Question

A unit vector perpendicular to the plane containing \( \vec{A} = i + j - 2k \) and \( \vec{B} = 2i - j + k \) is

• A. \( \frac{1}{\sqrt{26}} (-i + 3j - 4k) \)
• B. \( \frac{1}{\sqrt{19}} (-i + 3j - 3k) \)
• C. \( \frac{1}{\sqrt{35}} (-i + 5j - 3k) \)
• D. \( \frac{1}{\sqrt{35}} (i - j - 3k) \)

Hint:

The cross product of two vectors gives a vector perpendicular to both. Normalize the result to get the unit vector.

Answer 1

The Correct Option is D

Step 1: Finding the cross product. 
A vector perpendicular to the plane containing \( \vec{A} \) and \( \vec{B} \) can be found by calculating the cross product \( \vec{A} \times \vec{B} \). 

The cross product calculation yields: \[ \vec{A} \times \vec{B} = \hat{i}(1 + 2) - \hat{j}(1 + 4) + \hat{k}(-1 - 2) = 3\hat{i} - 5\hat{j} - 3\hat{k} \] 
Step 2: Finding the unit vector. 
The magnitude of \( \vec{A} \times \vec{B} \) is: \[ \sqrt{3^2 + (-5)^2 + (-3)^2} = \sqrt{35} \] Thus, the unit vector is: \[ \frac{1}{\sqrt{35}}(3\hat{i} - 5\hat{j} - 3\hat{k}) \]