Determine the x, y, z coordinates of point A

If F = {350i – 250j – 450k} and cable AB is 9 m long, determine the x, y, z coordinates of point A.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

We will first write where point A and point B is in Cartesian vector notation.

We can use our diagram to write the location of our points are like so:

$A:(-xi+yj+zk)$

(Our $i$ for point A is negative because point A lies on the negative x-axis side)

$B:(0i+0j+0k)$

We will now figure out the position vector, $r_{AB}$ which can be found by subtracting the corresponding components of A from B. (If this is unfamiliar to you, please read this detailed guide on expressing forces in Cartesian notation)

$r_{AB}\,=\,\left\{(0-(-x))i+(0-y)j+(0-z)k)\right\}$

$r_{AB}\,=\,\left\{xi-yj\,-\,zk\right\}$

From the question, we know the length of this rope is 9m. That means the magnitude of $r_{AB}\,=\,9$ m. The unit vector, denoted u, is each of $r_{AB}$ divided by the magnitude.

$u\,=\,\left(\dfrac{x}{9}i\,-\,\dfrac{y}{9}j\,-\,\dfrac{z}{9}k\right)$

We can also figure out the unit vector of F. Which is equal to:

$u\,=\,\dfrac{350i-250j-450k}{\sqrt{(350)^2+(-250)^2+(-450)^2}}$

$u\,=\,0.562i-0.401j-0.723k$

The key to this question is realizing that since force F is directed from point A to B, then both unit vectors must be equal. Therefore, we can write:

$\left(\dfrac{x}{9}i\,-\,\dfrac{y}{9}j\,-\,\dfrac{z}{9}k\right)\,=\,0.562i-0.401j-0.723k$

We can now solve for each term.

$\dfrac{x}{9}\,=\,0.562$

$x\,=\,5.058$ m

$-\dfrac{y}{9}\,=\,-0.401$

$y\,=\,3.609$ m

$-\dfrac{z}{9}\,=\,-0.723$

$z\,=\,6.507$ m

$x\,=\,5.058$ m
$y\,=\,3.609$ m
$z\,=\,6.507$ m