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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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Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 3

The Dot Product

Vectors

Campbell University

Oregon State University

Baylor University

Boston College

Lectures

02:56

In mathematics, a vector (from the Latin word "vehere" meaning "to carry") is a geometric entity that has magnitude (or length) and direction. Vectors can be added to other vectors according to vector algebra. Vectors play an important role in physics, engineering, and mathematics.

11:08

In mathematics, a vector (from the Latin word "vehere" which means "to carry") is a geometric object that has a magnitude (or length) and direction. A vector can be thought of as an arrow in Euclidean space, drawn from the origin of the space to a point, and denoted by a letter. The magnitude of the vector is the distance from the origin to the point, and the direction is the angle between the direction of the vector and the axis, measured counterclockwise.

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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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