Question

Find the lengths of the sides of the triangle PQR.

P(4, −2, −3),    Q(8, 0, 1),    R(10, −4, −3) what kind of triangle is it?

 

Answer 1

The points are P = (4, -2, -3), Q = (8, 0, 1) and R  = (10, -4, -3) 

The point 

b)

Let P (x1, y1, z1) = (4, -2, -3) and Q (x2, y2, z2) = (8, 0, 1)

Lenth of the Line Segment =  √ [(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2] ----------------------> (1)

Substitute  P (x1, y1, z1) = (4, -2, -3) and Q (x2, y2, z2) = (8, 0, 1) in equation (1)

PQ  =  √ [ (8 - 4)^2 + (0 + 2)^2 + (1 + 3)^2 ]

       =   √ [ (4)^2 + (2)^2 + (4)^2 ]

       =   √ [ 16 + 4 + 16 ]

       =   √ [ 36 ]

       =   6

Let Q (x1, y1, z1) = (8, 0, 1) and R (x2, y2, z2) = (10, -4, -3) 

Substitute  Q (x1, y1, z1) = (8, 0, 1) and R (x2, y2, z2) = (10, -4, -3)  in equation (1)

QR  =  √ [ (10 - 8)^2 + (-4 - 0)^2 + (-3 - 1)^2 ]

       =  √ [ (2)^2 + (-4)^2 + (4)^2 ]

       =   √ [ 4 + 16 + 16 ]

       =   √ [ 36 ]

       =   6

Let R (x1, y1, z1) = (10, -4, -3) and P (x2, y2, z2) =  (4, -2, -3)

Substitute  R (x1, y1, z1) = (10, -4, -3) and P (x2, y2, z2) =  (4, -2, -3)  in equation (1)

RP  =  √ [ (4 - 10)^2 + (-2 + 4)^2 + (-3 + 3)^2 ]

       =  √ [ (- 6)^2 + (2)^2 + (0)^2 ]

       =  √ [ 36 + 4 + 0 ]

       =  √40

RP^2  <  PR^2 + QR^2

(√40 )^2  <  6^2 + 6^2

40  <  36 + 36

40  <  72

Hence, it is a acute angled triangle

Acute angled triangle.