Find the direction cosines of the unit vector perpendcular to the plane `vecr.(2hati+2hatj-3hatk)=5 and vecr.(3hati-3hatj+5hatk)=3`

Explanation

Equation of given planes are :
`vecr.(2hati+2hatj-3hatk) = 5` and `vecr.(3hati-3hatj+5hatk) = 3`
If `vecn_(1)` and `n_(2)` be the normal to the planes `vecr_(1).vecn_(1) = d_(1)` and `vecr_(2)-vecn_(2)= d_(2)` then angle is given by
`cos theta =|(vecn_(1).vecn_(2))/(|vecn_(1).vecn_(2)|)|"........"(1)`
Here, `vecn_(1)=2hati+2hatj-3hatk` and `vecn_(2) = 3hati-3hatj+5hatk`
`:. vecn_(1).vecn_(2)=(2hati+2hatj-3hatk).(3hati-3hatj+5hatk)`
`= 6 - 6 - 15 = - 15`
`|vecn_(1)| = sqrt((2)^(2)+(2)^(2)+(-3)^(2)) = sqrt(4+4+9) = sqrt(17)`
`|vecn_(2)| = sqrt((3)^(2) - (-3)^(2) +(5)^(2)) = sqrt(9+9+25) = sqrt(43)`
`:. cos theta = |(1-5)/(sqrt(7)sqrt(43))| rArr cos theta = (15)/(sqrt(731))`
`rArr theta = cos^(-1)((15)/(sqrt(731)))`