show that the points (-2 ,3, 5) (1 ,2 ,3) and (7, 0, 1) are collinear.
Answers
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If three points are collinear, then they lie on a line.
Firstly let us calculate distance between the 3 points
i.e. PQ, QR and PR
Calculating PQ
P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 2, y1 = 3, z1 = 5
x2 = 1, y2 = 2, z2 = 3
Distance PQ = √[(1 – (-2))2 + (2 – 3)2 + (3 – 5)2]
= √[(3)2 + (-1)2 + (-2)2]
= √[9 + 1 + 4]
= √14
Calculating QR
Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 1, y1 = 2, z1 = 3
x2 = 7, y2 = 0, z2 = – 1
Distance QR = √[(7 – 1)2 + (0 – 2)2 + (-1 – 3)2]
= √[(6)2 + (-2)2 + (-4)2]
= √[36 + 4 + 16]
= √56
= 2√14
Calculating PR
P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 2, y1 = 3, z1 = 5
x2 = 7, y2 = 0, z2 = – 1
Distance PR = √[(7 – (-2))2 + (0 – 3)2 + (-1 – 5)2]
= √[(9)2 + (-3)2 + (-6)2]
= √[81 + 9 + 36]
= √126
= 3√14
Thus, PQ = √14, QR = 2√14 and PR = 3√14
So, PQ + QR = √14 + 2√14
= 3√14
= PR
∴ The points P, Q and R are collinear
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