Question

What are the coordinates of point B on AC Such that the ratio of AB to AC is 5:6

Answer 1

Answer:

$\vec B = \left(x_{A}+\frac{5}{6}\cdot (x_{C}-x_{A}), y_{A}+\frac{5}{6}\cdot (y_{C}-y_{A}), z_{A}+\frac{5}{6}\cdot (z_{C}-z_{A})\right)$

Step-by-step explanation:

Let suppose that A, B, and C have the following points with respect to origin in the Euclidean space:

$\vec A = (x_{A}, y_{A}, z_{A})$

$\vec B = (x_{B}, y_{B}, z_{B})$

$\vec C = (x_{C}, y_{C}, z_{C})$

Besides, let consider that locations of A and B are currently known. The ratio is:

$\frac{AB}{AC} = \frac{5}{6}$

$AB = \frac{5}{6} \cdot AC$

Vectorially speaking, expression can be rewritten in the following terms:

$\overrightarrow{AB} = \frac{5}{6}\cdot \overrightarrow{AC}$

$(x_{B}-x_{A}, y_{B}-y_{A}, z_{B}-z_{A}) = \frac{5}{6}\cdot (x_{C}-x_{A}, y_{C}-y_{A}, z_{C}-z_{A})$

Now, each side of the equation is summed vectorially by $\vec A$ and coordinates of point B are finally found:

$\vec B = \left(x_{A}+\frac{5}{6}\cdot (x_{C}-x_{A}), y_{A}+\frac{5}{6}\cdot (y_{C}-y_{A}), z_{A}+\frac{5}{6}\cdot (z_{C}-z_{A})\right)$