Question

Calculate the components of a vector of magnitude unity which is at right angles to the vectors 2i+j-4k and 3i+j-k

Answer 1

Any vector which is perpendicular to two vectors can be defined as the cross product of the two vectors. i.e.

C = A x B

=> C = (2i+j-4k)(3i+j-k)

= 2(i x j) - 2(i x k) + 3(j x i) -(j x k) - 12(k x i) -4(k x j)

= 2k + 2j - 3k - i - 12j +4i

=>C = 3i -10j -k

Now the unit vector along C can be given as

C^ = $\frac{C}{|C|}$

|C| = √(3² + 10² + 1²)

=> |C| = √110

Hence

C^ = (3i - 10j-k)/√110

So the unit vector perpendicular to both the given vectors is given by $\frac{3i}{\sqrt{110} }-\frac{10j}{\sqrt{110} }-\frac{k}{\sqrt{110} }$