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Find the distance of the point (-2,3,-4)...

Find the distance of the point `(-2,3,-4)` from the line `(x+2)/3=(2y+3)/4=(3z+4)/5` measured parallel to the plane `4"x"+12"y"-3"z"+1=0.`

Text Solution

Verified by Experts

The correct Answer is:
`9`
Line through point `P(-2, 3, -4)` and parallel to the given line `(x+2)/(3)= (2y+3)/(4)= (3z+4)/(5)` is
`" "(x+2)/(3)= (y+ (3)/(2))/(2)= (z+ (4)/(3))/((5)/(3))= lamda`
Any point on this line is `Q[3lamda-2, 2lamda-(3)/(2), (5)/(3)lamda-(4)/(3)]`
Direction ratios of `PQ` are `[3lamda, (4lamda-9)/(2), (5lamda+8)/(3)]`
Now PQ is parallel to the given plane
`" "4x+12y-3z+1=0`
Hence, line is perpendicular to the normal to the plane. Thus,
`" "4(3lamda)+12((4lamda-9)/(2))-3((5lamda+8)/(3))=0`
or `" "lamda=2`
`rArr" "Q(4, (5)/(2), 2)`
`rArr" "PQ= sqrt((6)^(2)+ ((5)/(2)-3)^(2)+ (6)^(2))= (17)/(2)`
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