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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 4x + 12y - 3z + 1 = 0 .

VENNA RAMANJANEYA REDDY
18 Points
7 years ago
Hi,

Let P be the given point P(x1,y1,z1)= (-2,3,-4)
Let Q(x2,y2,z2) be the point on the given line (x+2)/3 = (2y+3)/4 = (3z+4)/5
Let (x+2)/3 = (2y+3)/4 = (3z+4)/5 = k
So Q(x2,y2,z3) = (3k-2, (4k-3)/2, (5k-4)/3 )

From the problem statement, the line PQ should be parallel to the given plane equation 4x + 1 2 y - 3 z + 1 = 0.----(i)
If this line PQ is parallel to the plane then its direction vector must be perpendicular to the plane's normal vector.
The direction vector normal or perpendicular to the plane 4x + 12y - 3z + 1= 0 is D (4, 12, -3)
If two vectors are perpendicular then their scalar or dot product is zero.
Therefore D and Q are perpendicular to each other.
=> 4(3k-2) + 12 * ( 4k-3 )/2  - 3 * (5k-4) / 3 = 0
=> k = 2

So Q(x2,y2,z2) = ( 4,5/2, 2)
The distance between PQ = =1 7 / 2.