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

$\sqrt{43}$

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 $(-3, 7, 2)$ and $(2, 4, -1)$.Here,$x_{1} = -3, y_{1} = 7, z_{1} = 2$ and $x_{2} = 2, y_{2} = 4, z_{2} = -1$.Substituting these values into the distance formula:$PQ = \sqrt{(2 - (-3))^{2} + (4 - 7)^{2} + (-1 - 2)^{2}}$$PQ = \sqrt{(2 + 3)^{2} + (-3)^{2} + (-3)^{2}}$$PQ = \sqrt{(5)^{2} + 9 + 9}$$PQ = \sqrt{25 + 9 + 9}$$PQ = \sqrt{43}$

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

Similar Questions

Find the equation of the set of points $P$ such that $PA^{2} + PB^{2} = 2k^{2}$,where $A$ and $B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$,respectively.

Find the coordinates of a point on the $y$-axis which is at a distance of $5\sqrt{2}$ from the point $P(3, -2, 5)$.

The distance between the points $(1, 3, 2)$ and $(2, 1, 3)$ is:

The distance of the point $P(a, b, c)$ from the $x$-axis is

The distance of the point $(a, b, c)$ from the $z$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module