Question

1. Let P be any point on the line: 1:= (4,8, -1)+(2,0,–4),TER. Let Q be any point on the line: 12: .X-7_1-2_2+1 -6 Find the Cartesian equation of the plane formed by all the possible midpoints of the line segment PQ.

Answer 1

Both equations are given in different forms, so at first, convert second line to parametric form:

i.e.

12:7= (7,2, -1) +r(-6,2,2)

now consider both the lines:

Any point P on the line $l_1$ is given as;

P=(4+23, 8, -1-45)

Also any point Q is given as :

2 Q =(7-67, 2+28,-1+2 27

Now from midpoint formula, midpoint (M) of any points P and Q will be given as:

M - ( 땅 카트, 어릴령, 벌려

simplifying;

M (11+22 11+28-67, 0+27 -2-42.+27 2 2

.....(1)

1 = ( + 2-37, 5+, -1-2t + x

So the equation (1) represents locus of all points which are midpoints of P and Q. So this will be a parametrized equation of the required plane. Hence inorder to convert this to cartesian form, lets substitute x,y, and z for each element in M:

i.e.

..

x = 5 t t-3r y = 5tr B - - 1-2ttr © mo

So from equation (B), $r=y-5$ . So substitute this value of r in equation (A)

5 sc = 들 + t-3Cg-5 2 5 2

or

2. 43g 15년들

Now substitute the above value of t and r in equation (C) to obtain the equation of plane in cartesian form:

3 t ur 3=-1-2(SC434-1552) + (y-5) =-1-20c-by+30+5+4-5

On simplifying;

$z=-2x-5y+29$.

So equation of plane is:

$z+2x+5y=29$.