Find a unit vector that is perpendicular to both of the vectors 4i+2j-3k and 2i-3j+k?

Answer 1

The unit vector is (0.347833... , 0.496904... , 0.795046... )

It was found as follows. Take the cross-product of (2, - 3, 1) and (4, 2, - 3) to find a vector perpendicular to both. It's

7i + 10j +16k.

[ This resulting vector was a final edit, due to several mistakes. In penance, I'll check orthogonality. The dot products are:

1. With (2, - 3, 1), 14 - 30 + 16 = 0.
2. With (4, 2, - 3), 28 + 20 - 48 = 0.

So it IS perpendicular to the original two vectors. ]

To normalise this vector to produce a UNIT vector in the same direction, one must divide by its modulus, that is by the square root of the sum of the squares of the i, j, and k components.

The sum of the squares of the components is 405, and ITS square root is 20.1246118... .

So the requested normalized unit vector is:

(0.347833... , 0.496904... , 0.795046... )

CHECK: To the 9 significant figures shown on my calculator, the squares of these components indeed sum to 1.

Live long and prosper

Answer 2

First, take the cross product, v, of both these vectors
to get a vector perpendicular to both:
It is the determinant of

i j k
4 2 -3
2 -3 1

which works out to -7i -10j -16k. But -v also
satisfies the conditions of the problem.
So we want a unit vector in the opposite direction of v.
Thus we must divide by its length which works out to
be √405. So the final answer is
1/√405(7i + 10 j + 16k).
So
Sin

Answer 3

To find a vector perpendicular to those two vectors, you need to take a cross product of those vectors...

Once you have the resulting vector...you then divide each term by the magnitude of the vector...this will result in a unit perpendicular vector