Finding the 3d vector coordinates

Question

I have a problem with calculating the coordinates of a $3$ dimensional vector. In my coursebook there is given a picture of a pyramid, which you don't have to see for answering my question.

So there is the point $T = (3,3,8)$ and $B = (6,6,0)$. Together they form a line. Now there is the point $P$ such, that $BP=\frac{1}{4}BT$.

How can I find the coordinates of point $P$? I do know that you can just take the average when you want to find a point right in the middle of $2$ other points. But here I tried this:

$\frac{3}{4} \big( (3,3,8) + (6,6,0) \big)= \frac{3}{4} (9,9,8) = (6.75,6.75, 6)$. The answer book says that point $P$ is $(5.25,5.25,2)$.

Anyone knows how to calculate the point $P$?

Answer 1

Add $\frac34$ of the difference between $B$ and $T$ to $T$. So $P$ has coordinates $$(3,3,8)+\frac34((6,6,0)-(3,3,8))=(5.25,5.25,2).$$ In terms of vectors, $\overrightarrow{OP}=\overrightarrow{OT}+\frac{3}{4}\overrightarrow{TB}=\overrightarrow{OT}+\frac{3}{4}(\overrightarrow{OB}-\overrightarrow{OT})$.

Answer 2

Since $BP$ is $\frac{1}{4}$ BT, we can take the middle point of $BT$ and then take the middle point of our result to the edges like this:

$(\frac{6+3}{2},\frac{6+3}{2},\frac{0+8}{2})=(4.5,4.5,4)$

Then take the middle point again:

$(\frac{6+4.5}{2},\frac{6+4.5}{2},\frac{4+0}{2})=(5.25,5.25,2)$

This works since you basically divided the line segment into two and then divided the second section in half.

Answer 3

To get the coordinates of the vector from $B$ to $T$, you subtract the coordinates of $B$ from the coordinates of $T$: $$ T - B = (3,3,8) - (6,6,0) = {}? $$

If you consider the coordinates of the points to be the coordinates of position vectors, this is all perfectly ordinary vector arithmetic.

To get the coordinates of the vector from $B$ to $P$ you would subtract coordinates again, except we don’t know what the coordinates of $P$ are yet so this is as far as we can go: $$ P - B = {}? $$

But you know $BP= \frac14 BT.$ So if $P$ is in the same direction from $B$ as $T$ is, the two vectors whose formulas are written above are related in the same way: $$ P - B = \frac14 (T - B). $$

You can add $B$ to both sides: $$ P = B + \frac14 (T - B). $$

And now you just need to plug in the known values on the right.