Find the distance to the point (7,3,2) from the plane 2x+y+z=10?

Answer 1

For the following point and plane

P(u,v,w)
ax + by + cz + d = 0

The formula for the distance from the point to the plane is given by:

Distance = | au + bv + cw + d | / √(a² + b² + c²)

For the point and plane in question we have
P(7,3,2)
2x + y + z - 10 = 0
Note that the equation has to equal zero so I subtracted 10 from both sides of the equation.

Distance = | 2*7 + 1*3 + 1*2 - 10 | / √(2² + 1² + 1²)
Distance = | 14 + 3 + 2 - 10 | / √(4 + 1 + 1)
Distance = 9 / √6 = 9√6 / 6 = 3√6 / 2

Answer 2

◙ first let us take P=(7,3,2) as a new origin, now equation of the plane looks:
2(x+7) +(y+3) +(z+2) =10; or; 2x +y +z=-9;
♠ now we find distance D of the plane from the new origin normalizing new equation: |N| = sqrt(2^2 +1^2 +1^2) = sqrt(6); thus equation is:
♣ 2x/sqrt6 +y/sqrt6 +z/sqrt6 =-9/sqrt6, hence D=-9/sqrt6; minus shows that P is on the other side of the plane with respect to the old origin.

Answer 3

2(7)+3+2=10
14+3+2=10
19=10
1.9