Section 4.1

Areas, Volumes, and Cross Products

> read(`C:\\classes\\2002-2003\\spring\\math215\\Lectures\\LAprocs.map`);

Warning, the protected names norm and trace have been redefined and unprotected

Warning, the name changecoords has been redefined

Warning, the name arrow has been redefined

>

The Area of a Parallelogram

A parallelogram determined by two non-zero and non-parallel vectors a =[ a[1] , a[2] ] and b = [ b[1] , b[2] ] in R^2 .

> plotparellelogram();

[Maple Plot]

>

What is the area of this parallelogram?

Answer

Area of the parallelogram is given as,

(1) Area = (base)(altitude) = ||a|| h = ||a|| ||b|| sin( theta ) = ||a|| ||b|| sqrt(1-cos(theta)^2) .

Note: (page 24) (a b) = ||a|| ||b|| cos(theta) .

Square both sides of the equation (1) to get,

Area^2 = `||a||`^2 `||b||`^2 ( 1-cos(theta)^2 ) = `||a||`^2 `||b||`^2 - `||a||`^2 `||b||`^2 cos(theta)^2 = `||a||`^2 `||b||`^2 - (a*b)^2 .

Notice:

`||a||`^2 = a[1]^2+a[2]^2

`||b||`^2 = b[1]^2+b[2]^2

(a*b)^2 = a[1]*b[1]+a[2]*b[2]

Thus, we have

Area^2 = a[1]^2+a[2]^2 + b[1]^2+b[2]^2 + a[1]*b[1]+a[2]*b[2] = (a[1]*b[2]-a[2]*b[1])^2

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Area of the parallelogram formed by two non-zero and non-parallel vectors

a =[ a[1] , a[2] ] and b = [ b[1] , b[2] ] in R^2 is given by Area =| a[1]*b[2]-a[2]*b[1] |. The

number inside the absolute value signs is known as the determinant of the matrix,

A := matrix([[a[1], a[2]], [b[1], b[2]]])

det(A) = a[1]*b[2]-a[2]*b[1] .

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Example 1: Find the area of the parallelogram given, where

>

> a := matrix(2,1,[3,1.5]); b := matrix(2,1,[1,2]);

a := matrix([[3], [1.5]])

b := matrix([[1], [2]])

> plotparellelogram2();

[Maple Plot]

> A := augment(a,b);

A := matrix([[3, 1], [1.5, 2]])

> `Area` = det(A);

Area = 4.5

>

Cross Products

Given two independent vectors v = [ v[1], v[2], v[3] ] and u= [ u[1], u[2], u[3] ] in R^3 . We want to find a vector

p = [ p[1], p[2], p[3] ] in R^3 that is perpendicular to both vectors v and u .

> transpose(matrix(3,1,[p[1],p[2],p[3]])) * matrix(3,1,[u[1],u[2],u[3]]) = 0; transpose(matrix(3,1,[p[1],p[2],p[3]])) * matrix(3,1,[v[1],v[2],v[3]]) = 0;

matrix([[p[1], p[2], p[3]]])*matrix([[u[1]], [u[2]]...

matrix([[p[1], p[2], p[3]]])*matrix([[v[1]], [v[2]]...

>

> evalm(transpose(matrix(3,1,[p[1],p[2],p[3]])) &* matrix(3,1,[u[1],u[2],u[3]])) = 0; evalm(transpose(matrix(3,1,[p[1],p[2],p[3]])) &* matrix(3,1,[v[1],v[2],v[3]])) = 0;

matrix([[p[1]*u[1]+p[2]*u[2]+p[3]*u[3]]]) = 0

matrix([[p[1]*v[1]+p[2]*v[2]+p[3]*v[3]]]) = 0

>

> matrix(2,3,[v[1],v[2],v[3],u[1],u[2],u[3]]) * matrix(3,1,[p[1],p[2],p[3]])= 0 ;

matrix([[v[1], v[2], v[3]], [u[1], u[2], u[3]]])*ma...

>

Find the nullspace.

> matrix(2,4,[v[1],v[2],v[3],0,u[1],u[2],u[3],0]);

matrix([[v[1], v[2], v[3], 0], [u[1], u[2], u[3], 0...

> AUG := rref(matrix(2,4,[v[1],v[2],v[3],0,u[1],u[2],u[3],0]));

> p[1] = (v[3]*u[2] - v[2]*u[3])/(u[1]*v[2]-v[1]*u[2]) * p[3]; p[2] = (u[1]*v[3] - u[3]*v[1])/(u[1]*v[2]-v[1]*u[2]); p[3] = p[3];

p[1] = (v[3]*u[2]-v[2]*u[3])/(u[1]*v[2]-v[1]*u[2])*...

p[2] = (u[1]*v[3]-u[3]*v[1])/(u[1]*v[2]-v[1]*u[2])

p[3] = p[3]

>

Let, p[3] = v[2]*u[1]-v[1]*u[2] . Then we have,

> p[1] = v[3]*u[2] - v[2]*u[3]; p[2] = u[1]*v[3] - u[3]*v[1]; p[3] =v[2]*u[1]-v[1]*u[2];

p[1] = v[3]*u[2]-v[2]*u[3]

p[2] = u[1]*v[3]-u[3]*v[1]

p[3] = u[1]*v[2]-v[1]*u[2]

> p = det(matrix(2,2,[u[2],u[3],v[2],v[3]]))*i - det(matrix(2,2,[u[1],u[3],v[1],v[3]]))*j + det(matrix(2,2,[u[1],u[2],v[1],v[2]]))*k;

p = (v[3]*u[2]-v[2]*u[3])*i-(u[1]*v[3]-u[3]*v[1])*j...

Where i=[1,0,0], j=[0,1,0] and k=[0,0,1].

> p = Det(matrix(2,2,[u[2],u[3],v[2],v[3]]))*i - Det(matrix(2,2,[u[1],u[3],v[1],v[3]]))*j + Det(matrix(2,2,[u[1],u[2],v[1],v[2]]))*k;

p = Det(matrix([[u[2], u[3]], [v[2], v[3]]]))*i-Det...

>

The vector p is known as the cross product of the two vectors u and v , i.e. p = u x v.

Use the determinat of the symbolic matrix ,

> p = Det(matrix(3,3,[i,j,k,u[1],u[2],u[3],v[1],v[2],v[3]]));

p = Det(matrix([[i, j, k], [u[1], u[2], u[3]], [v[1...

>

Example 2: Find a vector perpendicular to,

> v := matrix(3,1,[-1, 3, -2]); u :=matrix(3,1,[3, -2, 1]);

v := matrix([[-1], [3], [-2]])

u := matrix([[3], [-2], [1]])

> v1 := [-1, 3, -2]: u1 := [3, -2, 1]:

>

> A := transpose(augment(matrix(3,1,[i,j,k]),u,v));

A := matrix([[i, j, k], [3, -2, 1], [-1, 3, -2]])

>

> `u x v` = det(A);

`u x v` = i+5*j+7*k

>

> A := transpose(augment(matrix(3,1,[i,j,k]),v,u));

A := matrix([[i, j, k], [-1, 3, -2], [3, -2, 1]])

>

> `v x u` = det(A);

`v x u` = -i-5*j-7*k

>

Right hand Rule

> plotcrossproduct(v1,u1);

[Maple Plot]

>

******************************************************************

Let v = [ v[1], v[2], v[3] ] and u= [ u[1], u[2], u[3] ] in R^3 . Then the area of the parallelogram in R^3

formed by u and v is given as || u x v || or || v x u ||.

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Example 3: Find the area of the parallelogram formed by the vectors,

> v := matrix(3,1,[1,4,6]); u :=matrix(3,1,[6,3,1]);

v := matrix([[1], [4], [6]])

u := matrix([[6], [3], [1]])

>

> A := transpose(augment(matrix(3,1,[i,j,k]),u,v));

A := matrix([[i, j, k], [6, 3, 1], [1, 4, 6]])

>

> `u x v` = det(A); p := det(A):

`u x v` = 14*i-35*j+21*k

>

> `||u x v||` = sqrt(coeff(p,i)^2 + coeff(p,j)^2 + coeff(p,k)^2);

`||u x v||` = 7*sqrt(38)

>

Example 1 (revisted): Find the area of the parallelogram given, where

>

> a := matrix(3,1,[3,1.5,0]); b := matrix(3,1,[1,2,0]);

a := matrix([[3], [1.5], [0]])

b := matrix([[1], [2], [0]])

>

> A := transpose(augment(matrix(3,1,[i,j,k]),a,b));

A := matrix([[i, j, k], [3, 1.5, 0], [1, 2, 0]])

>

> `a x b` = det(A); p := det(A):

`a x b` = 4.5*k

>

> `||a x b||` = sqrt(coeff(p,i)^2 + coeff(p,j)^2 + coeff(p,k)^2);

`||a x b||` = 4.500000000

>

Properties of the cross product

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Let a and b be vectors in R^3 .

1.) b x c = -(c x b)

2.) a x (b x c) is generally different from (a x b) x c

3.) a x (b+c) = a x b + a x c

(a+b) x c = (a x c) + (b x c)

4.) b (b x c) = (b x c) c = 0 Since (b x c) is perpendicular to b and c.

5.) || b x c || = Area of the parallelogram determined by b and c.

6.) a ( b x c) = (a x b) c = ( + or -) Volume of the box determined by a, b, and c.

7.) a x (b x c) = (a c) b - (a b) c

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Exercise

1, 3, 5-7, 10-15, 19-21, 25, 26, 30-34.