Question

A straight line passes through the points \( (10,8,6) \) and \( (13,9,4) \). A unit vector parallel to this line is

• A. \( \frac{1}{\sqrt{17}}(3\hat{i} + 2\hat{j} + 2\hat{k}) \)
• B. \( \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k}) \)
• C. \( \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} + 2\hat{k}) \)
• D. \( \frac{1}{\sqrt{11}}(3\hat{i} + \hat{j} + 2\hat{k}) \)
• E. \( \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k}) \)

Hint:

Unit vector = direction vector divided by its magnitude.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A vector parallel to the line passing through two points A and B can be found by calculating the displacement vector \(\vec{AB}\) or \(\vec{BA}\). A unit vector is a vector with a magnitude of 1. To find the unit vector in the direction of a given vector, we divide the vector by its magnitude.
Step 2: Key Formula or Approach:
1. Let A = (10, 8, 6) and B = (13, 9, 4). 2. Find the direction vector \(\vec{d} = \vec{AB} = B - A\). 3. Calculate the magnitude of this vector, \(|\vec{d}|\). 4. The unit vector is \(\hat{d} = \frac{\vec{d}}{|\vec{d}|}\).
Step 3: Detailed Explanation:
1. Find the direction vector. Let the two points be \(P_1 = (10, 8, 6)\) and \(P_2 = (13, 9, 4)\). The direction vector \(\vec{d}\) parallel to the line is given by \(\vec{P_1P_2}\): \[ \vec{d} = \langle 13-10, 9-8, 4-6 \rangle \] \[ \vec{d} = \langle 3, 1, -2 \rangle \] In vector notation, this is \(\vec{d} = 3\hat{i} + \hat{j} - 2\hat{k}\). 2. Calculate the magnitude of the direction vector. \[ |\vec{d}| = \sqrt{3^2 + 1^2 + (-2)^2} \] \[ |\vec{d}| = \sqrt{9 + 1 + 4} = \sqrt{14} \] 3. Find the unit vector. The unit vector \(\hat{d}\) is found by dividing the vector \(\vec{d}\) by its magnitude \(|\vec{d}|\): \[ \hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{3\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{14}} \] \[ \hat{d} = \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k}) \] Step 4: Final Answer:
The unit vector parallel to the line is \(\frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k})\).