Engineering
Mathematics
Distance Formulae and Its Application
theorem in space
Distance from a Plane
Question

Find the point in yz−plane which is equidistant from the points A(3,2,−1),B(1,−1,0) and C(2,1,2).

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Solution
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Given points are A(3,2,−1), B(1,−1,0) and C(2,1,2).

We know that x coordinate of every point in yz− plane is zero. 

Any point in the yz− plane is of the form P(0,y,z).

 Now, according to question
∣AP∣ = ∣BP∣ = ∣CP∣  ..........  Given 
⇒  AP= BP2 
⇒  (0 − 3)+ (y − 2)+ (z + 1)= (0 − 1)+ (y + 1)+ (z − 0)2

⇒  3y − z − 6 = 0  ....... (i)

and BP= CP2.
⇒  (0 − 1)+ (y + 1)+ (z − 0)= (0 − 2)+ (y − 1)+ (z − 2)2

⇒   4y + 4z − 7 = 0 ......... (ii)
On solving, (i) and (ii) we get 

y=3116 and z=316.
Hence, the required point is (0,3116,316).