to the same line. If your calculated vector isn't in the options, try multiplying it by -1.

Answer 1

The Correct Option is C

Concept: A vector parallel to a line is formed by the direction ratios \( (l, m, n) \) found in the denominators of the symmetric form. To find a unit vector, we divide the direction vector by its magnitude.

Step 1: Identify the direction vector.
From the denominators of \( \frac{x - 2}{3} = \frac{y + 3}{-1} = \frac{z - 2}{-4} \): \( \vec{v} = 3\hat{i} - \hat{j} - 4\hat{k} \).

Step 2: Calculate the magnitude of the vector.
\[ |\vec{v}| = \sqrt{(3)^2 + (-1)^2 + (-4)^2} \] \[ |\vec{v}| = \sqrt{9 + 1 + 16} = \sqrt{26} \]

Step 3: Find the unit vector.
\[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{26}}(3\hat{i} - \hat{j} - 4\hat{k}) \]