Show that the line `vecr=(2hati-2hatj+3hatk)+lambda(hati-hatj+4hatk)` is parallel to the plane `vecr.(hati+5hatj+hatk)=5.`
Also, find the distance between the given line and the given plane.

Explanation

The given line is in the fom `vecr=veca+lambdavecb` and the given plane is in the form `vecr.vecn=q`, where
`veca=(2hati-2hatj+3hatk), vecb=(hati-hatj+4hatk), vecn=(hati+5hatj+hatk)` and q=5.
We know that the line `vecr=veca+lambdavecb` is parallel to the plane `vecr.vecn=q` only when `vecb.vecn=0`
Here, `vecb.vecn=(hati-hatj+4hatk). (hati+5hatj+hatk)=(1 xx 1)+(-1) xx 5 + 4 xx 1=0`.
Hence, the given line is parallel to the given plane.
Distance between the given line and the given plane
`=|veca.vecn-q|/|vecn| = |(2hati-2hatj+3hatk).(hati+5hatj+hatk)-5|/|hati+5hatj+hatk|`
`=|(2 xx 1)+(-2) xx 5+(3 xx 1)-5|/sqrt(1^(2)+5^(2)+1^(2))`
`=|2-10+3-5|/sqrt(27) = 10/(3 sqrt(3))` units.
Hence, the distance between the given line and the given plane is `10/(3sqrt(3))` units.