These problems are done with the CAS. See Hand Solutions for details.

A cross product of vectors with numerical components is computed. The cross product is orthogonal to the original vectors.

a = sym([4 3 -2]), b = sym([2 -1 1])

c = cross(a,b)

a_dot_c = dot(a,c)

b_dot_c = dot(b,c)

A cross product of vectors with symbolic components is computed.

syms t

a = [t cos(t) sin(t)], b = [1 -sin(t) cos(t)]

c = simplify(cross(a,b))

a = [1 0 -2], b = [0 1 1], c = cross(a,b)

figure

quiver3(0,0,0, 1,0,-2); grid on; hold on

quiver3(0,0,0, 0,1,1)

quiver3(0,0,0, 2,-1,1)

axis equal; axis([0 2 -1 1 -2 1])

xlabel('x'); ylabel('y'); zlabel('z')

title('SET8, 821/8')

view(31, 26)

Recall that . The figure provides the data and we deduce that uv is directed out of the page.

mag_u = 4, mag_v = 5, theta = sym(pi/4)

mag_u = 4

mag_v = 5

mag_u_cross_v = mag_u * mag_v * sin(theta)

We can reason from the figure or just assign plausible numerical values to the components of a and b.

So for c, we see that .

a = 3*[cosd(60) sind(60), 0]

b = [0 0 2]

c = cross(a,b)

Here we find two unit vectors orthogonal to a given pair of vectors.

u = sym([0 1 -1]), v = sym([1 1 0])

c = cross(u,v)

w = c/norm(c)

neg_w = -w

The area of the parallelogram is the magnitude of the cross product of a pair of edge vectors that span a vertex.

P = sym([1 0 2]), Q = sym([3 3 3]), R = sym([7 5 8]), S = sym([5 2 7])

PQ = Q-P, PS = S-P

c = cross(PQ,PS)

V = norm(c)

V_appx = double(V) % in cm^3

V_appx = 16.4012

%

M = [P; Q; R; S]

x = M(:,1), y = M(:,2), z = M(:,3)

x = [x; x(1)]; y = [y; y(1)]; z = [z; z(1)]; % Return to starting vertex!

figure

fill3(x,y,z, 'y'); grid on; hold on

plot3(x,y,z, 'LineWidth', 3)

xlabel('x'); ylabel('y'); zlabel('z')

title('SET8, 821/28')

A normal vector to and the area of a triangle are determined.

P = sym([2 -3 4]), Q = sym([-1, -2 2]), R = sym([3 1 -3])

PQ = Q-P, PR = R-P

n = cross(PQ, PR) % normal (perpendicular) vector to the triangle with vertices P, Q, R

A = norm(n)/2 % area of triangle

A_appx = double(A) % in cm^2

A_appx = 13.2193

The volume of a parallelepiped (sheared box) is computed by taking the absolute value of the scalar triple product of a triple of vectors that span a vertex.

P = sym([3 0 1]), Q = sym([-1 2 5]), R = sym([5 1 -1]), S = sym([0 4 2])

PQ = Q-P, PR = R-P, PS = S-P

c = cross(PR, PS)

V = abs(dot(PQ,c)) % in cm^3

We see that the magnitude of the cross product uv has a maximum of 15 at . It has a minimum value of 0 at . The cross product uv points in the direction of k (out of the page) for and in the direction of -k (into the page) for or .

syms theta

u = 3*[cos(theta), sin(theta) 0], v = sym([0 5 0])

c = cross(u,v)

mag_c = simplify(norm(c))

figure

fplot(mag_c, [-pi pi], 'LineWidth', 3); grid on; hold on

axis([-4 4 -5 20])

xlabel('Angle \theta'); ylabel('15 | cos(\theta) |')

title('SET8, 822/42: || u x v ||')

%

figure

fplot(15*cos(theta), [-pi pi], 'g', 'LineWidth', 3); grid on; hold on

plot([-4 4], [0 0], 'm', 'LineWidth', 3)

axis([-4 4 -20 20])

xlabel('Angle \theta'); ylabel('15 cos(\theta)')

title('SET8, 822/42: 15 cos(\theta)')