Vector Product

$$\begin{aligned}[t]\Vect{a}\times \Vect{b} &= \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix} \\ &= \begin{pmatrix} (1)(3) - (-1)(2) \\ (-1)(1) - (2)(3) \\ (2)(2) - (1)(1) \end{pmatrix} \\ &= \begin{pmatrix} 3 - (-2) \\ -1 - 6 \\ 4 - 1 \end{pmatrix} \\ &= \begin{pmatrix} 5 \\ -7 \\ 3 \end{pmatrix}\end{aligned}$$

$$\begin{aligned}[t]\Vect{b}\times \Vect{c} &= \begin{pmatrix} b_2c_3 - b_3c_2 \\ b_3c_1 - b_1c_3 \\ b_1c_2 - b_2c_1 \end{pmatrix} \\ &= \begin{pmatrix} (2)(4) - (3)(0) \\ (3)(2) - (1)(4) \\ (1)(0) - (2)(2) \end{pmatrix} \\ &= \begin{pmatrix} 8 - 0 \\ 6 - 4 \\ 0 - 4 \end{pmatrix} \\ &= \begin{pmatrix} 8 \\ 2 \\ -4 \end{pmatrix}\end{aligned}$$

$$\begin{aligned}[t]\Vect{c}\times \Vect{d} &= \begin{pmatrix} c_2d_3 - c_3d_2 \\ c_3d_1 - c_1d_3 \\ c_1d_2 - c_2d_1 \end{pmatrix} \\ &= \begin{pmatrix} (0)(-2) - (2)(3) \\ (2)(1) - (-1)(-2) \\ (-1)(3) - (0)(1) \end{pmatrix} \\ &= \begin{pmatrix} 0 - 6 \\ 2 - 2 \\ -3 - 0 \end{pmatrix} \\ &= \begin{pmatrix} -6 \\ 0 \\ -3 \end{pmatrix}\end{aligned}$$

$$\begin{aligned}[t]\Vect{u}\times \Vect{v} &= \begin{pmatrix} u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1 \end{pmatrix} \\ &= \begin{pmatrix} (-1)(-4) - (0)(1) \\ (0)(3) - (2)(-4) \\ (2)(1) - (-1)(3) \end{pmatrix} \\ &= \begin{pmatrix} 4 - 0 \\ 0 - (-8) \\ 2 - (-3) \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ 8 \\ 5 \end{pmatrix}\end{aligned}$$

1. Find \(\Vect{a} \times \Vect{b}\) $$ \begin{aligned}[t] \Vect{a}\times \Vect{b} &= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix} \\ &= \begin{pmatrix} (2)(-1) - (3)(3) \\ (3)(-1) - (1)(-1) \\ (1)(3) - (2)(-1) \end{pmatrix} \\ &= \begin{pmatrix} -2 - 9 \\ -3 - (-1) \\ 3 - (-2) \end{pmatrix} \\ &= \begin{pmatrix} -11 \\ -2 \\ 5 \end{pmatrix} \end{aligned} $$
2. Determine the scalar products Using the result from part 1: $$ \begin{aligned}[t] \Vect{a} \cdot (\Vect{a} \times \Vect{b}) &= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -11 \\ -2 \\ 5 \end{pmatrix} \\ &= (1)(-11) + (2)(-2) + (3)(5) \\ &= -11 - 4 + 15 \\ &= 0 \end{aligned} $$ And for the second one: $$ \begin{aligned}[t] \Vect{b} \cdot (\Vect{a} \times \Vect{b}) &= \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -11 \\ -2 \\ 5 \end{pmatrix} \\ &= (-1)(-11) + (3)(-2) + (-1)(5) \\ &= 11 - 6 - 5 \\ &= 0 \end{aligned} $$
3. Explanation The results are both zero. This is because, by definition, the vector product \(\Vect{a} \times \Vect{b}\) produces a vector that is orthogonal (perpendicular) to both \(\Vect{a}\) and \(\Vect{b}\). The scalar product of two orthogonal vectors is always zero. The calculations in part 2 verify this fundamental property.

The base unit vectors are \(\Vect{i}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\), \(\Vect{j}=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\), and \(\Vect{k}=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\).

1. Cross product of a vector with itself $$ \Vect{i} \times \Vect{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} (0)(0)-(0)(0) \\ (0)(1)-(1)(0) \\ (1)(0)-(0)(1) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \Vect{0} $$ Similarly, \(\Vect{j} \times \Vect{j} = \Vect{0}\) and \(\Vect{k} \times \Vect{k} = \Vect{0}\).
2. Cross product of two different base vectors $$ \Vect{i} \times \Vect{j} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} (0)(0)-(0)(1) \\ (0)(0)-(1)(0) \\ (1)(1)-(0)(0) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \Vect{k} $$ $$ \Vect{j} \times \Vect{i} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} (1)(0)-(0)(0) \\ (0)(1)-(0)(0) \\ (0)(0)-(1)(1) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} = -\Vect{k} $$

We apply the definition of the vector product of \(\Vect{a}\) with itself.
Let \(\Vect{b} = \Vect{a}\), so \(b_1=a_1\), \(b_2=a_2\), and \(b_3=a_3\).$$\begin{aligned}[t]\Vect{a}\times \Vect{a} &= \begin{pmatrix} a_2a_3 - a_3a_2 \\ a_3a_1 - a_1a_3 \\ a_1a_2 - a_2a_1 \end{pmatrix} \\ &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ &= \Vect{0}\end{aligned}$$Since the multiplication of real numbers is commutative (\(ab=ba\)), each component simplifies to zero.

$$\begin{aligned} \Vect{a}\times \Vect{b} &= \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\\ &=\begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix}\\ &= \begin{pmatrix} -(a_3b_2 - a_2b_3) \\ -(a_1b_3 - a_3b_1) \\ -(a_2b_1 - a_1b_2) \end{pmatrix} \\ &= -\begin{pmatrix} a_3b_2 - a_2b_3 \\ a_1b_3 - a_3b_1 \\ a_2b_1 - a_1b_2 \end{pmatrix} \\ &= -\left( \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\times \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \right)\\ &= -\left( \Vect{b}\times \Vect{a}\right) \end{aligned} $$Therefore, we have shown that \(\Vect{a} \times \Vect{b} = -(\Vect{b} \times \Vect{a})\).

We will prove this property by expanding both sides of the equation by their components.

• Left-Hand Side (LHS): First, we find the sum \(\Vect{b}+\Vect{c}\): $$ \Vect{b}+\Vect{c} = \begin{pmatrix} b_1+c_1 \\ b_2+c_2 \\ b_3+c_3 \end{pmatrix} $$ Now we calculate \(\Vect{a} \times (\Vect{b}+\Vect{c})\): $$ \begin{aligned} \Vect{a} \times (\Vect{b}+\Vect{c}) &= \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times \begin{pmatrix} b_1+c_1 \\ b_2+c_2 \\ b_3+c_3 \end{pmatrix} \\ &= \begin{pmatrix} a_2(b_3+c_3) - a_3(b_2+c_2) \\ a_3(b_1+c_1) - a_1(b_3+c_3) \\ a_1(b_2+c_2) - a_2(b_1+c_1) \end{pmatrix} \\ &= \begin{pmatrix} a_2b_3 + a_2c_3 - a_3b_2 - a_3c_2 \\ a_3b_1 + a_3c_1 - a_1b_3 - a_1c_3 \\ a_1b_2 + a_1c_2 - a_2b_1 - a_2c_1 \end{pmatrix} \end{aligned} $$
• Right-Hand Side (RHS): First, we find the two separate cross products: $$ \Vect{a} \times \Vect{b} = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix} \quad \text{and} \quad \Vect{a} \times \Vect{c} = \begin{pmatrix} a_2c_3 - a_3c_2 \\ a_3c_1 - a_1c_3 \\ a_1c_2 - a_2c_1 \end{pmatrix} $$ Now we add them together: $$ \begin{aligned} (\Vect{a} \times \Vect{b}) + (\Vect{a} \times \Vect{c}) &= \begin{pmatrix} (a_2b_3 - a_3b_2) + (a_2c_3 - a_3c_2) \\ (a_3b_1 - a_1b_3) + (a_3c_1 - a_1c_3) \\ (a_1b_2 - a_2b_1) + (a_1c_2 - a_2c_1) \end{pmatrix} \\ &= \begin{pmatrix} a_2b_3 + a_2c_3 - a_3b_2 - a_3c_2 \\ a_3b_1 + a_3c_1 - a_1b_3 - a_1c_3 \\ a_1b_2 + a_1c_2 - a_2b_1 - a_2c_1 \end{pmatrix} \end{aligned} $$

Since the resulting vectors for the LHS and RHS are identical, we have proven that \(\Vect{a} \times (\Vect{b}+\Vect{c}) = (\Vect{a} \times \Vect{b}) + (\Vect{a} \times \Vect{c})\).

Yes. By pointing the fingers of the right hand in the direction of \(\Vect{a}\) (along the negative x-axis) and curling them toward the direction of \(\Vect{b}\) (along the positive z-axis), the thumb points in the direction of the positive y-axis.

Yes. By pointing the fingers of the right hand in the direction of \(\Vect{v}\) (along the positive x-axis) and curling them toward the direction of \(\Vect{w}\) (along the negative z-axis), the thumb correctly points in the direction of the positive y-axis.

No. By pointing the fingers of the right hand in the direction of \(\Vect{v}\) (along the positive x-axis) and curling them toward the direction of \(\Vect{w}\) (along the negative z-axis), the thumb points in the direction of the positive y-axis.

No. By pointing the fingers of the right hand in the direction of \(\Vect{a}\) (along the negative x-axis) and curling them toward the direction of \(\Vect{b}\) (along the positive z-axis), the thumb points in the direction of the positive y-axis.

The area of a parallelogram defined by two vectors is equal to the magnitude of their vector product: Area \(= |\Vect{AB} \times \Vect{AC}|\).

1. Determine the component vectors $$ \Vect{AB} = \begin{pmatrix} 3-1 \\ 2-1 \\ 1-1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} $$ $$ \Vect{AC} = \begin{pmatrix} 1-1 \\ 3-1 \\ 3-1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} $$
2. Calculate the vector product \(\Vect{AB} \times \Vect{AC}\) $$ \begin{aligned}[t] \Vect{AB} \times \Vect{AC} &= \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} (1)(2) - (0)(2) \\ (0)(0) - (2)(2) \\ (2)(2) - (1)(0) \end{pmatrix} \\ &= \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} \end{aligned} $$
3. Calculate the magnitude of the resulting vector $$ \begin{aligned}[t] \text{Area} &= |\Vect{AB} \times \Vect{AC}|\\ &= \left\| \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} \right\| \\ &= \sqrt{2^2 + (-4)^2 + 4^2} \\ &= \sqrt{4 + 16 + 16} \\ &= \sqrt{36} \\ &= 6 \end{aligned} $$

The area of the parallelogram is 6 square units.

The area of a parallelogram defined by two vectors is equal to the magnitude of their vector product: Area \(= |\Vect{AB} \times \Vect{AC}|\).

1. Determine the component vectors $$ \Vect{AB} = \begin{pmatrix} 3-1 \\ 2-1 \\ 1-1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} $$ $$ \Vect{AC} = \begin{pmatrix} 1-1 \\ 3-1 \\ 3-1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} $$
2. Calculate the vector product \(\Vect{AB} \times \Vect{AC}\) $$ \begin{aligned}[t] \Vect{AB} \times \Vect{AC} &= \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} (1)(2) - (0)(2) \\ (0)(0) - (2)(2) \\ (2)(2) - (1)(0) \end{pmatrix} \\ &= \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} \end{aligned} $$
3. Calculate the magnitude of the resulting vector $$ \begin{aligned}[t] \text{Area} &= |\Vect{AB} \times \Vect{AC}|\\ &= \left\| \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} \right\| \\ &= \sqrt{2^2 + (-4)^2 + 4^2} \\ &= \sqrt{4 + 16 + 16} \\ &= \sqrt{36} \\ &= 6 \end{aligned} $$

The area of the parallelogram is 6 square units.

The area of a triangle defined by two vectors is half the magnitude of their vector product, since this triangle is half of the parallelogram built on those vectors: Area \(= \frac{1}{2}|\Vect{AB} \times \Vect{AC}|\).

1. Determine the component vectors $$ \Vect{AB} = \begin{pmatrix} 3-1 \\ 2-1 \\ 1-1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} $$ $$ \Vect{AC} = \begin{pmatrix} 1-1 \\ 3-1 \\ 3-1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} $$
2. Calculate the vector product \(\Vect{AB} \times \Vect{AC}\) $$ \begin{aligned}[t] \Vect{AB} \times \Vect{AC} &= \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} (1)(2) - (0)(2) \\ (0)(0) - (2)(2) \\ (2)(2) - (1)(0) \end{pmatrix} \\ &= \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} \end{aligned} $$
3. Calculate the magnitude and the area The magnitude of the cross product vector is: $$ \begin{aligned}[t] |\Vect{AB} \times \Vect{AC}| &= \left\| \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} \right\| \\ &= \sqrt{2^2 + (-4)^2 + 4^2} \\ &= \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \end{aligned} $$ The area of the triangle is half of this magnitude: $$ \text{Area of Triangle} = \frac{1}{2} |\Vect{AB} \times \Vect{AC}| = \frac{1}{2} \times 6 = 3 $$

The area of the triangle ABC is 3 square units.

The area of a triangle defined by two vectors is half the magnitude of their vector product, since this triangle is half of the parallelogram built on those vectors: Area \(= \frac{1}{2}|\Vect{AB} \times \Vect{AC}|\).

1. Determine the component vectors Since A is the origin, the components of the vectors are simply the coordinates of the points B and C. $$ \Vect{AB} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} $$ $$ \Vect{AC} = \begin{pmatrix} 1 \\ 2 \\ 6 \end{pmatrix} $$
2. Calculate the vector product \(\Vect{AB} \times \Vect{AC}\) $$ \begin{aligned}[t] \Vect{AB} \times \Vect{AC} &= \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 6 \end{pmatrix} \\ &= \begin{pmatrix} (2)(6) - (3)(2) \\ (3)(1) - (-1)(6) \\ (-1)(2) - (2)(1) \end{pmatrix} \\ &= \begin{pmatrix} 12 - 6 \\ 3 - (-6) \\ -2 - 2 \end{pmatrix} \\ &= \begin{pmatrix} 6 \\ 9 \\ -4 \end{pmatrix} \end{aligned} $$
3. Calculate the magnitude and the area The magnitude of the cross product vector is: $$ \begin{aligned}[t] |\Vect{AB} \times \Vect{AC}| &= \left\| \begin{pmatrix} 6 \\ 9 \\ -4 \end{pmatrix} \right\| \\ &= \sqrt{6^2 + 9^2 + (-4)^2} \\ &= \sqrt{36 + 81 + 16} = \sqrt{133} \end{aligned} $$ The area of the triangle is half of this magnitude: $$ \text{Area of Triangle} = \frac{1}{2} |\Vect{AB} \times \Vect{AC}| = \frac{\sqrt{133}}{2} $$

The area of the triangle ABC is \(\frac{\sqrt{133}}{2}\) square units.