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$C$ 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

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  • $\begingroup$ Is $c.d$ the dot product? ($c\cdot d$) $\endgroup$
    – BLAZE
    Nov 27, 2015 at 12:54
  • $\begingroup$ yes it is the dot product of the 2 vectors $\endgroup$
    – miu meow meow
    Nov 27, 2015 at 12:56
  • $\begingroup$ 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. $\endgroup$
    – Justpassingby
    Nov 27, 2015 at 13:03
  • $\begingroup$ @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. $\endgroup$
    – BLAZE
    Nov 27, 2015 at 13:05

1 Answer 1

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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:

$$ \vec a=|CD|=\sqrt{|\vec c|^2+|\vec d|^2-2|c||d|\cos \theta}=\sqrt{|\vec c|^2+|\vec d|^2-2( \vec c \cdot \vec d)}=\sqrt{5^2+7^2-2(4)}=\sqrt{66} $$

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  • $\begingroup$ 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) $\endgroup$
    – miu meow meow
    Nov 27, 2015 at 12:52
  • $\begingroup$ I've changed from an Hint to an answer. You can see the exact form. $\endgroup$
    – Emilio Novati
    Nov 27, 2015 at 14:51

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