Perpendicular distance from point to a plane

PirateFan308
Messages
91
Reaction score
0

Homework Statement


Find the perpendicular distance from the point (1,2,3) to the plane x-2y-z=1
One method: find the equation of the line throughout (1,2,3) perpendicular to the plane. Find the intersection of this line and the plane


The Attempt at a Solution


I know the vector (1,-2,-1) is a vector perpendicular to the plane
I'm not sure how to find the distance between where the vector intersects the
plane to the point.

I know what to do if the vector was a line, but I am confused as to how to change the vector into a line.
 
Physics news on Phys.org
I think the line for the vector (1,-2,-1) that goes through P (1,2,3) is:
x=1+t y=2-2t z=3-t

Is this correct?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top