Question

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured parallel to a line

Answer 1

Answer:

The distance is:  

$\displaystyle\frac{3\sqrt{142}}{10}$

Step-by-step explanation:

We re-write the equation of the line in the format:

$\displaystyle\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}$

Notice we divided the fraction of y by 2/2, and the fraction of z by 3/3.

In that equation, the director vector of the line is built with the denominators of the equation of the line, thus:

$\displaystyle\vec{v}=\left< 3, 2, \frac{5}{3}\right>$

Then the parametric equations of the line along that vector and passing through the point (-2, 3, -4) are:

$x=-2+3t\\y=3+2t\\\displaystyle z=-4+\frac{5}{3}t$

We plug them into the equation of the plane to get the intersection of that line and the plane, since that intersection is the image on the plane of the point (-2, 3, -4)  parallel to the given line:

$\displaystyle x+y+z=3\to -2+3t+3+2t-4+\frac{5}{3}t=3$

Then we solve that equation for t, to get:

$\displaystyle \frac{20}{3}t-3=3\to t=\frac{9}{10}$

Then plugging that value of t into the parametric equations of the line we get the coordinates of the intersection:

$\displaystyle x=-2+3\left(\frac{9}{10}\right)=\frac{7}{10}\\\displaystyle y=3+2\left(\frac{9}{10}\right)=\frac{24}{5} \\\displaystyle z=-4+\frac{5}{3}\left(\frac{9}{10}\right)=-\frac{5}{2}$

Then to find the distance we just use the distance formula:

$\displaystyle d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

So we get:

$\displaystyle d=\sqrt{\left(-2-\frac{7}{10}\right)^2+\left(3-\frac{24}{5}\right)^2+\left(-4 +\frac{5}{2}\right)^2}=\frac{3\sqrt{142}}{10}$