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I am trying to answer part $d)$ by using my answer to part $c)$. From what I can see, the only possible way to do this is to find the lenght of $AB$ and $OB$, and, using the angle in part $c)$, apply the sine rule to find angle $OAB$. Are there are any quicker ways to find angle $OAB$ by using my answer to part $c)$?

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  • $\begingroup$ Find vector b, take dot product, find $cos(OAB)$ $\endgroup$
    – Nikunj
    Commented Jan 25, 2016 at 22:00
  • $\begingroup$ We have to use the answer to part $c)$. $\endgroup$
    – user307397
    Commented Jan 25, 2016 at 22:03
  • $\begingroup$ Then you can apply sine law in triangle OAB $\endgroup$
    – Nikunj
    Commented Jan 25, 2016 at 22:26

2 Answers 2

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You don't need to use the answer to part C. The vector from $A$ to $B$ is $$\mathbf{u}=\langle6-2,4-4,2-6\rangle=\langle4,0,-4\rangle$$ and the vector from $A$ to $O$ is $$\mathbf{v}=\langle-2,-4,-6\rangle$$ The angle between these is $$\theta=\arccos \left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)=\arccos \left(\frac{16}{4\sqrt{2}\times2\sqrt{14}}\right)=\arccos\left(\frac{1}{\sqrt{7}}\right)\approx67.8^\circ$$

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  • $\begingroup$ That's more like it! Given a "+1" now. Although it doesn't meet the criteria of using the angle from part (c)... $\endgroup$
    – tomi
    Commented Jan 25, 2016 at 22:22
  • $\begingroup$ Initially it seemed to me that the question used the word "hence" to indicated that part D could be answered in the same fashion as part C. $\endgroup$
    – cheesyfluff
    Commented Jan 25, 2016 at 22:28
  • $\begingroup$ They would use "similarly" to indicate that. "Hence" in Maths questions always means "using the result from the previous section" $\endgroup$
    – tomi
    Commented Jan 25, 2016 at 22:49
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The questioner is right to suspect that it must be quite easy:

Length $OA$ = length $OB$.

Triangle $AOB$ is isosceles.

Now use the angle you have. So it is "hence"

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  • $\begingroup$ Yeah, that was my initial idea too but I thought that for 2 marks that would have been too much work, and so I wondered if there were any other quicker methods. $\endgroup$
    – user307397
    Commented Jan 25, 2016 at 22:03
  • $\begingroup$ Thanks a lot, but did you have to calcuiate $OA$ and $OB$ to know the triangle was isosceles, or did you find out otherwise? $\endgroup$
    – user307397
    Commented Jan 25, 2016 at 22:15
  • $\begingroup$ Point $A$ is at $(2,4,6)$ while point $B$ is at $(6,4,2)$ which makes it obvious that they have the same lengths. They have the same components, except the $x$ and $z$ are switched. $\endgroup$
    – cheesyfluff
    Commented Jan 25, 2016 at 22:17
  • $\begingroup$ I agree with @cheesyfluff that it is obvious. It asks you to use the components of $\vec {OA}$ and $\vec {OB}$ for part (c) so they expect you to have spotted that that they are equal length. $\endgroup$
    – tomi
    Commented Jan 25, 2016 at 22:20
  • $\begingroup$ Thanks, I hadn't actually found the co-ordinates, and I thought that perhaps you used some clever trick. $\endgroup$
    – user307397
    Commented Jan 25, 2016 at 22:48

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