Vector Algebra · Lesson 9 of 9

Chapter Summary and Practice

“Connect the whole chapter and practise choosing between geometric constructions, components, projections and vector products.”

Learning Objectives

• Connect vector definitions, direction descriptions and component calculations. • Choose an efficient method for segment division, angles, projections and areas. • Use dot and cross products with their correct scalar or vector meanings. • Check results using direction, units, limiting cases and geometric interpretation. • Solve mixed problems that combine several chapter concepts.

One subject viewed in several ways

A vector can be seen as a directed arrow, a displacement between points, or a component triple. These descriptions carry the same information in different forms. A useful solution often begins by choosing the form that makes the required relationship easiest to see.

Magnitude is a scalar and is never negative. Equality of vectors requires both equal magnitude and equal direction; equal lengths alone are insufficient. The zero vector has no unique direction, and this matters whenever a formula requires an angle, a unit direction or division by a magnitude.

Question being asked	Useful method	What the answer represents
Combine or reverse displacements	Triangle law, parallelogram law, component addition or subtraction	A vector
Find a unit direction or specified length	Divide by the magnitude, then scale if required	A vector
Locate a point on an extended or internal segment	Section formula with checked ratio and orientation	A position vector
Find an angle or test perpendicularity	Dot product and cosine	A scalar or angle
Measure a part along a direction	Scalar or vector projection	Specify which is required
Find a normal direction or an area	Cross product, then normalise or take magnitude	A vector for a normal; a scalar for area

Reconnecting the fundamental formulas

The formulas are most useful when their origins are remembered. Components arise by adding three perpendicular movements; the section formula follows by comparing segment vectors. The two products then measure different aspects of the angle: alignment through cosine and transverse spread through sine.

Components, lengths and directionLaTeX

a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, ∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}, \overset{a}{^} = \frac{a}{∣ a ∣} = l \hat{i} + m \hat{j} + n \hat{k}

Normalisation requires a nonzero vector. Here

(l, m, n)

are direction cosines and

l^{2} + m^{2} + n^{2} = 1

Points and section formulaLaTeX

P Q = q - p, r_{internal} = \frac{n p + m q}{m + n}, r_{external} = \frac{m q - n p}{m - n}

The distance ratio is

P R : R Q = m : n

. Internal division has

m, n > 0

; external division also requires

m \neq = n

. The midpoint is

(p + q) /2

The two productsLaTeX

a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = ∣ a ∣∣ b ∣ cos θ, ∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ

The dot product is a scalar and the cross product is a vector whose direction is determined by the right-hand rule. Angle interpretations require nonzero inputs.

Projection and geometric sizeLaTeX

proj_{b} a = \frac{a \cdot b}{∣ b ∣ ^{2}} b, A_{parallelogram} = ∣ a \times b ∣, A_{triangle} = \frac{1}{2} ∣ a \times b ∣

Projection onto

b

requires

b \neq = 0

. Area formulas use adjacent side vectors.

The dot product is commutative; the cross product changes sign when its inputs are reversed. Both distribute over addition. Useful length checks include $∣ a \cdot b ∣ \leq ∣ a ∣∣ b ∣$ and $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ . Opposite directions can give equality in the first inequality, but two nonzero vectors need the same direction for equality in the second.

Mixed examples with a reason for each method

A problem may combine several kinds of answer: a resultant vector, a unit direction and an area, for example. Keep these goals distinct and reuse intermediate results where appropriate. Check each stage against the geometry instead of treating every formula as an isolated rule.

Example — A diagonal direction and an area

Problem
The adjacent side vectors of a parallelogram are $a = 2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $b = \hat{i} - 2 \hat{j} - 3 \hat{k}$ . Find a unit vector along the diagonal from their common tail, and find the area.

1.The diagonal vector is the sum $d = a + b = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$ .
2.Its magnitude is $9 + 36 + 4 = 7$ , so the required unit vector is $(3 \hat{i} - 6 \hat{j} + 2 \hat{k}) /7$ .
3.The area uses a different operation: $a \times b = 22 \hat{i} + 11 \hat{j}$ .
4.Taking its magnitude gives $484 + 121 = 115$ square units. A unit direction and an area cannot be interchanged, even though they came from the same two vectors.

Example — An angled walking route

Problem
A girl walks 4 km west and then 3 km in a direction 30 degrees east of north. Find her resultant displacement.

1.Choose $\hat{i}$ east and $\hat{j}$ north. The first vector is $- 4 \hat{i}$ .
2.The second vector has east component $3 sin 3 0^{\circ} = 3/2$ and north component $3 cos 3 0^{\circ} = 33 /2$ .
3.The resultant is $- (5/2) \hat{i} + (33 /2) \hat{j}$ km, of magnitude $25/4 + 27/4 = 13$ km.
4.It points northwest. Its angle west of north satisfies $tan ϕ = 5/ (33)$ . The distance walked is 7 km, which is different from this displacement magnitude.

Example — Equal inclinations from perpendicular vectors

Problem
Let $a, b, c$ be mutually perpendicular vectors of the same nonzero magnitude $s$ . Show that $a + b + c$ is equally inclined to all three.

1.Let $v = a + b + c$ . Perpendicularity eliminates the mixed terms, so $∣ v ∣^{2} = 3 s^{2}$ and $∣ v ∣ = s 3$ .
2.The dot product $v \cdot a = s^{2}$ , and likewise for $b$ and $c$ .
3.Each angle therefore has cosine $s^{2} / (s 3 s) = 1/ 3$ . All three angles are equal.
4.In the coordinate-basis case, a vector equally inclined to all three positive axes has direction cosines all $1/ 3$ , or all $- 1/ 3$ for the opposite orientation. Equal angles should not be confused with independently choosing three signs.

Example — A unit-vector condition with an unknown

Problem
The dot product of $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of $2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $λ \hat{i} + 2 \hat{j} + 3 \hat{k}$ equals 1. Find $λ$ .

1.The sum is $s = (λ + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}$ , which is nonzero for every real $λ$ .
2.The unit vector is $s / (λ + 2)^{2} + 40$ . The stated dot product gives $(λ + 6) / (λ + 2)^{2} + 40 = 1$ .
3.The left numerator must be positive. Squaring gives $(λ + 6)^{2} = (λ + 2)^{2} + 40$ , hence $8 λ = 8$ and $λ = 1$ .
4.Check in the original equation: $7/ 9 + 40 = 1$ . The candidate satisfies the unsquared equation.

Example — Combining basis-vector operations

Problem
Evaluate $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ .

1.Evaluate the bracketed cross products first: $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{i} \times \hat{k} = - \hat{j}$ and $\hat{i} \times \hat{j} = \hat{k}$ .
2.The expression becomes $\hat{i} \cdot \hat{i} - \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{k}$ .
3.Each self-dot product is 1, so the result is $1 - 1 + 1 = 1$ . The middle minus sign comes from reversing the cyclic cross-product order.

Checking a solution before accepting it

A correct-looking algebraic expression may still answer the wrong question. Check whether the result should be a scalar, vector, angle, position or area. Next inspect its sign, direction, magnitude and units; then use a simple special case to see whether the formula behaves as expected.

Common mistake

Keep these distinctions visible: equality of magnitudes versus equality of vectors; distance versus displacement; initial versus terminal position; dot versus cross product; signed projection versus projection length; and a normal vector versus an area. Never normalise $0$ or a zero cross product.

For angles, check that a computed cosine lies in $[- 1, 1]$ and retain the supplementary possibility when using sine alone. For internal division, check that the point is between the endpoints. For areas, use side vectors from a common vertex and take the magnitude; a zero answer indicates a degenerate figure.

How vector language developed

The word vector comes from a Latin word meaning “to carry”. Wessel and Argand helped connect directed segments with complex numbers, and Hamilton used the term vector in his work on quaternions. Earlier work on combining forces, including Stevin’s, already used geometric addition. Gibbs and Heaviside later developed the scalar and vector operations into the vector analysis used today; Tait also helped establish its applications.

Putting the chapter together

In the mixed practice, decide the method before beginning the arithmetic. Name the required directed vectors, state any nonzero condition, and explain the final geometric conclusion. A short independent check often reveals more than repeating the same calculation.

Quick check

For nonzero $a, b$ , which pair of conditions is impossible?

Quick check

Two unit vectors have a unit-vector sum. Their angle is

Quick check

If nonzero vectors satisfy $∣ a \cdot b ∣ = ∣ a \times b ∣$ , their possible angles are

Quick check

A triangle has collinear vertices. Its area from the cross-product formula is

Quick check

For nonzero $a, b$ , $∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2}$ means

Quick check

If $P R : R Q = 1 : 2$ externally for position vectors $p, q$ , then $r$ is

Quick check

Which statement is valid for arbitrary vectors?

Practice Problems

Write a unit vector in the XY-plane making an angle of 30 degrees with the positive x-axis. Explain how changing the angle through a full turn covers all planar unit directions.
Find the scalar components, magnitude and direction cosines of the vector from $P (x_{1}, y_{1}, z_{1})$ to $Q (x_{2}, y_{2}, z_{2})$ , stating when the direction cosines exist.
A person walks 5 km south and then 4 km at 60 degrees east of north. Find the resultant vector and its magnitude using east and north as positive axes.
Does $∣ a ∣ = ∣ b ∣$ imply $a = b$ ? Give a counterexample. Then give an example of collinear unequal vectors.
Find all real $x$ such that $x (\hat{i} + \hat{j} + \hat{k})$ is a unit vector. Describe the two directions.
Find a vector of magnitude 5 in the direction of the resultant of $\hat{i} + 2 \hat{j} - \hat{k}$ and $2 \hat{i} - \hat{j} + 3 \hat{k}$ .
For $a = 2 \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} - \hat{j} + 2 \hat{k}$ , find a unit vector along $a - 2 b$ .
Show that $A (1, - 2, - 8)$ , $B (5, 0, - 2)$ and $C (11, 3, 7)$ are collinear and find the ratio in which $B$ divides $A C$ .
For $P (1, 2, 0)$ and $Q (4, 5, 3)$ , find the point $R$ dividing $P Q$ externally in ratio 1:2. Show that $P$ is the midpoint of $R Q$ .
For adjacent parallelogram sides $\hat{i} + 2 \hat{j} + \hat{k}$ and $2 \hat{i} - \hat{j} + \hat{k}$ , find a unit vector along the diagonal from their common tail and the area.
Find the direction cosines of a vector equally inclined to all three positive axes. Explain why the three signs cannot be chosen independently if the angles are equal.
For $a = \hat{i} + \hat{j}$ and $b = \hat{j} + \hat{k}$ , find a vector perpendicular to both with magnitude 6. Give both possible answers.
The dot product of $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along $(λ + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}$ is 1. Solve for $λ$ and check the original equation.
For three mutually perpendicular vectors of equal nonzero magnitude, prove that their sum is equally inclined to all three.
Prove that $(a + b) \cdot (a + b) = ∣ a ∣^{2} + ∣ b ∣^{2}$ holds exactly when $a \cdot b = 0$ . Explain the right-angle interpretation for nonzero inputs.
State the full interval of angles for which two nonzero vectors have nonnegative dot product. Include the endpoint cases.
Two unit vectors have a sum of magnitude one. Find their angle and explain why the angle is obtuse.
Evaluate $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ , showing the orientation signs.
For two nonzero vectors, find all angles satisfying $∣ a \cdot b ∣ = ∣ a \times b ∣$ . Explain why two answers are possible.
Resolve $2 \hat{i} + 3 \hat{j} + \hat{k}$ into a vector parallel to $\hat{i} + \hat{j}$ and a perpendicular vector. Verify their sum and perpendicularity.
Find the triangle area for $A (1, 1, 1)$ , $B (2, 3, 1)$ and $C (3, 1, 2)$ . Then explain what changing the vertex order does to the cross product and to the area.
If $a + b + c = 0$ , derive a formula for the sum of the three pairwise dot products in terms of their squared magnitudes.

Key Takeaways

Use the geometric, positional or component description that fits the question. Direction and magnitude together determine a vector; zero vectors require care in angle and normalisation formulas. Position differences and section formulas describe segments and division points. Dot products handle angles, projections, perpendicularity and squared lengths. Cross products handle oriented normals and, through magnitude, geometric areas. Every calculation should end with a result of the correct kind and a geometric check.

Previous · Lesson 8

Calculating Cross Products and Areas

End of chapter

One subject viewed in several ways

Reconnecting the fundamental formulas

Mixed examples with a reason for each method

Checking a solution before accepting it

Putting the chapter together

Quiz

Practice Problems

Key Takeaways