Find the distance between the point and the plane. (4, 3, 2) x - y + 5z = 8 Step 1 The distance between a plane and a point Q not in the plane is given by the following formula |v• n| ||n|| D where P is a point in the plane, n is normal to the plane, and v is the vector from P to Q. Let P = (8, 0, 0) and let Q = (4, 3, 2). Find v. v = (-4 V -4 3 V 3, 2 V 2) Step 2 Find an appropriate normal vector n. n = (1 |-1 , 5 5 ) |-1 Step 3 Use your results from Part 1 and Part 2 to find D, the distance between the given point and plane. D =

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