Solve Vector Equation 2x-(xdoti)i for x

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In summary, xdot i is just a so and not a function. So if x= ai+ bj then "x dot i" is just a so and not a function. The (xi) at the end is actually x cross i, I didnt read it carefully emough.
  • #1
square_imp
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Hi all,

I would like some help with a tricky vector equation, I need to solve for x.

2x-(xdoti)i = i + 5j - (xi)

The dot is a dot product and not multiply. i am not sure how to deal with the dot product, I am little confused. Any help would be excellent. Thanks. :smile:
 
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  • #2
if
[tex]\vec{x}=x_i \hat{i} + x_j \hat{j}[/tex]
and
[tex]\vec{y}=y_i \hat{i} + y_j \hat{j}[/tex]
then
[tex]x \dotprod y = x_i y_i + x_j y_j[/tex]
 
  • #3
square_imp said:
Hi all,
I would like some help with a tricky vector equation, I need to solve for x.
2x-(xdoti)i = i + 5j - (xi)
The dot is a dot product and not multiply. i am not sure how to deal with the dot product, I am little confused. Any help would be excellent. Thanks. :smile:
I'm a little confused myself! (xdoti) implies that x is a vector but then "xi" is not clear. If x= ai+ bj then "x dot i" is just a so "2a- (x dot i)i would be
2ai+ 2bj- ai= ai+ 2bj. But then I have no idea what (xi) means.
 
  • #4
The (xi) at the end is actually x cross i, I didnt read it carefully emough. Now, I have had another crack at it working out the dot and cross product sections in the equation, this is my working:

2x-(ai + bj).(i) = i +5j -((ai + bj) cross i)

substituting the x for ai(unit vector) and bj(unit vector)
gives:

2x-ai = i + 5j - bk
x = [(a+1)i + 5j -bk]/2

is this working correct? The bit I am most unsure about is the working out of the cross product using matrices. Thanks for the help, sort of clicked my brain into gear.
 
  • #5
Write x as ai+ bj+ ck. Then 2x= 2ai+ 2bj+ 2ck and (x dot i)i= ai so 2x- (x dot i)i= (2a-a)i+ 2bj+ 2ck= ai+ 2bj+ 2ck. x cross i= cj- bk.

2x-(xdoti)i = i + 5j - (x cross i) is
ai+2bj+ 2ck= i+ (5-c)j- bk

We must have a=1, 1b=5- c and 2c= -b.

Can you solve those?
 
  • #6
Thanks guys, got it all sorted I think. Realised my mistake with the matrix. Got x to be: (i + 2j + k) which when substitutued back into the original formula all works out. Thanks again. :biggrin:
 

FAQ: Solve Vector Equation 2x-(xdoti)i for x

What is a vector equation?

A vector equation is an equation that involves vectors, which are quantities that have both magnitude and direction. In a vector equation, the variables represent vectors instead of regular numbers or variables.

What is a scalar?

A scalar is a quantity that only has magnitude and no direction. It is represented by a regular number or variable in a vector equation. In the equation 2x-(xdoti)i for x, 2 and x are scalars.

How do I solve a vector equation?

To solve a vector equation, you need to manipulate the equation to isolate the vector variable on one side of the equation. This can be done by using vector addition, scalar multiplication, and other vector operations. In the equation 2x-(xdoti)i for x, you can solve for x by dividing both sides by 2 and then multiplying by (i dot i), which is equal to 1.

What is the dot product of two vectors?

The dot product of two vectors is a scalar value that represents the projection of one vector onto another. It is calculated by multiplying the magnitudes of the two vectors and the cosine of the angle between them. In the equation 2x-(xdoti)i for x, (xdoti) represents the dot product of x and i.

Why is solving vector equations important?

Solving vector equations is important because it allows us to model and solve real-world problems that involve quantities with both magnitude and direction. It is also a fundamental concept in mathematics and physics, and is used in fields such as engineering, computer graphics, and navigation.

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