Question

How do you determine a distance between P(3,-2,5) and the plane 2x+4y-z=2?

Answer 1

distance of the point ( x1,y1,z1) from the plane ax+by+cz +d= 0 is given by |(ax1+by1+cz1 +d)/sqrt(a^2 +b^2 +c^2))|

Answer 2

thus the reqd distance here is
| 2*3 +4(-2) - 5 -2) / sqrt(2^+4^2 +(-1)^2) |
=| -9 /sqrt(21) |
=| 9 /sqrt(21) |

Answer 3

Thank you! I completely forgot the formula... Thanks for answering :)

Answer 4

welcome dear...