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Question

Find the vector equation of the line passing through (1,2,3) and parallel to the planes r(^i^j+2^k)=5 and r(3^i+^j+^k)=6.

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Solution

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r.(^i^j+2^k)=5 Plane I
r.(3^i+^j+^k)=6 Plane II
Let direction ratios of required line are a,b,c
Given Direction ratio of normal from plane 1 in (1,1,2)
Direction ratio of normal from plane 2 in (3,1,1)
Since
Required line is 1 to normal of plane I
using condition of perpendicularity
a1a2+b1b2+c1c2=0
ab+2c=0....(1)
Same Required line is to normal of plane II
using condition
a1a2+b1b2+c1c2=0
3a+b+c=0....(2)
Solving eq (1) & (2)
a12=b61=c1+3
a3=b5=c4=λ (say)
a=3λ b=5λ & c=4λ
Direction vector to line b=λ(3^i+5^j+4^k)
Required equation of line r=a+λb
r=(^i+2^j+3^k)+λ(3^i+5^j+4^k)
Hence vector equation of line is r=(^i+2^j+3^k)+λ(3^i+5^j+4^k)

1178377_785831_ans_2a70f3c58b2648f1ae7690dc52529dc3.png

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