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Find an equation of the plane.

The plane that passes through the point $ (3, 5, -1) $ and contains the line $ x = 4 - t , y = 2t - 1 , z = -3t $

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$$8 x+y-2 z=31$$

03:05

Wen Zheng

Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 5

Equations of Lines and Planes

Vectors

Johns Hopkins University

Missouri State University

Campbell University

Idaho State University

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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Find an equation of the pl…

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Find the equation of the p…

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in the question they're asking to find the equation of the plane that passes to a 0.35 minus one and contains the line having coordinates X equal to four minus t. Why equal to two? T minus one is equal to minus treaty. Mhm. So in order to find the equation of the plane first, we have to find the two points in the plane given a point in the plane. That is Let's suppose the point is PX coordinates are 35 -1. And from the coordinates of the lying that is containing in the plane we can find another point in the plane and let us name the point is Q. And its coordinates would be Mhm. The constant terms In the three coordinates that is four minus one and zero. Mhm. And in order to find yeah yeah. In order to find the direction vector of the line containing the points P and Q. We have to find Yeah. The direction vector as follows, where PQ vector is equal to four minus three, icap Plus -1 -5. Jacob plus zero minus minus one. Kick up. Therefore this is equal to Icap minus six. Jacob last kick up and the normal vector to the plane is equal to the coefficient of the Variables in the three coordinates That is -1, two and -3. Mhm. This can be written in victor, former minus icap plus to Jacob -3 K Cup. No this is the direction vector of the points containing in the plane and this is the normal vector of the plane. In order to find though equation of the plane. You have to find across productive, speak you victor and in victor that is equal to Yeah he'll be Icap, Jacob and Jacob and the values of PQ vectors are 1 -6, 1 and in Vectors AR -1: -3. Therefore calculating the cross product from the values are the two victims. We get 16 I gap plus to Jacob -4 K Cup. This is the cross product of directors in the form of victor. And in order to find the equation we have to find this vector in terms of X, Y and Z coordinates. So We have to write the equation at 16 x minus any point that is containing in the plane. Let's suppose P. That is given in the question 3, 5 and -1 So X -3 plus two. Why minus five minus four zero plus one is equal to zero. This is the required equation of the plane. So after solving this equation redirect the answer as eight x less way minus two. That Is equal to 31. And this is the required answer of the given question

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