*Notes to a video lecture on http://www.unizor.com*

__Vectors+ 01__

*Problem A*

Given a two-dimensional vector

*with known coordinates*

**A***and another two-dimensional vector*

**(a**_{1},a_{2})*with known coordinates*

**B***.*

**(b**_{1},b_{2})Find the coordinates

*of vector*

**(x,y)***, if it's known that*

**X****scalar product**

*and*

**A·X = c****scalar product**

*.*

**B·X = d***Solution A*

We have a system of two linear equations with two unknowns

**a**_{1}·x + a_{2}·y = c

**b**_{1}·x + b_{2}·y = d*Answer A*

**x = (c·b**_{2}− d·a_{2}) /**Δ**

*y = (d·a*Δ_{1}− c·b_{1}) /where Δ is

**determinant**of the matrix of coefficients

**Δ**

*= a*_{1}·b_{2}− a_{2}·b_{1}*Note A1*

The obvious necessary and sufficient condition for the problem to have a unique solution is that

**Δ**, which means that vectors

*≠0**and*

**A***must not be collinear.*

**B***Note A2*

This problem can be easily extended to three-dimensional or any

*-dimensional case with*

**N***-dimensional vectors*

**N**

**A(a**_{1},a_{2},...,a_{N})

**B(b**_{1},b_{2},...,b_{N})

**X(x**_{1},x_{2},...,x_{N})and scalar products

**A·X =****Σ**

_{i∈[1,N]}

*a*_{i}·x_{i}

**B·X =****Σ**

_{i∈[1,N]}

*b*_{i}·x_{i}The solution would lead to a system of three or, generally,

*linear equations with as many unknowns - coordinates of unknown vector*

**N***.*

**X**Similar condition on the determinant of a matrix of coefficients not equal to zero should be specified to have a unique solution.

*Problem B*

Determine a three-dimensional vector

**X(x**_{1},x_{2},x_{3})if its known that

**X****⨯**and

*A = B*

**A****⊥**

*X*where

*and*

**A(a**_{1},a_{2},a_{3})*are given non-collinear vectors and*

**B(b**_{1},b_{2},b_{3})**⨯**signifies a

**vector (cross) product**of two three-dimensional vectors.

*Solution B*

First of all, let's analyze the geometry of these three vectors

*,*

**A***and*

**B***.*

**X**Since

**X****⨯**, vector

*A = B**, as follows from the definition of the*

**B****vector product**, must be perpendicular to two others:

**B****⊥**

*X*

**B****⊥**

*A*Since it's given that

**A****⊥**, all three vectors are mutually perpendicular to each other.

*X*In particular, both

*and*

**A***lie in the plane perpendicular to vector*

**X***.*

**B**The magnitude of vector

*equals to a product of magnitudes of*

**B***and*

**A***times a sine of an angle from*

**X***to*

**A***.*

**X**The direction of vector

*is along the perpendicular to a plane defined by vectors*

**B***and*

**A***.*

**X**Without the condition

**A****⊥**we have the freedom to change the magnitude of

*X**and rotate it within the same plane where it was, changing inverse-proportionally the sine of the angle from*

**X***to*

**A***and getting the same vector*

**X**

**B=X****⨯**. So, this last condition is important to get to the unique solution.

*A*With the condition

**A****⊥**the sine of an angle from

*X**to*

**A***is*

**X***and all we need to do to obtain vector*

**1***is to determine its magnitude |*

**X***| from an obvious equation*

**X**|

*|·|*

**A***| = |*

**X***|*

**B**since the direction of

*is known - it's perpendicular to a plane defined by*

**B***and*

**A***.*

**B**Knowing the magnitude |

*|=|*

**X***|/|*

**B***| and line perpendicular to a plane defined by*

**A***and*

**A***, along which vector*

**B***should be positioned, all that remains to establish is to chose one of two directions along this line. This is an elementary choice based on the right hand rule of forming a*

**X****vector product**.

Let's approach this problem in coordinate form.

Perpendicularity

**A****⊥**can be expressed in terms of scalar product

*B**, that is*

**A·B=0**

**a**_{1}·b_{1}+ a_{2}·b_{2}+ a_{3}·b_{3}= 0We will use it later on.

As derived in the

*Math 4 Teens*course on

*UNIZOR.COM*site (see lecture

*Vectors*-

*Vector Product*-

*Coordinate Form*), since

**X****⨯**,

*A=B*

**b**_{1}= x_{2}·a_{3}− x_{3}·a_{2}

**b**_{2}= x_{3}·a_{1}− x_{1}·a_{3}

**b**_{3}= x_{1}·a_{2}− x_{2}·a_{1}This is a system of three linear equations with three unknowns.

Can we solve it and be done?

NO!

Our geometric analysis showed that without condition

**A****⊥**the problem does not have a unique solution, and this condition is not part of our system of 3 linear equations with 3 unknowns.

*X*What's wrong with solving the above system of 3 equations?

THESE THREE EQUATIONS ARE LINEARY DEPENDENT and, therefore, we really have only two independent equations, not three.

To determine this, we can rewrite our three equations in a canonical form

**x**_{1}·0 + x_{2}·a_{3}+ x_{3}·(−a_{2}) = b_{1}

**x**_{1}·(−a_{3}) + x_{2}·0 + x_{3}·a_{1}= b_{2}

**x**_{1}·a_{2}+ x_{2}·(−a_{1}) + x_{3}·0 = b_{3}and calculate the determinant of a matrix of coefficients

**Ω**:

0 | a_{3} | −a_{2} |

−a_{3} | 0 | a_{1} |

a_{2} | −a_{1} | 0 |

*det(*

**Ω**

*)*

**=**

= 0·0·0 + a

− a

− a= 0·0·0 + a

_{1}·a_{2}·a_{3}− a_{1}·a_{2}·a_{3}−− a

_{2}·0·a_{2}− a_{1}·0·a_{1}−− a

_{3}·0·a_{3}0 = 0Condition

**A****⊥**will be the really independent third equation

*X*

**a**_{1}·x_{1}+a_{2}·x_{2}+a_{3}·x_{3}= 0Let's assume that

*, in which case we can find*

**a**_{3}≠0*and*

**x**_{1}*in terms of*

**x**_{2}*from the first two equations of the system above.*

**x**_{3}

**b**_{1}+ x_{3}·a_{2}= x_{2}·a_{3}

**−b**_{2}+ x_{3}·a_{1}= x_{1}·a_{3}from which follows

[

*]*

**b**_{1}+ x_{3}·a_{2}

**/a**_{3}= x_{2}[

*]*

**−b**_{2}+ x_{3}·a_{1}

**/a**_{3}= x_{1}Substitute this into an expression

**a**_{1}·x_{1}+a_{2}·x_{2}+a_{3}·x_{3}= 0getting

*[*

**a**_{1}·*]*

**−b**_{2}+ x_{3}·a_{1}*[*

**/a**

+ a_{3}++ a

_{2}·*]*

**b**_{1}+ x_{3}·a_{2}

**/a**

+ a_{3}++ a

_{3}·x_{3}= 0Resolve it for

*:*

**x**_{3}

**x**

= a_{3}·(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}) == a

_{1}·b_{2}− a_{2}·b_{1}or

**x _{3} = (a_{1}·b_{2} − a_{2}·b_{1})/ |A|^{2}**

*is a modulus (magnitude, length) of vector*

**|A|***.*

**A**From this we can find other coordinates of vector

*.*

**X***(*

**x**

= b

= (b

− a

= (b

_{2}·a_{3}= b_{1}+ x_{3}·a_{2}== b

_{1}+ a_{2}·(a_{1}·b_{2}−a_{2}·b_{1})/|A|^{2}== (b

_{1}·|A|^{2}+ a_{2}·a_{1}·b_{2}−− a

_{2}·a_{2}·b_{1})/|A|^{2}== (b

_{1}·a_{1}^{2}+ b_{1}·a_{2}^{2}+ b_{1}·a_{3}^{2}+ a_{2}·a_{1}·b_{2}− a_{2}·a_{2}·b_{1})/|A|^{2}=*b*cancels

_{1}·a_{2}^{2}*−a*)

_{2}·a_{2}·b_{1}*[*

= (b

+ a

== (b

_{1}·a_{1}^{2}+ b_{1}·a_{3}^{2}++ a

_{2}·a_{1}·b_{2})/|A|^{2}==

*]*

**a**

+ b_{1}·(a_{1}·b_{1}+a_{2}·b_{2}) ++ b

_{1}·a_{3}^{2}*(since*

**/|A|**

^{2}=*a*

a)

_{1}·b_{1}+a_{2}·b_{2}+a_{3}·b_{3}=0,a

_{1}·b_{1}+a_{2}·b_{2}=−a_{3}·b_{3}*[*

==

*]*

**a**_{1}·(−a_{3}·b_{3}) + b_{1}·a_{3}^{2}

**/|A|**

= a^{2}== a

_{3}·(b_{1}·a_{3}− a_{1}·b_{3})/|A|^{2}Therefore,

**x _{2} = (a_{3}·b_{1} − a_{1}·b_{3})/ |A|^{2}**

**x _{1} = (a_{2}·b_{3} − a_{3}·b_{2})/ |A|^{2}**

Obviously, we could've come to the same result if we considered the mutual perpendicularity of all three vectors

*,*

**A***and*

**B***. This factor alone is sufficient to state that vector*

**X***is collinear with vector*

**A⨯B***.*

**X**As we know (see the reference above, when we considered

*), vector*

**B=X⨯A***has coordinates*

**Y=A⨯B**

**y**_{1}= a_{2}·b_{3}− a_{3}·b_{2}

**y**_{2}= a_{3}·b_{1}− a_{1}·b_{3}

**y**_{3}= a_{1}·b_{2}− a_{2}·1_{2}which exactly what numerators in the above expressions for X-components are.

The magnitude of vector

*is*

**Y***.*

**|Y|=|A|·|B|**Since it's given that

*, all we need to find vector*

**|B|=|A|·|X|***is to take vector*

**X***and to alter its length to satisfy the condition*

**Y***, which means to divide it by*

**|B|=|A|·|X|***.*

**|A|**
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