Question

If a b c are three non coplanar find the point of intersection of the line passing through the points 2a+3b+c,3a+4b-2c,a-2b+3c and a-6b+6c

Answer 1

Answer:

As, a, b and c are three non coplanar vectors, and line passes through th points having direction ratios, (2,3,1), (3,4,-2), (1,-2,3), (1,-6,6).

Equation of line passing through two points (a,b,c) and (p,q,r) is given by

$\frac{x-a}{p-a}=\frac{y-b}{q-b}=\frac{z-c}{r-c}$

Equation of line passing through (2,3,1), (3,4,-2) is,

$\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-1}{-3}=m$

because, 3-2=1, 4-3=1,-2-1=-3

Equation of line passing through (1,-2,3), (1,-6,6) is,

$\frac{x-1}{0}=\frac{y-2}{-4}=\frac{z-3}{3}=n$

because, 1-1=0, -6+2=-4, 6-3=3

x=m+2, y=m+3, z= -3 m+1

x=1, y=-4 n+2,z=3 n+3

m+2=1

→m=1-2

m= -1

→→→m +3=-4 n+2

-1+3=-4 n+2

→-4 n=2-2

→→ - n=0

n=0

When we substitute the value of m, and n in variable z ,z=3 n+3, and z=-3m+1, the two z variable of two lines  are not same.

So,the two lines will not intersect.