Author Topic: Division by Ratios Vectors (Read 1432 times) Tweet Share

squance · « **on:** March 16, 2008, 06:22:09 pm »

um im stuck on this question....

Let ABC be a triangle and let D divide the segment BC in the ratio 3:5. Express the vector AD as a combination of vector AB and vector AC.

Im don't know how to solve it.....
The answer is 1/8 (5b + 3c) where b = vector AB and c = vector AC

can someone please give me some pointers???

Collin Li · « **Reply #1 on:** March 16, 2008, 06:25:29 pm »

$\vec{BD} = \frac{3}{8}\vec{BC}$

$\vec{BC} = \vec{AC} - \vec{AB}$

$\implies \vec{BD} = \frac{3}{8}(\vec{AC} - \vec{AB})$

So,

AD = AB + BD (for some reason this line isn't showing up in LaTeX)
$\vec{AD} = \vec{AB} + \vec{BD}$

$= \vec{AB} + \frac{3}{8}(\vec{AC} - \vec{AB})$

$= \frac{3}{8}\vec{AC} + \frac{5}{8}\vec{AB}$

squance · « **Reply #2 on:** March 16, 2008, 06:33:41 pm »

btw..is my diagram above correct?

Collin Li · « **Reply #3 on:** March 16, 2008, 06:34:04 pm »

Yeah!

squance · « **Reply #4 on:** March 16, 2008, 06:35:57 pm »

Thanks so much...I will try to understand it

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Author Topic: Division by Ratios Vectors (Read 1432 times) Tweet Share

squance

Division by Ratios Vectors

Collin Li

Re: Division by Ratios Vectors

squance

Re: Division by Ratios Vectors

Collin Li

Re: Division by Ratios Vectors

squance

Re: Division by Ratios Vectors

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