# Vector Problems

• Jan 22nd 2010, 05:42 PM
Paymemoney
Vector Problems
Hi
Need help on the following questions:
1)
http://img202.imageshack.us/img202/8459/vector.jpg
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
• Jan 22nd 2010, 05:50 PM
VonNemo19
Quote:

Originally Posted by Paymemoney
Hi
Need help on the following questions:
1)
http://img202.imageshack.us/img202/8459/vector.jpg
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.
• Jan 22nd 2010, 06:30 PM
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?
• Jan 22nd 2010, 07:42 PM
Prove It
Quote:

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?

• Jan 22nd 2010, 08:19 PM
Paymemoney
well i looked at the diagram and counted the squares for a and b.

oops i forgot to add that $\vec{OA}=a$and $\vec{OB}=b$
• Jan 22nd 2010, 08:53 PM
Prove It
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.
• Jan 23rd 2010, 03:19 AM
HallsofIvy
Quote:

Originally Posted by Paymemoney
Hi
Need help on the following questions:
1)
http://img202.imageshack.us/img202/8459/vector.jpg
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?

Quote:

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
• Jan 23rd 2010, 03:45 AM
Paymemoney
Quote:

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

$(\frac{2+2-4}{\sqrt6})(\frac{i-j-4k}{\sqrt6})$
• Jan 23rd 2010, 04:21 AM
HallsofIvy
Quote:

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

Quote:

$(\frac{2+2-4}{\sqrt6})(\frac{i-j-4k}{\sqrt6})$
• Jan 23rd 2010, 02:55 PM
Paymemoney
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.