Find the distance from the point (0, 4, 3) to the plane 5x +1y +2z

distance =

Step-by-step explanation:

Without going through and deriving this (my blackboard doesn't have space and a textbook, teacher, or online tutorial is probably better for that), the equation for the distance from some point on the plane to some point in space is as follows:

•Note that the d in the equation should actually be D but doesn't allow me to do uppercase letters on mobile it seems... Anyway, d in the equation represents the value which the equation of the plane is equal to. You only gave the plane equation so I assumed d=0 in my calculations.

•x, y, and z all come from the point which you want to find the distance to. So (x,y,z)=(0,4,3) here.

•a, b, and c all come from the components of the vector normal to the surface. For planes, it simply somes to be the coefficients in front of each part. So (a,b,c)=(5,1,2) here.

With that you just need to plug the information in and simplify for your answer. My work is in the attachment, comment with questions or if anything seems off.

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )