Question

If \( \vec{OA} = 2i - j + k \), \( \vec{OB} = -3i - k \), and \( \vec{OC} = -2i + 2j - 3k \), then a unit vector perpendicular to the plane containing \( A, B, C \) is:

• A. \( \frac{8i - 4j + 2k}{2\sqrt{21}} \)
• B. \( \frac{6i + 2j + 3k}{7} \)
• C. \( \frac{9i + 2j + 6k}{11} \)
• D. \( \frac{8i + 2j + 5k}{\sqrt{93}} \)

Hint:

To find a unit vector perpendicular to a plane, calculate the cross product of two vectors lying in the plane, and then normalize the resulting vector.

Answer 1

The Correct Option is A

The normal vector to the plane is obtained by taking the cross product of the vectors \( \vec{OA} \) and \( \vec{OB} \). After calculating the cross product, we normalize the resulting vector to obtain the unit vector perpendicular to the plane. This results in \( \frac{8i - 4j + 2k}{2\sqrt{21}} \).