Question

Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.​

Answer 1

Answer:

determinant is 0 or area of triangle is 0

Answer 2

Answer:

$\huge \underbrace \mathtt \purple{answer}$

HI THERE,

⇨★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.

$\huge\boxed{\fcolorbox{blue}{orange}{hope it helps}}$