Distance from a point to a plane.

[Gregg]

Homework Statement

Find the distance of the point (2,4,7) from the plane determined by the points (3,4,4) , (4,5,7) and (6,8,9).

2 common vectors for the plane

\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 1 \\<br /> 3<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c}<br /> 3 \\<br /> 4 \\<br /> 5<br /> \end{array}<br /> \right)

The normal to the plane is then

\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 1 \\<br /> 3<br /> \end{array}<br /> \right)\times \left(<br /> \begin{array}{c}<br /> 3 \\<br /> 4 \\<br /> 5<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> -7 \\<br /> 4 \\<br /> 1<br /> \end{array}<br /> \right)

The vector from the point to a known point on the plane dot product with the normal vector divided by |n| gives

\frac{\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> -3<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{c}<br /> -7 \\<br /> 4 \\<br /> 1<br /> \end{array}<br /> \right)}{\sqrt{7^2+4^2+1^2}}=\frac{10}{\sqrt{66}} = \frac{5\sqrt{66}}{33}

The answer has just 3 in the denominator, not 33. Have I done this correctly?

 

[Dick]

That looks fine to me. May be a typo in the answers.