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

__Vectors+ 03 - Direction Cosines__

*Problem A*

Find three-dimensional vector

*by its magnitude (modulus, length)*

**V(v**_{1},v_{2},v_{3})*and three*

**l****direction cosines**of angles it forms with three axes of coordinates

*,*

**cos(α)***and*

**cos(β)***, where*

**cos(γ)***is an angle between*

**α***and*

**V***X*-axis,

*is an angle between*

**β***and*

**V***Y*-axis,

*is an angle between*

**γ***and*

**V***Z*-axis.

*Solution A*

Consider a unit vector along

*X*-axis

*.*

**i(1,0,0)**Its scalar (dot) product with vector

*is*

**V(v**_{1},v_{2},v_{3})

**V(v**_{1},v_{2},v_{3})·i(1,0,0) = v_{1}On the other hand, this scalar product is equal to

**V·i = |V|·|i|·cos(α) = l·cos(α)**Therefore,

**v**_{1}= l·cos(α)Analogously, considering scalar product of vector

*with unit vectors*

**V(v**_{1},v_{2},v_{3})*on*

**j(0,1,0)***Y*-axis, we'll get

**v**_{2}= l·cos(β)Similarly, considering scalar product of vector

*with unit vectors*

**V(v**_{1},v_{2},v_{3})*on*

**k(0,0,1)***Z*-axis, we'll get

**v**_{3}= l·cos(γ)*Problem B*

A triangle on a coordinate plane is given by coordinates of its three vertices

*,*

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

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

**C(c**_{1},c_{2})Find point

*on segment*

**X(x**_{1},x_{2})*that divides this segment in a ratio*

**AB***.*

**p/q***Note B*

Use vector algebra.

*Solution B*

Consider every segment on the picture above as a vector with a direction specified by the order of vertices used in notation.

For example,

*is a vector from vertex*

**CA***to vertex*

**C***.*

**A**As a vector, its coordinates are

*.*

**(a**_{1}−c_{1},a_{2}−c_{2})Also, notice that vectors

*and*

**AX***are collinear, hence the ratio between the length of segment*

**XB***to the length of*

**AX***is the same as the ratio of vectors*

**XB***and*

**AX**

**XB.**Using this approach, we can form the following equations

**CX = CA + AX**

**CX + XB = CB**

**q·AX = p·XB**The above system of three vector equations is a linear one and contains three unknown vectors:

*,*

**CX***and*

**AX***.*

**XB**Simple transformations will immediately yield the solution.

Resolving the second equation for

*and substituting into the first one result in a system of two equations*

**CX**

**CB − XB = CA + AX**

**q·AX = p·XB**Using the last equation to express

*in terms of*

**AX***as*

**XB***, we get a single linear equation with one unknown*

**AX = (p/q)·XB**

**CB − XB = CA + (p/q)·XB**The solution for

*is*

**XB**

**XB = (CB − CA) / (1+p/q)**Expressing the above equation in coordinate form allows to find coordinates of point

*.*

**X(x**_{1},x_{2})*[*

**b**_{1}−x_{1}=*]*

**(b**_{1}−c_{1}) − (a_{1}−c_{1})

**/ (1+p/q)**from which follows

*[*

**x**_{1}= b_{1}−*]*

**(b**_{1}−c_{1}) − (a_{1}−c_{1})

**/ (1+p/q)**or

*[*

**x**_{1}=*]*

**(p/q)·b**_{1}+a_{1}

**/ (1+p/q)**or

**x**_{1}= (q·a_{1}+ p·b_{1}) / (p+q)Analogously,

**x**_{2}= (q·a_{2}+ p·b_{2}) / (p+q)
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