Find the distance between the following pairs | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The distance between points $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ is given by the formula:$PQ = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y

Vedclass · Accepted Answer

$2 \sqrt{5}$

Vedclass · Answer

(A) The distance between points $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ is given by the formula:$PQ = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}}$Given points are $(2, -1, 3)$ and $(-2, 1, 3)$.Substituting the values into the formula:$Distance = \sqrt{(-2 - 2)^{2} + (1 - (-1))^{2} + (3 - 3)^{2}}$$= \sqrt{(-4)^{2} + (2)^{2} + (0)^{2}}$$= \sqrt{16 + 4 + 0}$$= \sqrt{20}$$= \sqrt{4 	imes 5} = 2 \sqrt{5}$

Find the distance between the following pairs of points: $(2, -1, 3)$ and $(-2, 1, 3)$.

Similar Questions

The coordinates of points $A, B, C, D$ are $(a, 2, 1), (1, -1, 1), (2, -3, 4)$ and $(a+1, a+2, a+3)$ respectively. If $AB = 5$ and $CD = 6$,then $a = $

The distance of the point $P(x, y, z)$ from the $yz$-plane is:

Find the distance between the following pairs of points: $(-3, 7, 2)$ and $(2, 4, -1)$.

$A$ point lies in the $yz$-plane,and the sum of its coordinates is $3$. If the distance of the point from the $xz$-plane is twice the distance of the point from the $xy$-plane,find its coordinates.

If $P(5, 4, a)$ and $Q(-1, 2, -2)$ are two points such that $PQ = 7$,then the values of $a$ are ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module