How to solve quadratic vector equation?

In summary: It's pretty easy to write exponents using the advanced menu. Click the Go advanced button, and use the X2 button.In summary, the author is trying to solve for a using a vector equation with a scalar c.
  • #1
Jeno
17
0

Homework Statement



(a)2 + a · b - c = 0

a and b are vectors which (in general) with different directions while c is a scalar. How to solve for a?​

The Attempt at a Solution



(a)2 + a · b - c = 0
(a + (1/2)b)2 - (1/4)b2 - c = 0
(a + (1/2)b)2 = (1/4)b2 + c​

In this case, if i carry the square from left hand side to right hand side, it appears that the left hand side turns into a vector while the right hand side remains as a scalar. I wonder why this happens and what would be the proper way to solve this equation.​
 
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  • #2
What does a2 mean if a is a vector?
 
  • #3
LCKurtz said:
What does a2 mean if a is a vector?

Hi, I think a2 should be a scalar with the magnitude of the square of the length of the vector a. Is there any problem with it? Do you know how to solve this equation?

Thank you.
 
  • #4
Jeno said:
Hi, I think a2 should be a scalar with the magnitude of the square of the length of the vector a. Is there any problem with it? Do you know how to solve this equation?

Thank you.
I'm pretty certain that what you have as a2 should be written as a · a. Assuming your equation is a · a + a · b - c = 0, you can use one of the definitions of the dot product to rewrite this equation as
|a|2 + |a| |b|cos(theta) - c = 0, where theta is the angle between a and b. This equation is quadratic in |a|.
 
  • #5
Given that a2 means a[itex]\cdot[/itex]a the equation doesn't have a unique solution and may have none. If you let:

[tex]\vec a = \langle x,y,z\rangle,\, \vec b = \langle u,v,w\rangle[/tex]

the equation becomes:

[tex]x^2+y^2+z^2+xu+yv+zw-c = 0[/tex]

which, depending on the various constants, might be a sphere or nothing.
 
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  • #6
Ya, i think a2 = a · a . Is there any other definition for a2?

So, if i don't know the direction of a, then i can't solve for a is it?

I think, in my situation, i can guess the direction for a. Let's say if i know that the direction of a is r (a vector with magnitude of r, such that r/r is a unit vector), so how can i solve for the equation without expressing it in the angle between them (theta), and just the vectors and scalar of b, c and r?

a=-(r/2r2)( +- (4r2c + (b · r)2 )1/2 + r · u )

This is the answer given in the book. It seems like after applying the formula to find the root of quadratic equation, then he multiply it with r2/r2. But it didn't sound quite correct to me if you consider the vector BAC CAB rule. The cross product term didn't appear in this equation, and in my situation, b can be in any direction, which doesn't restricted to be parallel to a.

which, depending on the various constants, might be a sphere or nothing.

Sorry, what do you mean by "nothing" here?

Thank you.
 
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  • #7
Jeno said:
Sorry, what do you mean by "nothing" here?

Thank you.

I mean there is no solution. For example, with the right choice for u,v,w, and c, the equation might end up something like

[tex](x-1)^2 + (y+2)^2+(z-3)^2 = -10[/tex]

having no real solutions.
 
  • #8
So is it similar to the normal quadratic equation where b2 - 4ac < 0 gives no real solution? ok, then i understand.

But i still don't understand how the book get the the expression above. I think the expression above should be valid, since it is written on the book, but i just don't know how to get that through a proper way.

Can anybody help? Thank you.
 
  • #9
What "expression above"? You showed no expression that you said came from the book.
 
  • #10
HallsofIvy said:
What "expression above"? You showed no expression that you said came from the book.

a=-(r/2r2)( +- (4r2c + (b · r)2 )1/2 + r · u )

This is the expression i meant. Thank you.
 
  • #11
Jeno said:
a=-(r/2r2)( +- (4r2c + (b · r)2 )1/2 + r · u )

This is the expression i meant. Thank you.

How is this equation related to the equation in post #1? The original equation involved vectors a and b, and a constant c. The equation you just showed has r and u, which aren't in the original equation.

Also, your last equation is very difficult to read, as it has what appear to be exponents that aren't shown as such. For example what you have written as 4r2c could be (4r)(2c) or might be 4r2c. And I'm guessing that (b · r)2 )1/2 is (b · r)2 )1/2.

It's pretty easy to write exponents using the advanced menu. Click the Go advanced button, and use the X2 button.
 
  • #12
Mark44 said:
How is this equation related to the equation in post #1? The original equation involved vectors a and b, and a constant c. The equation you just showed has r and u, which aren't in the original equation.

Also, your last equation is very difficult to read, as it has what appear to be exponents that aren't shown as such. For example what you have written as 4r2c could be (4r)(2c) or might be 4r2c. And I'm guessing that (b · r)2 )1/2 is (b · r)2 )1/2.

It's pretty easy to write exponents using the advanced menu. Click the Go advanced button, and use the X2 button.

Sorry, i have typed it wrong. The expression should be as follow:

a=-(r/2r2)( +- (4r2c + (b · r)2 )1/2 + r · b )

where r is a vector with direction of a and length of r.

Thank you.
 

Related to How to solve quadratic vector equation?

1. How do I identify a quadratic vector equation?

A quadratic vector equation can be identified by having at least one squared term and two or more vector variables.

2. What is the general form of a quadratic vector equation?

The general form of a quadratic vector equation is ax² + bx + cx + d = 0, where a, b, and c are vector coefficients and x is the vector variable.

3. How do I solve a quadratic vector equation using the quadratic formula?

To solve a quadratic vector equation using the quadratic formula, first rearrange the equation into the general form. Then, use the formula x = (-b ± √(b²-4ac)) / 2a, where a, b, and c are the vector coefficients in the equation. This will give you the values for x, which can be substituted back into the original equation to solve for the vector variables.

4. Can a quadratic vector equation have multiple solutions?

Yes, a quadratic vector equation can have multiple solutions. This is because the quadratic formula can give both a positive and negative value for x, which can each be used as a solution.

5. Are there any other methods for solving quadratic vector equations?

Yes, there are other methods for solving quadratic vector equations, such as factoring or completing the square. However, the quadratic formula is the most commonly used method and can be used for all types of quadratic equations.

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