Question
The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line (12)x=(13)y=(−16)z is :
Answer: Option A
:
A
Line parallel to the line x2=y2=z−6 and passing through (1,-2, 3) is
x−12=y+23=z−3−6=r
A = (2r + 1, 3r – 2, – 6r + 3) is a point on this line.
A lies in the plane x – y + z = 5
∴2r+1−3r+2−6r+3=5⇒r=17∴A=(97,−117,157)
Distance of A from (1, –2, 3) is
√(97−1)2+(−117+2)2+(157−3)2=17√4+9+36=77=1
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:
A
Line parallel to the line x2=y2=z−6 and passing through (1,-2, 3) is
x−12=y+23=z−3−6=r
A = (2r + 1, 3r – 2, – 6r + 3) is a point on this line.
A lies in the plane x – y + z = 5
∴2r+1−3r+2−6r+3=5⇒r=17∴A=(97,−117,157)
Distance of A from (1, –2, 3) is
√(97−1)2+(−117+2)2+(157−3)2=17√4+9+36=77=1
Was this answer helpful ?
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