# Vectors (finding magnitude)

$C$ and $D$ are points with position vectors $c$ and $d$ respectively. If the magnitude of $c=5$ and magnitude of $d=7$, and the dot product; $c\cdot d=4$. Find $|CD|$ (vector connecting $C$ and $D$)

I have tried $\cos \theta=\frac{4}{35}$ And substituted the angle into $a^2=7^2+5^2-2(7)(5)\cos \theta$ I got the wrong answer so i tried again but this time with $180-\theta$ for the angle but I still can't get an answer in exact form

• Is $c.d$ the dot product? ($c\cdot d$) Nov 27, 2015 at 12:54
• yes it is the dot product of the 2 vectors Nov 27, 2015 at 12:56
• Dear miu: this question looks like a course assignment; you are more likely to invite positive reactions if you give some indication of what you have tried already, or what you know about the possible approach. Nov 27, 2015 at 13:03
• @miu Welcome to Maths SE. 2 things: Use this to learn how to format via $\LaTeX$ and secondly, if someone gives an answer that is useful to you don't forget to tick it. Your answer is $\sqrt{66}$ which renders as $\sqrt{66}$. All the best. Nov 27, 2015 at 13:05

From the dot product: $$|\vec c| | \vec d|\cos \theta=\vec c \cdot \vec d$$ where $\theta$ is the angle between the two vectors. Than use the cosine rule:
$$\vec a=|CD|=\sqrt{|\vec c|^2+|\vec d|^2-2|c||d|\cos \theta}=\sqrt{|\vec c|^2+|\vec d|^2-2( \vec c \cdot \vec d)}=\sqrt{5^2+7^2-2(4)}=\sqrt{66}$$
• that is what i used but the answer was in exact form and it is $66^1/2$ (root 66 or 66 to the power of 1/2 but i cant seem to format it, sorry) Nov 27, 2015 at 12:52