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Chapter 10: Q 44. (page 846)

Use your answers from Exercise 14 to find the angle between the indicated planes in Exercises 44 and 45.
$7 x 鈭� 3 y + 5 z = 6$ and $2 x + 3 y 鈭� z = 1$

Short Answer

Expert verified

The angle between two planes is $胃 = \frac{蟺}{2}$

Step by step solution

01

Given information

The two planes $7 x - 3 y + 5 z = 6$ and $2 x + 3 y - z = 1$

02

Calculation

The goal is to determine the angle between the two planes shown.

The angle formed by two planes can be calculated as follows:

胃 = \cos^{- 1} (\frac{N_{1} 路 N_{2}}{||N_{1}|| ||N_{2}||})

03

Calculation

The normal vector of the plane $7 x - 3 y + 5 z = 6$ is $N_{1} = 鉄� 7, - 3, 5 鉄�$ and the normal vector of the plane $2 x + 3 y - z = 1$ is $N_{2} = 鉄� 2, 3, - 1 鉄�$

The angle between the two planes is:

$胃 = \cos^{- 1} (\frac{鉄� 7, - 3, 5 鉄� 路鉄� 2, 3, - 1 鉄�}{鈥� (7, - 3, 5) 鈭� 鉄� 2, 3, - 1 鉄� 鈥�})$ (Substitution)

$= \cos^{- 1} (\frac{7 (2) - 3 (3) + 5 (- 1)}{\sqrt{49 + 9 + 25} \sqrt{4 + 9 + 1}})$ (Dot Product)

$= \cos^{- 1} (\frac{14 - 9 - 5}{\sqrt{83} \sqrt{14}})$ (Simplify)

$= \cos^{- 1} (\frac{0}{\sqrt{83} \sqrt{14}})$

$= \cos^{- 1} (0)$

$= \frac{蟺}{2}$ (Rationalize)

The angle between two planes is $胃 = \frac{蟺}{2}$

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Most popular questions from this chapter

Consider the sequence of sums $\frac{1}{3}, \frac{1}{3} + \frac{1}{9}, \frac{1}{3} + \frac{1}{9} + \frac{1}{27}, \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}, 鈥�$

(a) What happens to the terms of this sequence of sums as k gets larger and larger?

(b) Find a sufficiently large value of k which will guarantee that every term past the kth term of this sequence of sums is in the interval (0.49999, 0.5).

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$v = <\frac{1}{5}, \frac{1}{3}>$

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54

$f (x) = \frac{2}{x + 1}$

what it means, in terms of limits, for a function to have a removable discontinuity, a jump discontinuity, or an infinite discontinuity at x = c

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