Question

Find the distance of (3, -2, 1) from the plane 2x−y+2z+3=0.

Answer 1

Answer:

$\frac{13}{3}$

Step-by-step explanation:

Distance =

$\frac{ax + \: by + cz + d}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } }$

Put the values...

$\frac{3*2 + (-2*-1) + \: 1*2 \: + \: 3}{ \sqrt{ {2}^{2} + {1}^{2} + {2}^{2} } }$

$\frac{13}{3}$

Here's Your Answer

Answer 2

Answer:

13/3 = Distance

Hope it helps you mate