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The distance of the point (1, –2, 3) fro...

The distance of the point `(1, –2, 3)` from the plane `x – y + z = 5` measured parallel to the line `x/2 = y/3 = -z/6` is :-

Text Solution

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The correct Answer is:
`1.00`

`P(1,-2,3) L -= x/2=y/3-z/6`
eq of line PQ is
`(x-1)/(2/7) = (y+2)/(3/7) = (z-3)/((-6)/7) = r`
If PQ = r
Then `Q -=(1+2/7 r, -2+3/7 r, 3-6/7 r)`
8 line on plane `x - y + z = 5`
`implies r = 1`
`:. PQ = 1`.
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