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Chapter 9: Problem 21

Find the distance between the points. \((1,-2,4), \quad(6,-2,-2)\)

Short Answer

Expert verified
The distance between the two points is \(\sqrt{61}\).

Step by step solution

01

Identify the coordinates

The first point has coordinates \(x_1=1\), \(y_1=-2\), and \(z_1=4\). The second point has coordinates \(x_2=6\), \(y_2=-2\), and \(z_2=-2\).
02

Implement the distance formula

Substitute the related coordinates in the distance formula: \(\sqrt{(6 - 1)^2 + (-2 - (-2))^2 + (-2 - 4)^2}\).
03

Simplify the expression

The expression simplifies to \(\sqrt{(5)^2 + (0)^2 + (-6)^2} = \sqrt{25 + 0 + 36}\).
04

Calculate the result

\(\sqrt{61}\) is the distance between the two points.

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Most popular questions from this chapter

In Exercises \(41-44,\) find the component form and magnitude of the vector \(u\) with the given initial and terminal points. Then find a unit vector in the direction of \(\mathbf{u}\). \(\frac{\text { Initial Point }}{(3,2,0)}\) \(\frac{\text { Terminal Point }}{(4,1,6)}\)

(a) find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), and (b) find the vector component of u orthogonal to v. $$ \mathbf{u}=\langle 1,0,4\rangle, \quad \mathbf{v}=\langle 3,0,2\rangle $$

Use vectors to prove that a parallelogram is a rectangle if and only if its diagonals are equal in length.

Find the component of \(u\) that is orthogonal to \(\mathbf{v},\) given \(\mathbf{w}_{\mathbf{1}}=\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). $$ \mathbf{u}=\langle 8,2,0\rangle, \quad \mathbf{v}=\langle 2,1,-1\rangle, \quad \operatorname{proj}_{\mathbf{v}} \mathbf{u}=\langle 6,3,-3\rangle $$

Use vectors to determine whether the points are collinear. (0,0,0),(1,3,-2),(2,-6,4)
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