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Question

If a, b, c are noncoplanar find the point of intersection of the line passing through the points 2a+3bc,3a+4b2c with the line joining the points a2b+3c,a6b+6c.

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Solution

r=a+λ(ba) (equation of a line)
So, the 2 intersecting lines are,
r1=(2a+3bc)+λ((3a+4b2c)(2a+3bc))=(2a+3bc)+λ(a+bc)
and r2=(a2b+3c)+μ((a6b+6c)(a2b+3c))=(a2b+3c)+μ(4b+3c)

At the point of intersection, r1=r2.
ie, (2a+3bc)+λ(a+bc)=(a2b+3c)+μ(4b+3c)

As a,b,c are non-coplanar, there could only be one linear combination of the vectors for each spanning vector. (characteristic of vector space)
ie, the coefficients of a,b,c are unique.

From equation of r2 we can see that whatever the value of μ, the coefficient of a is constant and is 1.
But for r1 only λ=(1) gives coefficient 1 for a.

So, the intersection point r=(2a+3bc)+(1)(a+bc)=(a+2b)

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