Question

How would you find the unit vector along the line joining point (2, 4, 4) to point (-3, 2, 2)?

Answer 1

`(1/sqrt33) ((-5), (-2), (-2))`

Explanation:

If point P is (2,4,4) and Q is (-3,2,2), the vector PQ would be (-5,-2,-2). To find the unit vector, divide vector PQ by its magnitude. `|| vec (PQ)||` would be `sqrt((-5)^2 +(-2)^2 + (-2)^2)) =sqrt33`. Hence unit vector would be
`(1/sqrt33) ((-5), (-2), (-2))`

Answer 2

Please see the explanation.

Explanation:

Given the points `(x_0, y_0, z_0) and (x_1, y_1, z_1)`

You make a vector in the direction from one point to another by subtracting each starting coordinate from its respective ending coordinate:

`barV = (x_1 - x_0)hati + (y_1 - y_0)hatj + (z_1 - z_0)hatk`

To make it a unit vector, `hatV`, you divide each component of the vector by the magnitude, `|barV|`:

`|barV| = sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)`

Therefore the general form for a unit vector is:

`hatV = (x_1 - x_0)/sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)hati + (y_1 - y_0)/sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)hatj + (z_1 - z_0)/sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)hatk`

For the given points `(2, 4, 4)` and `(-3, 2, 2)`

Compute the magnitude

`|barV| =sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2) =`

`sqrt((-3 - 2)^2 + (2 - 4)^2 + (2 - 4)^2) =`

`sqrt(33)`

We need to divide which it the same as multiplying by the reciprocal, `sqrt(33)/33`

Substitute into the unit vector general form:

`hatV = (-3 - 2)sqrt(33)/33hati + (2 - 4)sqrt(33)/33hatj + (2 - 4)sqrt(33)/33hatk`

Simplify:

`hatV = -(5sqrt(33))/33hati - (2sqrt(33))/33hatj - (2sqrt(33))/33hatk`