How would you find the minimum distance from the origin to the plane 4x + 12y + 3z = 52 .?

Question

possible solutions....



a. 1

b. 2

c. 3

d. 4

e. 5

f. 6

or is it none of these?

Answer 1

A normal from the plane to the origin has a distance given by
d = |{4(0) + 12(0) + 3(0) - 52}/sqr[4^2 + 12^2 + 3^2]|
= 52/sqr[16 + 144 + 9] = 52/13 =4
Answer d or 4

Answer 2

How would you find the minimum distance from the origin to the plane 4x + 12y + 3z = 52?

Rewrite the equation to set it equal to zero. Then apply the distance formula from a point (the origin) to a plane.

4x + 12y + 3z - 52 = 0

Distance = | 4*0 + 12*0 + 3*0 - 52 | / √(4² + 12² + 3²)

= 52 / √(16 + 144 + 9) = 52 / √169 = 52/13 = 4

The answer is d.
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