1. ## [SOLVED] Vectors Question...

Question:
In the diagram OABCDEFG is a cube which the lenght of each edge is $2$ units. Unit vectors i, j, k are parallel to $\vec{OA}$, $\vec{OC}$, $\vec{OD}$ respectively. The mid-points of $AB$ and $FG$ are $M$ and $N$ respectively.

(i) Express each of the vectors $\vec{ON}$ and $\vec{MG}$ in terms of i, j and k. [3 Marks]

(ii) Find the angle between the directions of $\vec{ON}$ and $\vec{MG}$, are correct to the nearest $0.1^o$.
[4 Marks]

Attempt:

(i) $\vec{ON} = i + 2j + 2k$
$\vec{MG} = -2i + j + 2k$

are they right? and will I be rewarded 3 marks for showing no steps? becuase I don't there are any steps needed!

(ii) Need Help!

2. Your answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.

(ii)

In maths, there's a fun thing called the scalar product (the "dot product" they call it at school).

It's defined like this:

a DOT b = |a| * |b| * cos(angle)

where a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.

So you can rearrange that to get

cos(angle) = [ a DOT b ] / [ |a| * |b| ]

So now all you need to know is what a DOT b is:

I'll show that with an example:

(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32

All you do is multiply the two "firsts" (x-components) together, then the two seconds, then the thirds and add them all up.

Work that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.

Enjoy.

3. Originally Posted by Fedex
Your answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.

(ii)

In maths, there's a fun thing called the scalar product (the "dot product" they call it at school).

It's defined like this:

a DOT b = |a| * |b| * cos(angle)

where a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.

So you can rearrange that to get

cos(angle) = [ a DOT b ] / [ |a| * |b| ]

So now all you need to know is what a DOT b is:

I'll show that with an example:

(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32

All you do is multiply the two "firsts" (x-components) together, then the two seconds, then the thirds and add them all up.

Work that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.

Enjoy.
Thanks for the explanation and the example

$a.b = \left|a\right|\times \left|b\right| \times \cos\theta$

$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right| \times \left|b\right|} \right)$

$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right) \times \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$

$\left|a\right|\times \left|b\right| = \left(\begin{array}{c}1\\2\\2\end{array}\right) \times \left(\begin{array}{c}2\\1\\2\end{array}\right) = (1 \times 2) + (2 \times 1) + (2 \times 2) = 8$

$\theta = \cos^{-1} = \frac{4}{8}$

$\theta = 60^o$

Are my steps correct? and is the answer correct?

4. Okay, a few mistakes there.

You wrote your two column vectors like this:

You SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the "cross" or "vector" product of them - that's something else entirely!)

So write it with a dot, yeah?

Also, when you worked out the magnitude of the vectors, you seem to have gone wrong.

The MAGNITUDE of a vector (its "length") is just a number, not a vector.

For for the vector a = (1,2,3)

|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]

Are you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.

5. Originally Posted by Fedex
Okay, a few mistakes there.

You wrote your two column vectors like this:

You SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the "cross" or "vector" product of them - that's something else entirely!)

So write it with a dot, yeah?

Also, when you worked out the magnitude of the vectors, you seem to have gone wrong.

The MAGNITUDE of a vector (its "length") is just a number, not a vector.

For for the vector a = (1,2,3)

|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]

Are you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.
Thanks again

$a.b = \left|a\right|.\left|b\right|.\cos\theta$

$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right|.\left|b\right|} \right)$

$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right). \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$

$\left|a\right| = \sqrt{(1^2 + 2^2 + 2^2} = 3$
$\left|b\right| = \sqrt{(-2^2 + 1^2 + 2^2)} = 1$

$\left|a\right|.\left|b\right| = 3\times1 = 3$

$\theta = \cos^{-1} \frac{4}{3}$ (Invalid)

Where did I go wrong?

6. What's (-2)^2 again?

Check out that part of your working for the magnitude of b.

I think that's your problem :-)

7. Originally Posted by Fedex
What's (-2)^2 again?

Check out that part of your working for the magnitude of b.

I think that's your problem :-)
$a.b = \left|a\right|.\left|b\right|.\cos\theta$

$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right|.\left|b\right|} \right)$

$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right). \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$

$\left|a\right| = \sqrt{(1^2 + 2^2 + 2^2} = 3$
$\left|b\right| = \sqrt{(-2^2 + 1^2 + 2^2)} = 3$

$\left|a\right|.\left|b\right| = 3\times3 = 9$

$\theta = \cos^{-1} \frac{4}{9} = 63.6^o$

8. Looks like it. I've not calculated the inverse cosine value myself though. I'll assume you managed to punch that into your calculator correctly.