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

B

=(x

A

+

6

5

⋅(x

C

−x

A

),y

A

+

6

5

⋅(y

C

−y

A

),z

A

+

6

5

⋅(z

C

−z

A

))

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})

A

=(x

A

,y

A

,z

A

)

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

B

=(x

B

,y

B

,z

B

)

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

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}

AC

AB

=

6

5

AB = \frac{5}{6} \cdot ACAB=

6

5

⋅AC

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

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

AB

=

6

5

⋅

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})(x

B

−x

A

,y

B

−y

A

,z

B

−z

A

)=

6

5

⋅(x

C

−x

A

,y

C

−y

A

,z

C

−z

A

)

Now, each side of the equation is summed vectorially by \vec A

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)

B

=(x

A

+

6

5

⋅(x

C

−x

A

),y

A

+

6

5

⋅(y

C

−y

A

),z

A

+

6

5

⋅(z

C

−z

A

))

kkkll

step-by-step explanation:

kljlklj

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: