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Chapter 10: Q 46. (page 801)

Find a vector in the direction of $<9, 0, - 6>$ and with magnitude 7.

Short Answer

Expert verified

The required vector is $\frac{7}{\sqrt{117}} <9, 0, - 6>$ .

Step by step solution

01

Step 1. Given information.

The given vector is:

$v = <9, 0, - 6>$ with magnitude 7.

02

Step 2. Find the norm of the vector.

$v = <9, 0, - 6> ||v|| = \sqrt{9^{2} + 0^{2} + {(- 6)}^{2}} = \sqrt{81 + 0 + 36} = \sqrt{117}$

Therefore, the norm of the given vector is $\sqrt{117}$ .

03

Step 3. Find the vector.

We know that the vector in the direction of vis $\frac{a}{||v||} v$ . Here ais the magnitude of the given vector.

The required vector is:

$\frac{a}{||v||} v = \frac{7}{\sqrt{117}} <9, 0, - 6>$

Therefore the vector in the direction of $v = <9, 0, - 6>$ and with a magnitude of 7 islocalid="1649504967884" $\frac{7}{\sqrt{117}} <9, 0, - 6>$ .

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Most popular questions from this chapter

Find $u + v a n d u - v$ also sketch $u, v, u + v, u - v$

role="math" localid="1649595165778" $u = (1, - 2) a n d v = (- 3, 6)$

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: ${鈭�}_{k = 0}^{n} 鈥� \frac{1}{k + 1} + {鈭�}_{k = 1}^{n} 鈥� k^{2}$ is equal to ${鈭�}_{k = 0}^{n} 鈥� \frac{k^{3} + k^{2} + 1}{k + 1}$ .

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

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

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

(f) True or False: ${鈭�}_{k = 0}^{n} 鈥� a_{k} = 鈭� a_{0} 鈭� a_{n} + {鈭�}_{k = 1}^{n 鈭� 1} 鈥� a_{k}$ .

(g) True or False: ${({鈭�}_{k = 1}^{10} 鈥� a_{k})}^{2} = {鈭�}_{k = 1}^{10} 鈥� a_{k}^{2}$ .

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

If u and v are vectors in ${鈩�}^{3}$ such that $u 路 v = 0$ and $u 脳 v = 0$ , what can we conclude about u and v?

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

Find the volume of the parallelepiped determined by vectors u, v and w. Do u, v and w form a right-handed triple?

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

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