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Question

Find the line through (2,1,3) and perpendicular to each of the lines r=i+jk+λ(2i2j+k) and r=2ij3k+μ(i+2j+2k)

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Solution

Point(2,1,3)r1=(ˆi+ˆjˆk)+λ(2ˆj2ˆj+ˆk)r2=(2ˆiˆj3ˆk)+μ(ˆi+2ˆj+2ˆk)equationofline:¯¯¯r=(2ˆiˆj+3ˆk)+t(b1ˆi+b2ˆj+b3ˆk)So,(b1ˆi+b2ˆj+b3ˆk).(2ˆi2ˆj+ˆk)=02b12b2+b3=0.............(1)(b1ˆi+b2ˆj+b3ˆk).(ˆi+2ˆj+2ˆk)=0b1+2b2+2b3=0............(2)b142=b241=b34+2b16=b23=b36b12=b21=b32so,¯¯¯r=(2ˆjˆj+3ˆk)++t(2ˆj+ˆj2ˆk)

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