Question

Which of the following is a unit vector perpendicular to the plane determined by the vectors A=2i+4j and B=i+j-k?

A. 1 / √6-(-2i + j-k) 6
B. 1 / √5 (i+2)
C. 1 / √6 1 -(-2i-j-k)
D. -2i + j-k

Answer 1

To find a unit vector perpendicular to the plane determined by the vectors A=2i+4j and B=i+j-k, we can take the cross product of the two vectors and then divide the result by its magnitude. The correct answer is C. 1 / √6 1 -(-2i-j-k).

Step-by-step explanation:

To find a unit vector perpendicular to the plane determined by the vectors A = 2i+4j and B = i+j-k, we can take the cross product of the two vectors. The cross product of two vectors is a vector that is perpendicular to both vectors in the plane.

Let's calculate the cross product:

A x B = (2i+4j) x (i+j-k) = (4k-(-4)i)-(2k-i+j) = -2i+6j+6k

Now, we need to find a unit vector in the same direction as -2i+6j+6k. To do this, we divide the vector by its magnitude:

Unit vector = (-2i+6j+6k) / |(-2i+6j+6k)| = (-2i+6j+6k) / sqrt((-2)^2+6^2+6^2) = (-2i+6j+6k) / sqrt(76)

So, the correct answer is C. 1 / √6 1 -(-2i-j-k)