If `vec(a) = 4 hat(i) + 3 hat(j) + 2 hat(k) and vec(b)= 3 hat(i)+ 2 hat(k), "find" |vec(b) xx 2 vec(a)|.`

Explanation

We have `vec(b) = ( 3 hat (i) + 2 hat(k) ) and 2 vec(a) = ( 8 hat(i) + 6 hat(j) + 4 hat(k))`
`:. ( vec(b) xx2 vec(a))=|(hat(i),hat(j), hat(k)),(3,0,2),(8,6,4)|`
`= (0-12) hat(i) + (16 - 12) hat(j) + (18-0) hat(k)`
`= (-12hat(i) + 4 hat(j) + 18 hat(k)).`
`:. |vec(b) xx 2 vec(a)| = |-12 hat(i) + 4 hat(j) + 18 hat(k)|`
` = sqrt((-12)^(2) + 4^(2) + (18)^(2))=sqrt(484 )= 22.`
Hence,`| vec(b) xx 2 vec(a)| = 22.`