Find the angle between the vectors `4hati-2hatj+4hatk and 3hati-6hatj-2hatk.`

Answer

`67.60`

Explanation

Let `vec(a)= 4hat(i) - 2hat(j) + 4hat(k) and vec(b)= 3hat(i)- 6hat(j)- 2hat(k)`
Let `theta` be the angle between the vectors `vec(a) and vec(b)`, then `cos theta= (vec(a).vec(b))/(|vec(a)||vec(b)|)` ...(i)
Now, `vec(a).vec(b)= 4 xx 3+ (-2) xx (-6) + 4 xx (-2)= 16`
and `|vec(a)|= sqrt(4^(2)+ (-2)^(2) + 4^(2))= 6`
and `|vec(b)|= sqrt(3^(2)+ (-6)^(2) + (-2)^(2))= 7`
Hence, from Eq (i), `cos theta= (16)/(6 xx7) = (8)/(21)`
`therefore theta= cos^(-1) ((8)/(21))= 67.60`