Vector Problems

**Paymemoney** · January 22nd 2010, 05:42 PM

Hi
Need help on the following questions:
1)

Find the terms of a and b:
i) $\vec{OD}$
ii) $\vec{DC}$
iii) $\vec{DE}$

2)In the following case find the vector resolutes of the first vector parallel to and perpendicular to the second vector.
$a=2i-2j+k$
$b=i+j+4k$

P.S

**VonNemo19** · January 22nd 2010, 05:50 PM

Originally Posted by Paymemoney

Hi
Need help on the following questions:
1)

Find the terms of a and b:
i) $\vec{OD}$
ii) $\vec{DC}$
iii) $\vec{DE}$

2)In the following case find the vector resolutes of the first vector parallel to and perpendicular to the second vector.
$a=2i-2j+k$
$b=i+j+4k$

P.S

Have you tried anything? Is your trouble with the concept of vectors in their component forms? Or, is it with the concept of when vector are cosidered to be orthoganal? If you were to show some work or ask a question, I could help you.

**Paymemoney** · January 22nd 2010, 06:30 PM

ok well i tired the first two and i got $\vec{OD}=2a+3b$ and $\vec{DC}=2a+4b$ .
I don't understand why this is incorrect, would you care to explain?

**Prove It** · January 22nd 2010, 07:42 PM

Originally Posted by Paymemoney

ok well i tired the first two and i got $\vec{OD}=2a+3b$ and $\vec{DC}=2a+4b$ .
I don't understand why this is incorrect, would you care to explain?

Please show HOW you got these answers...

**Paymemoney** · January 22nd 2010, 08:19 PM

well i looked at the diagram and counted the squares for a and b.

oops i forgot to add that $\vec{OA}=a$ and $\vec{OB}=b$

**Prove It** · January 22nd 2010, 08:53 PM

Well if you look at the grid, you'll see that 2 squares across = $\mathbf{b}$ and 1 square up = $\mathbf{a}$ .

Since $OD$ is 3 squares across, wouldn't it mean $1.5\mathbf{b}$ ? And then to go 2 squares up you would need to add $2\mathbf{a}$ ?

So $OD = 2\mathbf{a} + 1.5\mathbf{b}$ .

Now have a think about the others.

**HallsofIvy** · January 23rd 2010, 03:19 AM

Originally Posted by Paymemoney

Hi
Need help on the following questions:
1)

Find the terms of a and b:
i) $\vec{OD}$
ii) $\vec{DC}$
iii) $\vec{DE}$

What are a and b? Do you mean the coordinate vectors i and j? Or do you mean OA and OB?

2)In the following case find the vector resolutes of the first vector parallel to and perpendicular to the second vector.
$a=2i-2j+k$
$b=i+j+4k$

P.S

I gave you the formula here:
http://www.mathhelpforum.com/math-he...resolutes.html

**Paymemoney** · January 23rd 2010, 03:45 AM

Originally Posted by HallsofIvy

What are a and b? Do you mean the coordinate vectors i and j? Or do you mean OA and OB?

I gave you the formula here:
http://www.mathhelpforum.com/math-he...resolutes.html

I mean OA and OB.

yeh, but i get the wrong answer:
This is what i have done:
$(2i-2j+k) \cdot (\frac{i-j-4k}{\sqrt6})(\frac{i-j-4k}{\sqrt6})$

$(\frac{2+2-4}{\sqrt6})(\frac{i-j-4k}{\sqrt6})$

**HallsofIvy** · January 23rd 2010, 04:21 AM

Originally Posted by Paymemoney

I mean OA and OB.

yeh, but i get the wrong answer:
This is what i have done:
$(2i-2j+k) \cdot (\frac{i-j-4k}{\sqrt6})(\frac{i-j-4k}{\sqrt6})$

But you gave the second vector as i+j- 4k, not i- j- 4k

$(\frac{2+2-4}{\sqrt6})(\frac{i-j-4k}{\sqrt6})$

**Paymemoney** · January 23rd 2010, 02:55 PM

sorry about that, i made some typos.
Let me do it again, This is my full solution, according to the book's answers is incorrect.
$a=2i-2j+k$
$b=i+j+4k$

$=a \cdot \vec{b}(\vec{b})$
$=2i-2j+k \cdot (\frac{i+j+4k}{\sqrt6})(\frac{i+j+4k}{\sqrt6})$
$=\frac{4}{\sqrt6}(\frac{i+j+4k}{\sqrt6})$

$=\frac{4i}{6}+\frac{4j}{6}+\frac{16k}{6}$

$=\frac{4}{6}(i+j+4k)$

Now the answers says its $\frac{2}{9}(i+j+4k)$

I Hope this time i make sense.

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