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Vector Problem

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Hi guys, I have come across with some confusion of how to solve a question that asks you find the shortest distance from a point to a plane. 

For example, find the shortest distance from the point, P (1,2,-3) to the plane 3x +y -3z = -9. 

I've seen two methods to solve this question and they yield me with different answers. 

Method 1: dot multiplies the point and the normal vector and then divide by the magnitude of the normal vector. 

Method 2: first find the shorted line equation which is (1,2,-3) + landa (3,1,-3), and then substitute the x, y, and z values in the normal vector equation to find landa. Then multiply landa with the normal, and find its magnitude, which is the shortest distance. 

Can somebody please help me to clarify which method is correct? It would be great if you can also solve the example question too so I can follow the steps. Thanks!!

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First method is wrong as stated, because it does not take consideration of the D in Ax + By + Cz = D. To use projection, you need another point Q on the plane, find direction vector QP and then proceed via method 1. Second method is completely right.

First method: Use any point on plane such as Q(0, 0, 3). Find QP = (1,2,-3)-(0,0,3) = (1,2,-6). Find QP • n/|n| = (1,2,-6)•(3,1,-3) / sqrt(3^2 + 1^2+(-3)^2) = (3 + 2 + 18) / sqrt(19) = 23 / sqrt(19)

Second method. x = 1 + 3t, y = 2 + t, z = -3 - 3t (parametric form of a line, using t as the parameter). 3(1+3t) + (2+t) + -3(-3-3t) = -9 (set line and plane same x,y,z). 
t = -23/19. The point at which the perpendicular line intersect the plane is (1,2,-3) + (-23/19)•(3,1,-3), the difference between this point and P is (-23/19)(3,1,-3)
(-23/19)(3, 1, -3) has magnitude 23/19 * sqrt (3^2 + 1^2 + (-3)^2) = 23/19 * sqrt(19) = 23/sqrt(19)
As a bonus, you can prove to yourself that when t not simplified is t = a/b, then the distance is just a/sqrt(b). So if you get t = 40/20, distance would be 40/sqrt(20). This is a quick way to check answer.

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