Without using distance formula, find the distance from the point (0,0,1) to the plane $\displaystyle \;4y\;+\;3z\;=\;-12\;$
i dont have any idea about how to solve. :S
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Without using distance formula, find the distance from the point (0,0,1) to the plane $\displaystyle \;4y\;+\;3z\;=\;-12\;$
i dont have any idea about how to solve. :S
$\displaystyle (0,0,-4)$ is clearly on the plane and $\displaystyle (0,4/5,3/5)$ is the unit norm of the plane.Quote:
Originally Posted by Lafexlos
So $\displaystyle |((0,0,1)-(0,0,-4)) \cdot (0,4/5,3/5)| = 3$
is the length of the projection of the segment from $\displaystyle (0,0,-4)$ to $\displaystyle (0,0,1)$ onto the norm of the plane,
thus is the distance between $\displaystyle (0,0,1)$ to the plane.
Almost i get it but why are you dot producted with normal vector?
Edit: Ok. i get it now. thx for help. =)
