Find the angle between the line `vecr=hati+2hatj-hatk+lamda(hati-hatj+hatk)` and the plane `vecr*(2hati-hatj+hatk)=4`.

Explanation

The given line is `vecr= (veci+2vecj-veck)+lamda(veci-vecj+veck)`
Here, `vecb=veci-vecj+veck " "("type "vecr=veca+lamdavecb)`
The given plane is `vecr*(2veci-vecj+veck)=4 " " ("type "vecr*vecn=p)`
Here `vecn=vec(2i)-vecj+veck`
Now `costheta = (vecn*vecb)/(|vecn||vecb|)`
`" "` (If `theta` is the angle between the line and the normal to the plane)
`" "=((2veci-vecj+veck)*(veci-vecj+veck))/(sqrt(4+1+1)sqrt(1+1+1))`
`" "= (2+1+1)/(sqrt(6)sqrt(3))=(4)/(sqrt(2)*3)= (2sqrt2)/(3)`
`therefore" "theta= cos^(-1) ((2sqrt(2))/(3))`