Vector Algebra · Lesson 6 of 9

Projections and Dot Product Applications

“Use projections and dot-product identities to resolve vectors, calculate lengths and explain geometric inequalities.”

Learning Objectives

• Distinguish signed scalar projection, projection length and vector projection. • Resolve a vector into parts parallel and perpendicular to a direction. • Expand squared magnitudes using the dot product. • Explain Cauchy–Schwarz and the triangle inequality, including equality cases. • Use dot-product relations to solve multi-vector problems.

How much of a vector lies along a direction?

Imagine shining light perpendicular to a directed line so that a vector casts a shadow on it. The shadow records only the part along the line. Because the line has a positive direction, this part may be positive, negative or zero.

Definition

Signed scalar projection

For a unit vector $\overset{u}{^}$ along a directed line, the scalar projection of $a$ is $a \cdot \overset{u}{^} = ∣ a ∣ cos θ$ . It is a signed number. Its absolute value is the nonnegative length of the projected segment.

Projection onto a nonzero vectorLaTeX

comp_{b} a = \frac{a \cdot b}{∣ b ∣}, proj_{b} a = \frac{a \cdot b}{∣ b ∣ ^{2}} b

The first expression is a scalar. The second multiplies that scalar by the unit direction

b /∣ b ∣

, so it is a vector.

Positive and negative scalar projection— The dashed perpendicular finds the shadow. The projection vector follows the shadow arrow.

For a vector with positive projection, its projection vector points along $b$ ; for a negative projection, it points opposite to $b$ . If the vectors are perpendicular, both the scalar projection and projection vector are zero. If $a$ already lies on the projection line, its vector projection is $a$ itself, including when it points opposite to the chosen positive direction.

The familiar scalar components are projections too: $a_{1} = a \cdot \hat{i}$ , $a_{2} = a \cdot \hat{j}$ , $a_{3} = a \cdot \hat{k}$ . This explains why dividing these numbers by $∣ a ∣$ gives the cosines of the angles with the positive axes.

Example — A scalar projection and its vector

Problem
Project $a = 2 \hat{i} + 3 \hat{j} + 2 \hat{k}$ onto $b = \hat{i} + 2 \hat{j} + \hat{k}$ .

1.The dot product is $2 + 6 + 2 = 10$ and $∣ b ∣ = 6$ .
2.The signed scalar projection is $10/ 6 = 56 /3$ . It is positive.
3.The vector projection is $(10/6) (\hat{i} + 2 \hat{j} + \hat{k}) = (5/3) \hat{i} + (10/3) \hat{j} + (5/3) \hat{k}$ . This vector, unlike the scalar answer, includes direction.

Parallel and perpendicular parts

Once the parallel part is known, subtract it from the original vector to get the remaining part. This remainder is perpendicular to the chosen direction. The decomposition gives a geometric meaning to a scalar-product equation.

Resolving a vectorLaTeX

a = a_{∥} + a_{⊥}, a_{∥} = \frac{a \cdot b}{∣ b ∣ ^{2}} b, a_{⊥} = a - a_{∥}

For

b \neq = 0

, dotting the remainder with

b

gives

a \cdot b - (a \cdot b) = 0

Example — Resolving a vector into two parts

Problem
Write $β = 2 \hat{i} + \hat{j} - 3 \hat{k}$ as a sum of a vector parallel to $α = 3 \hat{i} - \hat{j}$ and a vector perpendicular to it.

1.Let the parallel part be $β_{1} = λ α$ . Then $β_{2} = β - λ α = (2 - 3 λ) \hat{i} + (1 + λ) \hat{j} - 3 \hat{k}$ .
2.Perpendicularity requires $α \cdot β_{2} = 3 (2 - 3 λ) - (1 + λ) = 5 - 10 λ = 0$ . Therefore $λ = 1/2$ .
3.Thus $β_{1} = (3/2) \hat{i} - (1/2) \hat{j}$ and $β_{2} = (1/2) \hat{i} + (3/2) \hat{j} - 3 \hat{k}$ .
4.Check that their sum is $β$ and that $3 (1/2) - 3/2 = 0$ for the perpendicularity test.

Lengths hidden inside dot products

The identity $v \cdot v = ∣ v ∣^{2}$ turns vector expressions into equations involving lengths. Distribute carefully, keeping both cross terms before combining them. This is the vector counterpart of expanding a square, with a dot product taking the place of ordinary multiplication.

Squared sum and differenceLaTeX

∣ a + b ∣^{2} ∣ a - b ∣^{2} = ∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2} = ∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2}

The cross terms combine because the dot product is commutative. Take a nonnegative square root only after finding the squared magnitude.

Example — Length without coordinates

Problem
Find $∣ a - b ∣$ if $∣ a ∣ = 2$ , $∣ b ∣ = 3$ and $a \cdot b = 4$ .

1.Expand the squared difference: $∣ a - b ∣^{2} = ∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2}$ .
2.Substituting gives $4 - 8 + 9 = 5$ .
3.Therefore $∣ a - b ∣ = 5$ . A magnitude is nonnegative, so there is no negative-square-root answer.

Example — Cancellation of cross terms

Problem
If $u$ is a unit vector and $(x - u) \cdot (x + u) = 8$ , find $∣ x ∣$ .

1.Expanding gives $x \cdot x + x \cdot u - u \cdot x - u \cdot u = 8$ .
2.The middle terms cancel by commutativity, and $u \cdot u = 1$ . Thus $∣ x ∣^{2} - 1 = 8$ .
3.Hence $∣ x ∣^{2} = 9$ and $∣ x ∣ = 3$ .

Why a product and a resultant have bounds

The cosine in a dot product can never have absolute value greater than one. This simple fact sets a universal bound on the scalar product. That bound, in turn, explains why taking a detour cannot produce a shorter route than the direct displacement.

Cauchy–Schwarz inequalityLaTeX

∣ a \cdot b ∣ \leq ∣ a ∣ ∣ b ∣

For nonzero inputs, divide by the two magnitudes to get

∣ cos θ ∣ \leq 1

. If an input is zero, both sides are zero.

Equality in this product bound holds for nonzero collinear vectors: they may point in the same or in opposite directions. The triangle inequality below has a more restrictive equality condition, because it bounds the magnitude of a sum.

Triangle inequalityLaTeX

∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣

Expand the square, use

a \cdot b \leq ∣ a ∣∣ b ∣

, and compare with

(∣ a ∣ + ∣ b ∣)^{2}

In detail, $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2} \leq ∣ a ∣^{2} + 2∣ a ∣∣ b ∣ + ∣ b ∣^{2}$ . Both compared quantities are nonnegative, so taking square roots preserves the inequality. For two nonzero vectors, equality requires $cos θ = 1$ : the directions must agree. If either vector is zero, equality also holds.

The direct route and the detour— The orange displacement AC is no longer than the two-edge route AB followed by BC.

Example — Collinearity from lengths

Problem
For $A (- 2, 3, 5)$ , $B (1, 2, 3)$ and $C (7, 0, - 1)$ , show that the points are collinear.

1.The successive vectors are $A B = (3, - 1, - 2)$ , $B C = (6, - 2, - 4)$ and $A C = (9, - 3, - 6)$ .
2.Their magnitudes are $14$ , $214$ and $314$ . Therefore $A C = A B + B C$ .
3.Equality in the triangle inequality means the two successive nonzero displacements point in the same direction. Thus the three points are collinear, with B between A and C.
4.The identity $A B + B C + C A = 0$ would hold for any three points and would not by itself establish this conclusion.

Several vectors in one relationship

When three vectors are related, expanding a squared sum may be more efficient than finding individual coordinates. Translate each stated perpendicularity into a zero dot product. Then group terms according to the information actually given.

Example — Perpendicular to the other two together

Problem
Let $∣ a ∣ = 3$ , $∣ b ∣ = 4$ , $∣ c ∣ = 5$ , and suppose each vector is perpendicular to the sum of the other two. Find $∣ a + b + c ∣$ .

1.The conditions say $a \cdot (b + c) = b \cdot (c + a) = c \cdot (a + b) = 0$ .
2.Expand the squared magnitude, grouping it as $∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + a \cdot (b + c) + b \cdot (c + a) + c \cdot (a + b)$ .
3.The grouped mixed terms vanish, leaving $9 + 16 + 25 = 50$ . Therefore the required magnitude is $52$ .

Example — A zero resultant constrains pairwise products

Problem
If $a + b + c = 0$ and the magnitudes are 3, 4 and 2, find $a \cdot b + b \cdot c + c \cdot a$ .

1.Dot the zero-resultant equation with itself. Its left side is zero.
2.Expand to get $0 = 9 + 16 + 4 + 2 (a \cdot b + b \cdot c + c \cdot a)$ .
3.Thus twice the required sum is $- 29$ , giving $- 29/2$ . This approach uses the complete relation rather than trying to determine all three products separately.

Common mistake

A signed scalar projection may be negative, but the length of a projected segment cannot be. Also, opposite nonzero vectors give equality in Cauchy–Schwarz but generally do not give equality in the triangle inequality.

Quick check

Project $\hat{i}$ as a signed scalar onto the direction $- \hat{i}$ . The answer is

Quick check

Which expression is the vector projection of a onto a nonzero b?

Quick check

If $∣ a ∣ = 2$ , $∣ b ∣ = 3$ and $a \cdot b = 1$ , then $∣ a + b ∣$ is

Quick check

For two nonzero vectors, equality in the triangle inequality occurs when they are

Quick check

If $a = a_{∥} + a_{⊥}$ is resolved relative to nonzero $b$ , which must hold?

Practice Problems

Find the signed scalar and vector projections of $\hat{i} - \hat{j}$ onto $\hat{i} + \hat{j}$ .
Find the scalar projection of $\hat{i} + 3 \hat{j} + 7 \hat{k}$ onto $7 \hat{i} - \hat{j} + 8 \hat{k}$ .
Resolve $3 \hat{i} + 4 \hat{j}$ into parts parallel and perpendicular to $\hat{i} + \hat{j}$ . Verify the perpendicular part using a dot product.
Given $∣ a ∣ = 4$ , $∣ b ∣ = 3$ and $a \cdot b = 6$ , find both $∣ a + b ∣$ and $∣ a - b ∣$ .
Prove the triangle inequality from the squared magnitude identity. Explain why opposite directions are not an equality case for two nonzero inputs.
Show that $A (1, 2, 7)$ , $B (2, 6, 3)$ and $C (3, 10, - 1)$ are collinear, and identify the point between the other two.
If three vectors have zero sum and squared magnitudes 4, 9 and 16, find the sum of their three pairwise dot products.

Key Takeaways

Scalar projection is signed; projection length is its absolute value; vector projection includes direction. Subtracting the parallel projection leaves a perpendicular remainder. A squared vector magnitude is a self-dot product, enabling useful algebraic expansions. Cauchy–Schwarz bounds a dot product; the triangle inequality bounds a resultant length. For nonzero inputs, equality in the triangle inequality requires the same direction.

Previous · Lesson 5

The Dot Product and Angles

Next · Lesson 7

The Cross Product and Orientation

How much of a vector lies along a direction?

Parallel and perpendicular parts

Lengths hidden inside dot products

Why a product and a resultant have bounds

Several vectors in one relationship

Quiz

Practice Problems

Key Takeaways