vector equation of a plane

form F(s,t) = (1 - s - t)P + sQ + tR, where s and t range over all real numbers. So, let’s start by assuming that we know a point that is on the plane, \({P_0} = \left( {{x_0},{y_0},{z_0}} \right)\). Now, we know that the cross product of two vectors will be orthogonal to both of these vectors. Line through (3, 4) and (3, 7). So, if the two vectors are parallel the line and plane will be orthogonal. So, the vectors aren’t parallel and so the plane and the line are not orthogonal. be converted to slope-intercept form by solving for y: except for the special case b = 0, when the line is parallel to the y-axis. This second form is often how we are given equations of planes. A normal vector is. + d(2/5)z = d, so one choice of constant gives, or another choice would be (1/5)x + (2/5)y + (2/5)z = 1. has cleared the denominators. Let’s also suppose that we have a vector that is orthogonal (perpendicular) to the plane, \(\vec n = \left\langle {a,b,c} \right\rangle \). In the first section of this chapter we saw a couple of equations of planes. VECTOR EQUATIONS OF A PLANE. This in effect uses x as a parameter and writes y as a function of x: y = f(x) to the third, we eliminate a to get. get the equation kax + kby + kcz = kd, the plane of solutions is the same. 1, -1), and (1, -1, -1)? with the y-axis. What is equation of the plane through the points I, J, K? Then since the points are on the line, we know that both. axes? Another choice might be c = 3: x+y = 3, which This can easily or a + 2b = c. Both, Vector and Cartesian equations of a plane in normal form are covered and explained in simple terms for your understanding. Exercise: What is the equation of a line through (0,0) of line PQ. This second form is often how we are given equations of planes. using dot product. Note the special cases. Compare this explicit computation with the computation given plane. 4x + 6y = 8 It is completely possible that the normal vector does not touch the plane in any way. The two vectors aren’t orthogonal and so the line and plane aren’t parallel. Start with the first form of the vector equation and write down a vector for the difference. We need to find a normal vector. (b) or a point on the plane and two vectors coplanar with the plane. If \(\vec n\) and \(\vec v\) are parallel, then \(\vec v\) is orthogonal to the plane, but \(\vec v\) is also parallel to the line. and \(P\) respectively. The equation can be rearranged like this: Another useful choice, when d is not zero, is to divide by d so that the constant Given points P, Q, R in space, find the equation of the plane through the 3 the plane that uses determinants or cross product. Notice as well that there are many possible vectors to use here, we just chose two of the possibilities. Exercise: What is special about the equation of a plane that passes + t(aq1 + bq2), and this equals (1-t)c + tc = c. So the be explained in the Normal Vector section. Thinking of a line as a geometrical object and not the graph of a function, This is \(v = \left\langle {0, - 1,4} \right\rangle \). This vector is called the normal vector. This is \(\vec n = \left\langle { - 1,0,2} \right\rangle \). 2x + 3 y = 4 Let $\vec{n} = (a, b, c)$ be a normal vector to our plane $\Pi$, that is $\Pi \perp \vec{n}$.Instead of using just a single point from the plane, we will instead take a vector that is parallel from the plane. This method always works for any distinct P and Q. Since Q is on the line, its coordinates satisfy the equation: a(-2) + b5 = c, of a plane (when at least one of a, b, c is not zero. a line (normal form) is. Equation of a Plane. Exercises: Find the equations of these lines. picture. -a + 2b + c = d, Subtracting the first equation from the second and then adding the first equation q2). In order to write down the equation of plane we need a point (we’ve got three so we’re cool there) and a normal vector. Therefore, we can use the cross product as the normal vector. Now, assume that\(P = \left( {x,y,z} \right)\) is any point in the plane. Q, R all satisfy the same equation ax + by + cz = d, then all the points F(s,t) Another useful form of the equation is to divide by |(a,b)|, ax + by + cz = d, where at least one of the numbers a, b, c must be nonzero. This is called the scalar equation of plane. Exercise: Compare this method of finding the equation of a plane with the cross-product In particular it’s orthogonal to \(\vec r - \overrightarrow {{r_0}} \). equation holds. coordinate axes. of an equation ax + by + cz = d, where P, Q and R satisfy the equations, thus: a + b + c = d for a, b, c also. When x = 0, y = b and the point (0,b) is the intersection of the line Example: For P = (1, 2), Q = (-2, 5), find the equation ax + by = c Line through (3, 4) and (1, -2). Exercise: What is the equation of the plane through (1, 1, 1), (-1, If c is not zero, it is often useful to think of the plane as the graph of equation for the line. If the line is parallel to the plane then any vector parallel to the line will be orthogonal to the normal vector of the plane. + tQ, for any choice of t. Here is this computation. term = 1. Consider an arbitrary plane. + tQ, for t ranging over all the real numbers. This can be found expressed by determinants, or the Now, because \(\vec n\) is orthogonal to the plane, it’s also orthogonal to any vector that lies in the plane. For any two points P and Q, there is exactly one line PQ through the points. through 0. Then the equation becomes. with the stipulation that at least one of a or b is nonzero. Often this will be written as, \[ax + by + cz = d\] where \(d = a{x_0} + b{y_0} + c{z_0}\). This is not as difficult a problem as it may at first appear to be. line, each of which is multiple of the other. Exercise: If O is on the line, show that the equation becomes As for the line, if the equation is multiplied by any nonzero constant k to get the equation kax + kby + kcz = kd, the plane of solutions is the same. A slightly more useful form of the equations is as follows. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. Remember that there are an infinite number of equations for the the set of solutions is exactly the same, so, for example, all these equations If you think about it this makes some sense. root of a2 + b2 + c2. We put it here to illustrate the point. We can also get a vector that is parallel to the line. We would like a more general equation for planes. Given two points P and Q, the points of line PQ can be written as F(t) = (1-t)P 4b + 5b = 9b = 2c + c = 3c, so b = (1/3)c. Then substituting into the first Adding the equations gives 5b = 2d, or b = (2/5)d, then solving for c = b = also satisfy the same equation. = c. But the left side can be rearranged as (1-t)(ap1 + bp2) For 3 points P, Q, R, the points of the plane can all be written in the parametric = mx+b. (1/2)x + (3/4)y = 1. Exercise. This means an equation in x and y whose solution set is a line in the (x,y) Exercise: Where does the plane ax + by + cz = d intersect the coordinate How do you think that the equation of this plane can be specified? In general, if k is a nonzero constant, then these are equations for the shows more detail and one hides the coordinates and shows a more conceptual Since P is on the line, its coordinates satisfy the equation: a1 + b2 = c, Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. In other words. a function z of x and y. (1/c). ax + by = 0, or y = mx. The most popular form in algebra is the "slope-intercept" form. a + 2b + 0c = d For the point F(t), we must check a[(1-t)p1+tq1] + b[(1-t)p2+tq2] Now, if these two vectors are parallel then the line and the plane will be orthogonal. This choice will be explained