Find unit vector and cross product

[Edwardo_Elric]

Homework Statement

Given the two vectors written in component-unit vector form below:
[tex]D = 3\hat{i} - \hat{j}[/tex]
[tex]E = 2\hat{i} + 4\hat{j}[/tex]
a.) Find the unit vector in the same direction as D
b.) Find the cross product of D x E
c.) Write the vector D in magnitude direction form

Homework Equations

The Attempt at a Solution

a.)
[tex]\hat{u} = \frac{\vec{D}}{D}[/tex]
[tex]\hat{u} = \frac{3\hat{i}}{10} - \frac{\hat{j}}{10}[/tex]



b.) Cz = DxEy - DyEx
Cz = [3][4] - [-1][2]
= 12 + 2
= 14

c.) D = sqrt(3^2 + 1)
= 3.2
theta = tan^-1(1/-3)
= 18.4
3.2... 18.4 degrees S of W

 

[learningphysics]

Homework Statement

Given the two vectors written in component-unit vector form below:
[tex]D = 3\hat{i} - \hat{j}[/tex]
[tex]E = 2\hat{i} + 4\hat{j}[/tex]
a.) Find the unit vector in the same direction as D
b.) Find the cross product of D x E
c.) Write the vector D in magnitude direction form

Homework Equations

The Attempt at a Solution

a.)
[tex]\hat{u} = \frac{\vec{D}}{D}[/tex]
[tex]\hat{u} = \frac{3\hat{i}}{10} - \frac{\hat{j}}{10}[/tex]

The denominators are wrong.

b.) Cz = DxEy - DyEx
Cz = [3][4] - [-1][2]
= 12 + 2
= 14

Yes, that's correct. I'd write the final answer as [tex]14\hat{k}[/tex]

c.) D = sqrt(3^2 + 1)
= 3.2
theta = tan^-1(1/-3)
= 18.4
3.2... 18.4 degrees S of W

The direction isn't right.

 

[Edwardo_Elric]

a.) [tex]\hat{u} = \frac{10}{3\hat{i}} - \frac{10}{\hat{j}}[/tex]

c.) D = 3.2, 18.4 S of W
?

 

[learningphysics]

a.) [tex]\hat{u} = \frac{10}{3\hat{i}} - \frac{10}{\hat{j}}[/tex]

No, what you had before was right except that you needed sqrt(10) instead of 10 in the denominator.

c.) D = 3.2, 18.4 S of W
?

I get the direction as 18.4 S of E

 

[Edwardo_Elric]

oh yeah i thot west is to the right
thanks