How to Find a Point when Given the Equation of the Line it's on, Another Point on the Line and the distance between the Two Points?

Question

(12th Grade Calculus Level) Let's say I'm given point A(1,2,4) and a line [x,y,z] = [4,3,9] + t[3,1,5]. I have to find point B which is on the line and is a distance of 5 units away from Point A. What is Point B?

This question is a subpart for the question in the image attached(click on link) where three points are given of a triangular prism in R^3 with a side length. I have to find the other three points of the triangular prism and to do this I must find the line of intersection between two planes and then use this line, the side length and the given point to find the other points in the triangular prism. If you could answer the question above or tell me a more efficient solution for someone who's taking 12th Grade Calculus in the USA, I would be happy. Link to Image

Answer 1

The distance of any point on you line, as measured from point A, is given by the magnitude (norm) of the vector $\mathbf{d} = (x,y,z) - (1,2,4)$. Given the equation of the line you thus need to calculate the magnitude of the vector $\mathbf{d} =(4-1,3-2,9-4) + t (3,1,5) = (3,1,5) + t(3,1,5)$. The magnitude of $\mathbf{d}$ (which is positive) is given by

$d = \sqrt{((3+3t)^2 + (1+t)^2 + (5+5t)^2)} = \sqrt{35(1+t)^2}$.

You are seeking points that are $d=5$ away from point A, such that you must solve the equation

$d = 5 = \sqrt{35(1+t)^2}$,

which we rearrange:

$(1+t)^2 = \frac{5}{7}$,

$1+t = \pm \sqrt{\frac{5}{7}}$

and hence $ t = -1 \pm \sqrt{\frac{5}{7}}$

You can then take these possible values of $t$ and substitute them into the equation of your line to find the two points that are a distance 5 away from A i.e.

$(x_B,y_B,z_B) = (4,3,9) + \left(-1 \pm \sqrt{\frac{5}{7}}\right) (3,1,5)$ or equivalently $(x_B,y_B,z_B) = (1,2,4) \pm \sqrt{\frac{5}{7}} (3,1,5)$