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Question

Find the vector and Cartesian equations of the line passing through the point (2,1,3) and perpendicular to the lines x11=y22=z33 and x3=y2=z5.

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Solution

Let the required line be parallel to the vector b given by, b=b1i+b2j+b3k
The position vector of the point (2,1,3) and parallel to vector b is
r=a+λb
r(2i+j+3k)+λ(b1i+b2j+b3k)--- (1)
The equation of the lines are
x11=y22=z33--- (2)
x3=y2=z5 ---- (3)
Line (1) and (2) are to each other.
b1+2b2+3b3=0--- (4)
Also, Line (1) and (3) are to each other.
3b1+2b2+5b3=0--- (5)
From equation (4) and (5), we obtain
b12(5)3(2)=b253(3)=b322(3)
b14=b214=b38
b12=b27=b34
Therefore direction ratios of b are 2,7,4.
b=2i7j+4k
Substituting b=2i+7j+4k in equation (1), we obtain
r(2i+j+3k)+λ(2i7j+4k)
This is the equation of the required line.

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