Shortest distance between a point and plane (vector calculus)

[Win_odd Dhamnekar]

Let \(\displaystyle Q = (x_0 , y_0 , z_0)\) be a point in \(\displaystyle R^3\) , and let P be a plane with normal form ax+ b y+ cz + d = 0 that does not contain Q. Then the distance D from Q to P is:

\(\displaystyle D=\frac{|ax_0 +by_0 +cz_0 +d|}{\sqrt{a^2+b^2+c^2}} \)

Now there is another equivalent formula which is as follows:

\(\displaystyle D= |(d-a)\cdot \hat{n}| \) or \(\displaystyle D=\frac{|(d-a)\cdot (b \times c)|}{|(b \times c)|} \) where d is any point not lying in the plane. \(\displaystyle \hat{n}\) is unit normal

to \(\displaystyle \vec{b},\vec{c}\)

If i assume b and c are free vectors in the plane, what is a in second formula using cross product of \(\displaystyle \vec {b}, \vec {c} \)? Is it a position vector?

 

[Win_odd Dhamnekar]

Let \(\displaystyle Q = (x_0 , y_0 , z_0)\) be a point in \(\displaystyle R^3\) , and let P be a plane with normal form ax+ b y+ cz + d = 0 that does not contain Q. Then the distance D from Q to P is:

\(\displaystyle D=\frac{|ax_0 +by_0 +cz_0 +d|}{\sqrt{a^2+b^2+c^2}} \)

Now there is another equivalent formula which is as follows:

\(\displaystyle D= |(d-a)\cdot \hat{n}| \) or \(\displaystyle D=\frac{|(d-a)\cdot (b \times c)|}{|(b \times c)|} \) where d is any point not lying in the plane. \(\displaystyle \hat{n}\) is unit normal

to \(\displaystyle \vec{b},\vec{c}\)

If i assume b and c are free vectors in the plane, what is a in second formula using cross product of \(\displaystyle \vec {b}, \vec {c} \)? Is it a position vector?

I got the answer to this question by doing some homework.