Question

If A=-4i+4j+4k and B=2i+2j+2k,find a unit vector of A×B. Also find angle between them?

Answer 1

Given , `vec A =-4i +4j+4k` and `vec B = 2i+2j+2k`

So, `|vec A | = sqrt(4^2 +4^2 +4^2)=4sqrt(3)`

and, `|vec B| = sqrt(2^2 +2^2 +2^2) = 2sqrt(3)`

So, `vec A × vec B = (-4i +4j+4k) × ( 2i+2j+2k)`

`=(-8k +8j-8k+8i+8j-8i)` `=16(j-k)`

So, unit vector along `vec A × vec B` is `(16(j-k))/sqrt(16^2 +16^2) = (16(j-k))/(16sqrt(2)) = (j-k)/sqrt(2)`

Now,angle between a vector and an unit vector is zero i.e located in the same direction.

If the angle between `vec A` and `vec B` is `theta`,we can say,

`vec A . vec B = |vec A| |vec B| cos theta`

so, `cos theta = (-8+8+8)/(8*3)=8/24=1/3`

so, `theta = cos ^-1(1/3)`