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Question
The lines x+1−10=y+3−1=z−41 and x+10−1=y+1−3=z−14 intersect at the point
The lines x+1−10=y+3−1=z−41 and x+10−1=y+1−3=z−14 intersect at the point
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Solution
The correct option is D (−11,−4,5)
For the first line, x=−10t−1,y=−t−3,z=t+4And for the second line, x=−s−10,y=−3s−1,z=4s+1where s and t are constants.Since, we need the intersection point, the corresponding values must be equal.⇒−10t−1=−s−10 or 10t−s=9⇒−t−3=−3s−1 or 3s−t=2⇒t+4=4s+1 or 4s−t=3If simultaneously solved the last two equations, we get s=1, and thus, t=1The point thus becomes x=−11,y=−4,z=5.
For the first line, x=−10t−1,y=−t−3,z=t+4
And for the second line, x=−s−10,y=−3s−1,z=4s+1
where s and t are constants.
Since, we need the intersection point, the corresponding values must be equal.
⇒−10t−1=−s−10 or 10t−s=9
⇒−t−3=−3s−1 or 3s−t=2
⇒t+4=4s+1 or 4s−t=3
If simultaneously solved the last two equations, we get s=1, and thus, t=1
The point thus becomes x=−11,y=−4,z=5.
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