If line joining points A and B having position vectors 6a − 4b + 4c and −4c respectively, and the line joining the points C and D having position vectors −a −2b −3c and a + 2b −5c intersect, then their point of intersection is

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Please find below the solution to the asked query:

Let Obe originOA=6a-4b+4c 6,-4,4OB=-4cAB=OB-OA=-4c-6a+4b-4cAB=-6a+4b-8cHence direction ratios of AB are -6,4,-8 i.e. 3,-2,4Hence equation of AB isx-63=y--4-2=z-44=kx-63=y+4-2=z-44=kHence any general point on AB will be x,y,z=3k+6,-2k-4,4k+4Similarly equation of CD will bex+11=y+22=z+3-1=λHence any general point on CD will be x,y,z=λ-1,2λ-2,-λ-3At point of intersection, general point of AB and CD will be samex=x and y=y3k+6=λ-1  and -2k-4=2λ-23k-λ=-7...i and 2λ+2k=-2 i.e. λ+k=-1....iiAdding equations we get3k-λ+λ+k=-7-14k=-8k=-2x,y,z=3k+6,-2k-4,4k+4=-6+6,4-4,-8+4=0,0,-4Hence they intersect at -4c

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  • 2
My answer is based on the assumption that a ,  b and c represent the unit vectors ( as you haven't specified what a , b and c are )  mutually perpendicular to each other along x , y and z axes
So, the correct answer is ( - 2 , - 4 ,  - 2 ) , this is the point of intersection of lines AB and CD
  • -1
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