Question

If the position vectors of the points \( A \) and \( B \) are \( 2i + 3j - k \) and \( i - j + 2k \) respectively, then the unit vector along \( \overrightarrow{BA} \) and in the direction of \( AB \) is:

• A. \( \frac{1}{\sqrt{14}} (3i + 2j + k) \)
• B. \( \frac{1}{\sqrt{26}} (-i - 4j + 3k) \)
• C. \( \frac{1}{\sqrt{26}} (-3i - 4j + k) \)
• D. \( \frac{1}{\sqrt{22}} (3i - 4j + 3k) \)

Hint:

To find a unit vector, first calculate the vector difference, then divide by the magnitude to normalize it.

Answer 1

The Correct Option is B

We are given the position vectors \( \vec{A} = 2i + 3j - k \) and \( \vec{B} = i - j + 2k \). Step 1: To find the direction of the vector \( \overrightarrow{BA} \), calculate \( \overrightarrow{BA} = \vec{A} - \vec{B} \): \[ \overrightarrow{BA} = (2i + 3j - k) - (i - j + 2k) = i + 4j - 3k \] Step 2: The magnitude of \( \overrightarrow{BA} \) is: \[ |\overrightarrow{BA}| = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \] Step 3: The unit vector in the direction of \( AB \) is: \[ \frac{\overrightarrow{BA}}{|\overrightarrow{BA}|} = \frac{1}{\sqrt{26}} (i + 4j - 3k) \] Thus, the correct answer is \( \frac{1}{\sqrt{26}} (-i - 4j + 3k) \).