1 Answer
Solution :-
Given:
The points(2, -1, 3) and (-1, 2, 1)
Let C(x, y, z) be any point on the given plane and C divides AB in ratio k: 1
Then, m = k and n = 1
A(2, -1, 3) and B(-1, 2, 1)
Coordinates of C are:
On comparing we get,
\(\dfrac{ [-k + 2] }{ [k + 1]}\) = x
\(\dfrac{ [2k – 1] }{ [k + 1]}\) = y
\(\dfrac{ [-k + 3]} { [k + 1]}\) = z
We know that x + y + z = 5
5(k + 1) = 4
5k + 5 = 4
5k = 4 – 5
5k = – 1
k = \( \dfrac{-1}{5}\)
We can say that, the plane divides AB externally in the ratio 1:5
Answered by Aaryan | 1 year agoRelated Questions
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