Engineering
Mathematics
Equation of Straight Line in 3D
Distance from a Plane
Question
The point in the xy− plane which is equidistant from (2,0,3),(0,3,2) and (0,0,1) is
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Solution
Let the points are A(2,0,3),B(0,3,2) and D(0,0,1)
We know that Z-coordinate of every point an xy-plane is zero so let p(x,y,0) be a point on xy-plane such that PA = PB = PC
Now, PA = PB
⇒ PA2 = PB2
⇒ (x − 2)2 + (y − 0)2 + (0 − 3)2
= (x − 0)2 + (y − 3)2 + (0 − 2)
⇒ 4x − 6y = 0
⇒ 2x − 3y = 0 ......(i)
and
PB = PC
⇒ PB2 = PC2
⇒ (x − 0)2 + (y − 3)2(0 − 2)2
= (x − 0)2 + (y − 0)2 + (0 − 1)2
⇒ −6y + 12 = 0
⇒ y = 2 ......(ii)
Putting y = 2 in equation (i),
we get x = 3
Hence, the required point is (3,2,0)