<Home>Teaching><MATH 32A (Fall 2021)>Week 2 Discussion Notes

Week 2 Discussion Notes

Table of Contents

Dot Product

The dot product of two vectors v⃗=⟨v1,v2,v3⟩\vec{v} = \ang{v_1, v_2, v_3} and w⃗=⟨w1,w2,w3⟩\vec{w} = \ang{w_1, w_2, w_3} is defined by

v⃗⋅w⃗=v1w1+v2w2+v3w3.\vec{v} \cdot \vec{w} = v_1w_1 + v_2w_2 + v_3w_3.

Properties

Commutativity

Notice that multiplication is commutative (i.e., ab=baab = ba), so

v⃗⋅w⃗=v1w1+v2w2+v3w3=w1v1+w2v2+w3v3=w⃗⋅v⃗.\begin{aligned} \vec{v} \cdot \vec{w} &= v_1w_1 + v_2w_2 + v_3w_3 \\ &= w_1v_1 + w_2v_2 + w_3v_3 \\ &= \vec{w} \cdot \vec{v}. \end{aligned}

This means the dot product is commutative or in other words, the order does not matter when you compute it.

Relationship to vector lengths

v⃗⋅v⃗=v12+v22+v32=∥v⃗∥2,\vec{v} \cdot \vec{v} = v_1^2 + v_2^2 + v_3^2 = \norm{\vec{v}}^2,

which looks almost like x⋅x=x2x \cdot x = x^2 for scalars.

Linearity

If u⃗,v⃗,w⃗\vec{u}, \vec{v}, \vec{w} are vectors and a,ba, b are scalars, then

u⃗⋅(av⃗+bw⃗)=au⃗⋅v⃗+bu⃗⋅w⃗,\vec{u} \cdot \p{a\vec{v} + b\vec{w}} = a\vec{u} \cdot \vec{v} + b\vec{u} \cdot \vec{w},

so you can distribute similar to regular addition and multiplication. Because the dot product is commutative, this is also true in the first slot of the dot product:

(av⃗+bw⃗)⋅u⃗=av⃗⋅u⃗+bw⃗⋅u⃗.\p{a\vec{v} + b\vec{w}} \cdot \vec{u} = a\vec{v} \cdot \vec{u} + b\vec{w} \cdot \vec{u}.

Cosine formula

If θ\theta is in the angle (chosen to be between 00 and π\pi) between the vectors v⃗\vec{v} and w⃗\vec{w}, then

v⃗⋅w⃗=∥v⃗∥∥w⃗∥cos⁡θ.\vec{v} \cdot \vec{w} = \norm{\vec{v}} \norm{\vec{w}} \cos\theta.

You can derive this using the law of cosines (you can ignore Case 2).

Relationship to orthogonality

We say that v⃗\vec{v} and w⃗\vec{w} are orthogonal (or perpendicular) if the angle between them is π2\frac{\pi}{2} (i.e., they make a right angle to each other). We can use the dot product to test for orthogonality, since if v⃗\vec{v} and w⃗\vec{w} are non-zero, then by the cosine formula,

v⃗⋅w⃗=∥v⃗∥∥w⃗∥cos⁡θ=0  ⟹  cos⁡θ=0.\vec{v} \cdot \vec{w} = \norm{\vec{v}} \norm{\vec{w}} \cos\theta = 0 \implies \cos\theta = 0.

Since θ\theta is between 00 and π\pi, this means that if v⃗⋅w⃗=0\vec{v} \cdot \vec{w} = 0, then θ=π2\theta = \frac{\pi}{2}. This means that if the dot product of two non-zero vectors is 00, then they are orthogonal to each other.

Projections

Given vectors u⃗\vec{u} and v⃗\vec{v}, we can draw this picture:

The green vector is the projection of u⃗\vec{u} along v⃗\vec{v} and is parallel to v⃗\vec{v}, so there exists a scalar λ\lambda such that

u⃗∥v⃗=λv⃗∥v⃗∥.\vec{u}_{\parallel\vec{v}} = \lambda \frac{\vec{v}}{\norm{\vec{v}}}.

I'm using a unit vector here because I care about the direction of v⃗\vec{v} (which is encapsulated in its unit vector) but I don't care about the length.

The red vector is perpendicular to v⃗\vec{v}, meaning u⃗⊥v⃗⋅v⃗=0\vec{u}_{\perp\vec{v}} \cdot \vec{v} = 0. From the picture, you can also see that

u⃗=u⃗∥v⃗+u⃗⊥v⃗=λv⃗∥v⃗∥+u⃗⊥v⃗,\vec{u} = \vec{u}_{\parallel\vec{v}} + \vec{u}_{\perp\vec{v}} = \lambda \frac{\vec{v}}{\norm{\vec{v}}} + \vec{u}_{\perp\vec{v}},

so to solve for λ\lambda, we can take the dot product with v⃗\vec{v} on both sides:

u⃗⋅v⃗=(λv⃗∥v⃗∥+u⃗⊥v⃗)⋅v⃗=λv⃗⋅v⃗∥v⃗∥+u⃗⊥v⃗⋅v⃗=λ∥v⃗∥  ⟹  λ=u⃗⋅v⃗∥v⃗∥.\begin{aligned} \vec{u} \cdot \vec{v} &= \p{\lambda \frac{\vec{v}}{\norm{\vec{v}}} + \vec{u}_{\perp\vec{v}}} \cdot \vec{v} \\ &= \lambda \frac{\vec{v} \cdot \vec{v}}{\norm{\vec{v}}} + \vec{u}_{\perp\vec{v}} \cdot \vec{v} \\ &= \lambda \norm{\vec{v}} \\ \implies \lambda &= \frac{\vec{u} \cdot \vec{v}}{\norm{\vec{v}}}. \end{aligned}

λ\lambda above is called the scalar component of u⃗\vec{u} along v⃗\vec{v}, and we can use this to write down the formula for the projection of u⃗\vec{u} along v⃗\vec{v}:

u⃗∥v⃗=(u⃗⋅v⃗∥v⃗∥)v⃗∥v⃗∥\vec{u}_{\parallel\vec{v}} = \p{\frac{\vec{u} \cdot \vec{v}}{\norm{\vec{v}}}} \frac{\vec{v}}{\norm{\vec{v}}}

Cross Product

Unlike the dot product, which gives you a scalar, the cross product of two vectors v⃗\vec{v} and w⃗\vec{w} give you another vector, denoted v⃗×w⃗\vec{v} \times \vec{w}. Geometrically, v⃗×w⃗\vec{v} \times \vec{w} is orthogonal to both v⃗\vec{v} and w⃗\vec{w}, and it follows the right-hand rule:

In the picture, v⃗\vec{v} points to the right, and w⃗\vec{w} points into the webpage. From the picture, you can already see that the cross product is not commutative.

Geometric Interpretation

Calculating the cross product is best understood by specifying its direction and magnitude. The direction is given by the right-hand rule, and the magnitude is given by the formula

∥v⃗×w⃗∥=∥v⃗∥∥w⃗∥∣sin⁡θ∣,\norm{\vec{v} \times \vec{w}} = \norm{\vec{v}} \norm{\vec{w}} \abs{\sin\theta},

there θ\theta is the angle between v⃗\vec{v} and w⃗\vec{w}. There's a nice geometric interpretation of this formula:

The area of a parallelogram is base×height\mathrm{base} \times \mathrm{height}. In the picture above, the base is ∥v⃗∥\norm{\vec{v}}, the length of v⃗\vec{v}, and the height is ∥w⃗∥sin⁡θ\norm{\vec{w}} \sin\theta, so ∥v⃗×w⃗∥\norm{\vec{v} \times \vec{w}} is the area of the parallelogram spanned by v⃗\vec{v} and w⃗\vec{w}.

Determinant Formula

If you want to calculate v⃗×w⃗\vec{v} \times \vec{w} algebraically, then you can use the determinant formula:

v⃗×w⃗=∣i^j^k^v1v2v3w1w2w3∣=∣v2v3w2w3∣i^−∣v1v3w1w3∣j^+∣v1v2w1w2∣k^=(v2w3−v3w2)i^−(v1w3−v3w1)j^+(v1w2−v2w1)k^.\begin{aligned} \vec{v} \times \vec{w} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} \\ &= \begin{vmatrix} v_2 & v_3 \\ w_2 & w_3 \end{vmatrix} \hat{i} - \begin{vmatrix} v_1 & v_3 \\ w_1 & w_3 \end{vmatrix} \hat{j} + \begin{vmatrix} v_1 & v_2 \\ w_1 & w_2 \end{vmatrix} \hat{k} \\ &= \p{v_2w_3 - v_3w_2} \hat{i} - \p{v_1w_3 - v_3w_1} \hat{j} + \p{v_1w_2 - v_2w_1} \hat{k}. \end{aligned}

Warning: Notice that there's a minus sign on j^\hat{j}.

Properties

Anticommutativity

From the picture above, you should be able to tell that w⃗×v⃗=−v⃗×w⃗\vec{w} \times \vec{v} = -\vec{v} \times \vec{w}.

Linearity

Like the dot product, you can distribute the cross product, but you have to make sure you keep things in order:

u⃗×(av⃗+bw⃗)=au⃗×v⃗+bu⃗×w⃗.\vec{u} \times \p{a\vec{v} + b\vec{w}} = a\vec{u} \times \vec{v} + b\vec{u} \times \vec{w}.

This works in the first slot too:

(av⃗+bw⃗)×u⃗=av⃗×u⃗+bw⃗×u⃗.\p{a\vec{v} + b\vec{w}} \times \vec{u} = a\vec{v} \times \vec{u} + b\vec{w} \times \vec{u}.

Relationship with parallel vectors

From either the formula for ∥v⃗×w⃗∥\norm{\vec{v} \times \vec{w}} or the determinant formula, you can show that

v⃗×v⃗=0⃗.\vec{v} \times \vec{v} = \vec{0}.

In fact, because the cross product is linear, given any scalar λ\lambda, you also get

v⃗×λv⃗=λ(v⃗×v⃗)=0⃗.\vec{v} \times \lambda\vec{v} = \lambda\p{\vec{v} \times \vec{v}} = \vec{0}.

This means that the cross product of two parallel vectors is always the zero vector 0⃗\vec{0}.