![]() The two planes may intersect in a line, or they may be parallel or even the same plane. Conversely, if we have two such equations, we have two planes. X = k, then the solution set of both equations togeteher is the line. This intuitive idea is made rigorous in all dimensions in a linear algebra course.Ī line in space cannot be given by one linear equation, since for any nonzero vector A, such an equation has a plane as a solution.īut a line is the intersection of two planes, so if we have two such planes, with two equations A. In each case, we can motivate this informally by saying that the space of solutions has dimension one less than the dimension of the containing space. X = h defines a line in the plane or a plane in 3-space. We have seen that one equation of the form A. Also, find a unit normal vector for line OA. How is this related to the example? Could you use the example to find the unit normal in this case?Įxercise on Lines in the Plane: Continuing with line m with equation 2x + 3y = 6, find the unit normal vectors for this line. Unit normal vectors: (1/3, 2/3, 2/3) and (-1/3, -2/3, -2/3)Įxercise: Find a unit normal vector for the plane with equation -2x -4y -4z = 0. ![]() ![]() Thus the vector (1/3)A is a unit normal vector for this plane. Thus for a plane (or a line), a normal vector can be divided by its length to get a unit normal vector.Įxample: For the equation, x + 2y + 2z = 9, the vector A = (1, 2, 2) is a normal vector. Any nonzero vector can be divided by its length to form a unit vector. X = 11.Ī unit vector is a vector of length 1. But since P is on the plane, if we set X = P, we must get the correct value of d. Solution: The equation must be (1, 2, 3). Find the equation of the plane through P = (1, -1, 4) with normal vector A. What is a normal vector for this line?Įxample: Finding a plane when the normal is known. How is this line related to m and OA? Finally, find an equation for line OA. Check that lines OA and m are perpendicular. On graph paper plot the line m with equation 2x + 3y = 6 and also plot the point A = (2,3). This also means that vector OA is orthogonal to the plane, so the line OA is perpendicular to the plane.Ĭareful: It is NOT true that for any point P in the plane, A is orthogonal to P (unless d = 0).Įxercise: Show that if A is a normal vector to a plane, and k is a nonzero constant, then kA is also a normal vector to the same plane.ĭebate: For any plane, is the 0 vector orthogonal to all the direction vectors of the plane?Įxercise on Lines in the Plane: The same reasoning works for lines. Thus the coefficient vector A is a normal vector to the plane. This means that vector A is orthogonal to the plane, meaning A is orthogonal to every direction vector of the plane.Ī nonzero vector that is orthogonal to direction vectors of the plane is called a normal vector to the plane. This means that the vector A is orthogonal to any vector PQ between points P and Q of the plane.īut the vector PQ can be thought of as a tangent vector or direction vector of the plane. If P and Q are in the plane with equation A. The equation of a line in the form ax + by + cz = d can be written as a dot product: The equation of a line in the form ax + by = c can be written as a dot product: Dot Product and Normals to Lines and Planes Dot Product and Normals to Lines and Planes
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