# Math Help - vector2

1. ## vector2

Let
-> -> -> ->
u = 3 i -2j + k

and
-> -> -> ->
v = 2i + j - k
-> ->
a)calculate the dot product of u and v

-> ->
a)calculate the cross product of u and v
-> ->
c)calculate 5 u - v

2. Originally Posted by Faz
Let
-> -> -> ->
u = 3 i -2j + k

and
-> -> -> ->
v = 2i + j - k
-> ->
a)calculate the dot product of u and v

-> ->
a)calculate the cross product of u and v
-> ->
c)calculate 5 u - v

hey mate,

In relation to (a), (b), for any two 3 Dimensional vectors of the form
v1 = a1*i + b1*j + c1*k, v2 = a2*i + b2*j + c2*k

Their Euclidean Inner or 'Dot' Product is calculated using either
(i) v1 . v2 = a1*a2 + b1*b2 + c1*c2
(ii) v1 . v2 = |v1|*|v2|*cos(y) where y is the angle between v1 and v2
Since the angle is unknown I would recommend using (i)

Their Cross Product is defined as,

v1 X v2 = det(A), where A is a 3x3 matrix where the first row of A = (i,j,k), the second row is v1 and the third is v2

For (c)
As before for a given three dimension vector of the form v1 (as defined previously)
Scalar Multiplication -
m*v1 = m(a1*i + b1*j + c1*k) = m*a1*i + m*b1*j + m*c*k
v1 + v2 = (a1 + a2)*i + (b1+b2)*j + (c1 + c2)*k

Hope this helps,

Let me know if you require any further assistance,

Regards,

David

3. ## vector2

hi

i got these answer plse check for me.

a)6i-2j-k
b)

i j k 1st row
3 -2 1 second row
2 1 -1 3rd row in matrix form

c)17i-9j-4k

thks.

4. Hello, Faz!

These are straight forward . . . exactly where is your difficulty?

Given: . $\begin{array}{ccccc}
\vec u &=& 3\vec i - 2\vec j + \vec k &=& \langle 3,\text{-}2,1\rangle \\
\vec v &=& 2\vec i + \vec j - \vec k &=& \langle 2,1,\text{-}1\rangle \end{array}$

a) Find: . $\vec u \cdot \vec v$

$\vec u\cdot \vec v \;=\;\langle3,\text{-}2,1\rangle\cdot\langle 2,1,\text{-}1\rangle \;=\;(3)(2) + (\text{-}2)(1) + (1)(\text{-}1) \;=\;6-2-1 \;=\;\boxed{3}$

b) Find: . $\vec u \times \vec v$

$\vec u \times \vec v \;=\;\begin{vmatrix}\vec i & \vec j & \vec k \\ 3 & \text{-}2 & 1 \\ 2 & 1 & \text{-}1 \end{vmatrix} \;=\;\vec i(2-1) - \vec j(\text{-}3-2) + \vec k(3+4)\;=\;\boxed{\vec i + 5\vec j + 7\vec k}$

c) Find: . $5\vec u - \vec v$

$5\vec u - \vec v \;=\;5(3\vec i - 2\vec j + \vec k) - (2\vec i + \vec j - \vec k)$

. . . . . $=\;15\vec i - 10\vec j + 5\vec k - 2\vec i - \vec j + \vec k$

. . . . . $= \;\boxed{13\vec i - 11\vec j + 6\vec k}$

5. ## vectors

Thank you guys...u are the best....i understand it very clearly.