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If $ r = \langle x, y, z \rangle $ , $ a = \langl…

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Problem 53 Hard Difficulty

Use a scalar projection to show that the distance from a point $ P_1 (x_1, y_1) $ to the line $ ax + by + c = 0 $ is
$$ \frac{\mid ax_1 + by_1 + c \mid}{\sqrt{a^2 + b^2 }} $$
Use this formula to find the distance from the point $ (-2, 3) $ to the line $ 3x - 4y + 5 = 0 $

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Chapter 12

Vectors and the Geometry of Space

Section 3

The Dot Product

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Video Transcript

The problem is use a scatter projection to show that the distance from a point- p, 1 x, 1 by 1, to the line a x plus b y plus c is equal to 0. Is a x, 1 plus b y 1 plus c over square root of a square plus b square use this formula to find the distance from the point negative 23 to the line 3 x, minus 4 y plus 5 is equal to 0 point. So first look at this graph here. If we choose some .000 and the line x, plus, b y plus c is equal to 0 point, then the scale production, absolute value of the scale projection of the locker, p, 0, p, 1 and 2 m is equal to the distance from pan to this line. Here the work to end is some lector that are perpendicular to it. To the line x, plus, b y plus c is equal to 0 point now, in this case, we can choose in as a d, this vector is a savanal to the line at x, plus b y plus c is equal to 0. Then we half so vector p. 0. P. 1 is equal to x, 1 minus x, 0 y 1 minus y 0. Then the distance d is equal to the scattered projection of loco p 1 unto this 1. So this is equal to p 0 p, 1 dot. N over magnitude of n, so this is equal to x, 1 x, 1 minus x, 0 times a plus y 1 minus y 0 times b over root of a square plus b square. This is equal to absolute value x, 1 times a times x, 1 minus plus b times y 1 minus a times x, 0 plus e times y 0 over retivea square, a b square since the .00 y 0 is underlying a x plus b y plus c. Is equal to 0, so we have a x, 0 plus b y 0 plus c is equal to 0, so a x 0 plus b y 0 is equal to negative c. So this is equal to absolute value, a x 1 plus b y 1 plus c over relative, a square plus b squared. So this is a formula here then we use this formula to find the distance from the point negative 23, which is the line 3 x, minus 4 y plus 5, is equal to 0, so the distance is equal to negative 2 times a. So this is negative. 6 and 3 times negative 4 negative 12 and plus 5 lutie 3, squared plus negative 4 square to the answer, is 13 over 5.

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Problem

If $ r = \langle x, y, z \rangle $ , $ a = \langl…

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Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 12

Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Caleb Elmore

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Caleb Elmore

Joseph Lentino

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Caleb Elmore

Joseph Lentino

James Stewart

Calculus: Early Transcendentals