Expert: Josh - 2/11/2004

Question
hmm. thanks a lot! is there any other simpler way to do this question ? i'm having a bit trouble understanding matrixes. can this be done using the dot product or cross product something like that?

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Followup To
Question -
can you help me on one problem?

how do you find the equation of the line that has teh following characteristics a) passes through point (2,5) b)makes an angle of 60 degrees to the line p = [-3,7] + t[1,6]
thank you!
Answer -
Hi Sue,

Let R be the rotation matrix.
R=[cos(w)  sin(w)]
 [-sin(w) cos(w)]
It is unitary and has determinant of 1.

The parametric equation p=[-3,7]'+t[1,6]' tells us that its gradient is 6. A straight line passing through (-3,7) and extends in the direction of vector (1,6) by the amount "t".
I assume that you understand this and know about matrix algebra. If not, write back and I'll talk some more.

I use [ , ]' to represent the transpose of a row vector.
i.e., it's suppose to be a column vector, but it's difficult to typeset here.

We multiply R by the basis u=[1,6]' to obtain another vector which deviates from u by +60,-60 degrees. Of course you have to substitute the angle w={+pi/3,-pi/3} radian respectively, into the matrix R, to get the two different matrices, R_pos and R_neg; where pi~=3.14159265...

So, let's say you've computed v_pos=R_pos*u, v_neg=R_neg*u.

The solutions are:
p=(2,5)+t*(v_pos) and
p=(2,5)+t*(v_neg)

Answer
Sue,

You need to understand what dot products and cross products are for. Dot product returns a scalar (dimensionless) value, it is used in vector projection and to decompose a vector into orthogonal components. The idea is similar to casting a shadow with a plank of wood as the sun moves across the sky.

Cross product gives a vector in 3D space which is orthogonal to the two input vectors. It is used in areas such as electromagnetics.

Matrix multiplication is the simplest way that I can think of, to obtain a vector with the specified gradient (which differs from u by 60 degrees, not 90 degrees, for instance). Let's look at an example together.

=====================================================
SECTION A: MATRIX MULTIPLICATION (case of 2x2)

Let A be a 2x2 matrix with entries
A =[a11 a12]
  [a21 a22]
Let x=[x1,x2]' be a column vector, with 2 rows and 1 column.

A.x  (here, "." stands for matrix multiplication) returns a 2x1 vector, just like x.

By definition, A.x=[a11*x1+a12*x2 a21*x1+a22*x2]'

Interpretation: The first element (or component) in this column vector is a11*x1+a12*x2 (w.r.t. the X-axis) and the second element in this column vector is a21*x1+a22*x2 (w.r.t. the Y-axis, if you like).

Okay, we are now ready to consider the case where the angle of separation is 60 degrees in the positive sense (measured counter-clockwise).
=====================================================
SECTION B: ROTATING THE VECTOR u by R.

PROCEDURE:
Let w=pi/3,
R_pos=[cos(pi/3)  sin(pi/3)] = [   1/2     sqrt(3)/2]
     [-sin(pi/3) cos(pi/3)]   [-sqrt(3)/2    1/2   ]
u=[1,6]
there's nothing difficult with calculating R_pos.u,

The result is a 2x1 column vector, v_pos, where
v_pos = R_pos.u
     =[(1/2)*1+(sqrt(3)/2)*6,-(sqrt(3)/2)*1+(1/2)*6]'

Its numerical value is roughly, v_pos=[5.6962 2.1340]'

Accordingly, one such straight line parametric equation is
p=[2 5]'+t*[(1+6*sqrt(3))/2 (-sqrt(3)+6)/2]'
It's better to leave it in exact form.
=====================================================
SECTION C: USE OF INNER (DOT) PRODUCT FOR VERIFICATION

Note: you can verify that the angle of separation between u and v_pos is 60 degrees or pi/3 radian.

Now use the inner product, which states that,
<u,v> OR u.v = |u|*|v|*cos(w)
Symbolically, let u=[u1 u2]' and v=[v1 v2]'.

If we normalize the vectors, u and v, we can write it as,
<u^,v^>=cos(w), where u^=u/|u| and v^=v/|v| by definition.
so, w [in radians]= cos ^(-1) (u1*v1+u2*v2)

Here are the number crunchings:
u=[u1 u2]'=[1 6]';
v=[v1 v2]'=[(1+6*sqrt(3))/2 (-sqrt(3)+6)/2]'~=[5.6962 2.1340]'

u^=[1 6]'/sqrt(1*1+6*6)=[1/sqrt(37) 6/sqrt(37)]
 ~=[0.1644 0.9864]'

v^=[5.6962 2.1340]'/sqrt(5.6962^2+2.1340^2)
 ~=[0.9364 0.3508]'

u^.v^ ~= (0.1644*0.9364 + 0.9864*0.3508) = cos(w)
Thus, w ~= inverse_cos(0.1644*0.9364 + 0.9864*0.3508)
        = 1.0472 or pi/3
This of course, is equivalent to 60 degrees.

Have a go at the case where the angle is negative 60 degrees, ie., w=-pi/3 and repeat the procedure given above in SECTION B.

Cheers.