Question

2. Show that the points (-1,0,7), (3, 2, 1) and (5, 3,-2) are collinear
verry very urgent plz ans​

Answer 1

HELLO DEAR,

GIVEN:- Three points,

P = ( -1 , 0 , 7 )

Q = ( 3 , 2 , 1 )

R = ( 5 , 3 , -2)

To show that given points are collinear.

SOLUTION:-

The the given points are collinear then , their direction ratio are proportional.

So,

direction ratio of PQ = { 3-(-1) , 2-0 , 1-7}

= {4 , 2 , -6}

Let, a1 = 4, b1 = 2 , c1 = -6

And,

direction ratio of QR ={5-3, 3-2, (-2-1)}

= { 2 , 1 , -3 }

Let a2 = 2 , b2 = 1 , c2 = -3

So,

a1 = 4 = 2.

a2 2

b1 = 2 = 2.

b2 1

c1 =(-6) = 2.

c2 (-3)

Therefore, the points are collinear.

I HOPE IT'S HELP YOU DEAR,

THANKS.

Answer 2

Given:

The coordinates of the points (-1,0,7), (3, 2, 1) and (5, 3,-2)

To prove:

The given points are collinear

Solution:

Let the points for the given coordinates is

A(-1,0,7),

B(3, 2, 1)

C(5, 3,-2)

If the points are collinear then the sum of the distance of the two lines must be equal to the distance of the third one.

Length of the line for the two coordinates is given by:

$\sqrt{(x2-x1)^{2}+(y2-y1)^{2}+(z2-z1)^{2}}$

Length of the line AB

AB = $\sqrt{(-1-3)^{2}+(0-2)^{2}+(7-1)^{2}}$

AB = $\sqrt{16+4+36}$

AB = $\sqrt{56}$

AB = $2\sqrt{14}$

Length of the line BC

BC = $\sqrt{(3-5)^{2}+(2-3)^{2}+(1+2)^{2}}$

BC = $\sqrt{4+1+9}$

BC = $\sqrt{14}$

Length of the line AC

AC = $\sqrt{(-1-5)^{2}+(0-3)^{2}+(7+2)^{2}}$

AC = $\sqrt{36+9+81}$

AC = $\sqrt{126}$

AC = $3\sqrt{14}$

Here,

AB + BC = AC

So, The given points are collinear.