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Question

Find the equation of the line passing through the point (1,3,2) and perpendicular to the lines x1=y2=z3 and x+23=y12=z+15

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Solution

Let the direction ratios of the required line be a, b, c. Since the required line is perpendicular to the given lines, therefore,

a+2b+3c=0--- (1)

3a+2b+5c=0---(2)
Solving (1) and (2), by cross multiplication, we get

a106=b95=c2+6=k
a=4k,b=14k,c=8k
Thus, the required line passing through P(-1, 3, -2) and

having the direction ratios a=4k,b=14k,c=8k is

x+14=y314=z+28 or x+12=y37=z+24

which is the required equation of the line.

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