Find the distance between the points P (2, -1, 3) and Q (-5, 2, 1)?
Given:
Point 1 = (2, -1, 3)
Point 2 = (-5, 2, 1)
Formula:
The distance between two points with coordinates (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is given by:
\(d = \sqrt{[(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2]}\)
Solution:
d = √[(2 - (-5))2 + ((-1) - 2)2 + (3 - 1)2 ]
= √(49 + 9 + 4)
= √62 units
Important Questions from Distance between points
Determine the shortest distance between the lines l1 and l2 whose vector equations are
\(\vec{r}=2 \hat{\imath}+\hat{\jmath}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \quad \ldots(1)\)
\(\vec{r}=3 \hat{\imath}+\hat{\jmath}-\hat{k}+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}) \quad \ldots(2)\)
The sum of distances from origin to (0, 5, 5) and (5, 8, 6) is:
The distance between the points (2, 3) and (4, 1) is.
The vertices of a triangle are \(A(2, 0, 0)\), \(B(0, 6, 0)\) and \(C(0, 0, 4)\). If \(AD\), \(BE\) and \(CF\) are the medians of the triangle, then what is \(AD^2 + BE^2 + CF^2\) equal to?
The distance of a point from the x-axis is called that point.
