Find `lambda` and u, if `(2hat(i)+6hat(j)+27hat(k))xx(hat(i)+lambdahat(j)+mu hat(k))=0`.

Explanation

Given , `(2hat(i)+6hat(j)27hat(k))xx(hat(i)+lambdahat(j)+mu hat(k))=0`
`rArr|(hat(i),hat(j),hat(k)),(2,6,27),(1,lambda,mu)|=0rArr hat(i)(6mu-27lambda)-hat(j)(2mu-27)+hat(k)(2lambda-6)=0hat(i)+0hat(j)+0hat(k)`
On comparing the corresponding components, we have `6mu-27 lambda=0,-2mu +27=0,2lambda -6=0 rArr 2mu =-9lambda,mu=(27)/(2) and lambda =3`
Hence, `lambda=3 and mu =(27)/(2)`.