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Chapter 10: Q 45. (page 846)

Use your answers from Exercise 14 to find the angle between the indicated planes in Exercises 44 and 45.
$鈭� x + 7 y 鈭� 2 z = 5$ and $3 x + 5 y 鈭� 4 z = 2$

Short Answer

Expert verified

The angle between two planes is $胃 = \cos^{- 1} (\frac{4 \sqrt{3}}{9})$

Step by step solution

01

Given information

The two planes $- x + 7 y - 2 z = 5$ and $3 x + 5 y - 4 z = 2$

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 $- x + 7 y - 2 z = 5$ is $N_{1} = 鉄� - 1, 7, - 2 鉄�$ and the normal vector of the plane $3 x + 5 y - 4 z = 2$ is $N_{2} = 鉄� 3, 5, - 4 鉄�$

The angle between the two planes is:

$胃 = \cos^{- 1} (\frac{鉄� - 1, 7, - 2 鉄� 路鉄� 3, 5, - 4 鉄�}{鈥� (- 1, 7, - 2 鉄� 鈥� 鈥� (3, 5, - 4 鉄� 鈥�})$ (Substitution)

$= \cos^{- 1} (\frac{- 1 (3) + 7 (5) - 2 (- 4)}{\sqrt{1 + 49 + 4} \sqrt{9 + 25 + 16}}) (Dot Product)$

$= \cos^{- 1} (\frac{40}{\sqrt{54} \sqrt{50}})$ (Simplify)

$= \cos^{- 1} (\frac{40}{3 \sqrt{6} 脳 5 \sqrt{2}})$

$= \cos^{- 1} (\frac{4}{3 \sqrt{3}})$

$= \cos^{- 1} (\frac{4 \sqrt{3}}{9})$ (Rationalize)

The angle between two planes is $胃 = \cos^{- 1} (\frac{4 \sqrt{3}}{9})$

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Most popular questions from this chapter

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

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

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

localid="1649405204459" role="math" $(u 脳 v) 路 w and u 路 (v 脳 w)$

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

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

Find a unit vector in the direction opposite to $<3, 4, 5>$ .

Find the norm of the vector. $v = (0, 0, 6)$

See all solutions

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