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Chapter 10: Q 27 (page 789)

Find the distance between (-1, 3, 5, 2, 0) and (0, 6, 1,-2, 3).

Short Answer

Expert verified

The required distance is $\sqrt{51}$

Step by step solution

01

Given information

The given points are (-1, 3, 5, 2, 0) and (0, 6, 1,-2, 3).

02

Finding the distance

The distance may be found as under:

$\sqrt{(0 - {(- 1))}^{2} + {(6 - 3)}^{2} + {(1 - 5)}^{2} + {(- 2 - 2)}^{2} + {(3 - 0)}^{2}}$

Simplifying:

$\sqrt{1 + 9 + 16 + 16 + 9}$

Further simplify:

$\sqrt{51}$

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Most popular questions from this chapter

In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.

$v = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$

If u, v and w are three vectors in $ℝ^{3}$ , which of the following products make sense and which do not?

localid="1649346164463" $(a) u \cdot (v \cdot w) (b) u \cdot (v \times w) (c) u \times (v \cdot w) (d) u \times (v \times w)$

If u and v are nonzero vectors in $ℝ^{3}$ , why do the equations role="math" localid="1649263352081" $u \cdot (u \times v) = 0$ and $v \cdot (u \times v) = 0$ tell us that the cross product is orthogonal to both u and v?

Find the norm of the vector. $v = (\frac{1}{3}, \frac{1}{3}, - \frac{1}{3})$

In Exercises 22–29 compute the indicated quantities when $u = (2, 1, - 3), v = (4, 0, 1), and w = (- 2, 6, 5)$

role="math" localid="1649400253452" $(u \times v) \times w and u \times (v \times w)$

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