Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured parallel to a line
(x + 2)/3 = (2y + 3)/4 = (3z + 4)/5

The distance is:  

Step-by-step explanation:

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

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:

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

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:

Then we solve that equation for t, to get:

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

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

So we get:

homozygous. one is homozygous, while two is heterozygous.

me! what did you put for this question?

step-by-step explanation: