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Chapter 10: Q 34 (page 789)

Find the equation of a sphere containing the point (1, 4, 7) and whose center is (-2, 3, 5).

Short Answer

Expert verified

The required equation is: ${(x + 2)}^{2} + {(y - 3)}^{2} + {(z - 5)}^{2} = 14$

Step by step solution

01

Given information

A sphere containing the point (1, 4, 7) and whose center is (-2, 3, 5).

02

Finding the radius

The radius will be the distance between the center and the point given.

Radius is:

$\sqrt{{(1 + 2)}^{2} + {(4 - 3)}^{2} + {(7 - 5)}^{2}}$

Simplify:

$\sqrt{14}$

03

Finding the equation

The equation of the sphere may be written as:

${(x + 2)}^{2} + {(y - 3)}^{2} + {(z - 5)}^{2} = {(\sqrt{14})}^{2}$

Simplify:

${(x + 2)}^{2} + {(y - 3)}^{2} + {(z - 5)}^{2} = 14$

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Most popular questions from this chapter

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

$u \times v and v \times u$

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The sum formulas in Theorem 4.4 can be applied only to sums whose starting index value is $k = 1$ .

(b) True or False: $\sum_{k = 0}^{n} \frac{1}{k + 1} + \sum_{k = 1}^{n} k^{2}$ is equal to $\sum_{k = 0}^{n} \frac{k^{3} + k^{2} + 1}{k + 1}$ .

(c) True or False: $\sum_{k = 1}^{n} \frac{1}{k + 1} + \sum_{k = 0}^{n} k^{2}$ is equal to $\sum_{k = 1}^{n} \frac{k^{3} + k^{2} + 1}{k + 1}$ .

(d) True or False: $(\sum_{k = 1}^{n} \frac{1}{k + 1}) (\sum_{k = 1}^{n} k^{2})$ is equal to $\sum_{k = 1}^{n} \frac{k^{2}}{k + 1}$ .

(e) True or False: $\sum_{k = 0}^{m} \sqrt{k} + \sum_{k = m}^{n} \sqrt{k}$ is equal to $\sum_{k = 0}^{n} \sqrt{k}$ .

(f) True or False: $\sum_{k = 0}^{n} a_{k} = - a_{0} - a_{n} + \sum_{k = 1}^{n - 1} a_{k}$ .

(g) True or False: ${(\sum_{k = 1}^{10} a_{k})}^{2} = \sum_{k = 1}^{10} a_{k}^{2}$ .

(h) True or False: $\sum_{k = 1}^{n} {(e^{x})}^{2} = \frac{e^{x} (e^{x} + 1) (2 e^{x} + 1)}{6}$ .

If u, v, and ware three mutually orthogonal vectors in $ℝ^{3}$ , explain why $u \times (v \times w) = 0$ .

In Exercises 24-27, find comp_uv, proj_uv, and the component of v orthogonal tou.

$u = <1, 2>, v = <3, 5>$

Use calculator graphs to make approximations for each of the limits in Exercises 67–74.

$\lim_{x \to 4} (3 - 4 x - 5 x^{2})$

See all solutions

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