Question

Find the distance of point (1,2,3) from line joining (-1,2,5)

Answer 1

Answer:

$\displaystyle \sf 2\sqrt{2} \ units$

Step-by-step explanation:

Given :

Two point A ( 1 , 2 , 3 ) and B ( - 1 , 2 , 5)

We are asked to find distance between them :

We know distance formula :

$\displaystyle \sf AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Putting values here we get :

$\displaystyle \sf AB=\sqrt{(-1-1)^2+(2-2)^2+(5-3)^2}$

$\displaystyle \sf AB=\sqrt{(-2)^2+(0)^2+(2)^2}$

$\displaystyle \sf AB=\sqrt{(4+4)}$

$\displaystyle \sf AB=\sqrt{4(2)}$

$\displaystyle \sf AB=2\sqrt{2} \ units$

Hence we get required answer.

Answer 2

GIVEN:

Points A = (1,2,3) and B = (-1,2,5)

TO FIND:

Distance between the points.

SOLUTION:

We know that,

Distance :

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

x1 = 1 ; x2 = -1 ; y1 = 2; y2 = 2; z1 = 3 ; z2 = 5

$= > \sqrt{ {( - 1 - 1)}^{2} + {(2 - 2)}^{2} + {(3 - 5)}^{2} } \\ \\ = > \sqrt{ {( - 2)}^{2} + 0 + {( - 2)}^{2} } \\ \\ = > \sqrt{4 + 4} \\ \\ = > \sqrt{8} \\ \\ = > 2 \sqrt{2}$

Therefore, the distance between the is 2√2.