The coefficients of the plane's equation provide a normal vector for the plane: \(\vecs{n}= 1,−2,1 \). To find vector \(\vecd{QP}\), we need a point in the plane. Any point will work, so set \(y=z=0\) to see that point \(Q=(5,0,0)\) lies in the plane. Find the component form of the vector from \(Q\) to \(P\):. Now, let's try out more problems to master the process of writing equations of planes. Example 1 Find the vector form of the equation of a plane given that both points, A = ( − 4, 2, 6) and B = ( 2, − 1, 3), lie on the plane. We also know that the vector, n =< 4, 4, − 1 >, is perpendicular to the plane. Solution Recall that the vector.
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This is called the scalar equation of plane. Often this will be written as, ax+by +cz = d a x + b y + c z = d. where d = ax0 +by0 +cz0 d = a x 0 + b y 0 + c z 0. This second form is often how we are given equations of planes. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane.. Following Definition 63, the equation of the plane in standard form is. 2 (x-1) + (y-1)+z = 0. The plane is sketched in Figure 10.55. Figure 10.55: Sketching the plane in Example 10.6.1. We have just demonstrated the fact that any three non-collinear points define a plane.
