Login/Sign Up
Home » Find the distance between the points : (i) R(1, -3, 4) and S(4, -2, -3) (ii) C(9, -12, -8) and the origin
Find the distance between the points : (i) R(1, -3, 4) and S(4, -2, -3) (ii) C(9, -12, -8) and the origin
CBSE | Class 11 | Exercise 26B | Maths | RS Aggarwal | Three Dimensional Geometry
Find the distance between the points : (i) R(1, -3, 4) and S(4, -2, -3) (ii) C(9, -12, -8) and the origin

Answers:

(i)

R(1, -3, 4) and S(4, -2, -3)

(x1,y1,z1) = (1, -3, 4)

(x2,y2,z2) = (4, -2, -3)

\begin{array}{l} D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ D = \sqrt {{{( 4 - 1)}^2} + {{(-2-(-3))}^2} + {{( - 3-4)}^2}} \\ D = \sqrt {9 + 1 + 49} \\ D = \sqrt {59} \end{array}

Distance between points R and S is √59 units.

(ii)

C(9, -12, -8) and the origin Coordinates of origin are (0, 0, 0)

(x1,y1,z1) = (9, -12, -8)

(x2,y2,z2) = (0, 0, 0)

\begin{array}{l} D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ D = \sqrt {{{( 0 - 9)}^2} + {{(0 - (-12))}^2} + {{( 0 - (-8))}^2}} \\ D = \sqrt {81 + 144 + 64} \\ D = \sqrt {289} \end{array}

The distance between points C and the origin is 17 units.

i

More Material