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Question

Find vector equation of line passing through point ^i+2^j+3^k and perpendicular to vector ^i+^j+^k and 2^i+^j+3^k

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Solution

Equation of line passing through ˆi+2ˆj3ˆk
Then, the equation is x1a=y2b=z3c
Line is to vector :(ˆi+ˆj+ˆk)×(2ˆi+ˆj+3ˆk)
=∣ ∣ ∣ˆiˆjˆk111213∣ ∣ ∣
=ˆi(31)ˆj(32)+ˆk(12)
=2ˆiˆjˆk
Hence,
Equation of line : x12=y21=z31.

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