Question

Show that the line r =( 2i+2j+3k)+lambda(i-j+4k) is parallel to the plane r.(i+5j+k)=5 find the distance between them

Answer 1

Answer:

10/3√3 sq. units

Step-by-step explanation:

Given line passes through point A 2i+2j+3k having dr's i-j+4k

Dr's of the normal to the given plane are i+5j+k.

For line to be parallel to the given plane, normal should to the plane should be perpendicular to the line and line shouldnt lie on the plane.

Clearly (i+5j+k).(i-j+4k) = 1 -5 + 4 = 0(hence perpendicular).

Also point A (2i+2j+3k).(i+5j+k)=15 not equal to 5 hence it doesn't lie on the plane.

Equation of plane in symmetrical form = x +5y + z-5=0

Perpendicular Distance of point A(a,b,c) from any plane lx+my+nz=d is given by

|la+mb+nz-d|/√l²+m²+n²

So, Distance of A from the plane is |2+5*2+3-5|/√1²+5²+1²

=|10|/√27 = 10/3√3 sq. units