Vector Algebra | ISC Class 12 Maths Notes

This note covers scalars and vectors, vector types, components, magnitude, direction cosines and ratios, vector addition and subtraction, scalar multiplication, unit vectors, position vectors, section formulae, dot products, scalar and vector projections, cross products, perpendicularity, collinearity, and areas of triangles and parallelograms.

What are scalars, vectors and directed line segments?

Definition: A scalar quantity has magnitude alone. A vector quantity has both magnitude and direction. Magnitude means the size or length of the quantity; it does not specify where the quantity points.

Distance and speed are scalars, whereas displacement and velocity are vectors. Distance measures the length travelled. Displacement describes the directed change from an initial position to a final position. Specifying a vector therefore requires attention to direction as well as size.

A directed line segment represents a vector geometrically. In A→B, A is the initial point, B is the terminal point, and the arrow points from A to B. The segment's length represents the vector's magnitude.

How is vector notation read?

We write a⃗ for a vector and |a⃗| for its magnitude. Vertical bars around a vector mean its length. The magnitude is non-negative: |a⃗| ≥ 0, where ≥ means greater than or equal to. A negative component does not mean a negative magnitude.

The notation A→B describes a directed segment; the notation AB, without an arrow, denotes its length when used as a distance. Reversing the arrow reverses the vector's direction. It leaves the distance between the endpoints unchanged.

The vectors considered here are free vectors. A free vector may be shifted parallel to itself without changing its magnitude or direction. This makes it possible to move a vector into position for addition while preserving the vector itself.

Keep the object and its measure distinct throughout a calculation. A vector answer needs a direction, often supplied through its components. A magnitude answer is a scalar and is reported as a non-negative value.

How do the different types of vectors compare?

Vectors can be classified by their length, direction, or initial point. These descriptions answer different questions. Two vectors may be collinear without being equal, and two equal vectors need not start at the same point.

Why are equality and equal magnitude different?

The zero vector is denoted by 0⃗. It has no definite direction; alternatively, it may be regarded as having any direction. It cannot be converted into a unit vector by dividing by its magnitude, because that would require division by zero.

For nonzero vectors, parallel or collinear directions may agree or oppose one another. The negative of A→B is B→A. Equality requires agreement in direction, so a vector and its negative are not equal unless both are zero.

The condition |a⃗| = |b⃗|, where b⃗ is a second vector, compares lengths alone. The stronger condition a⃗ = b⃗ compares the complete vectors. Their components, introduced next, provide a direct algebraic way of checking that stronger condition.

How are vectors expressed through components and position vectors?

Take three mutually perpendicular coordinate axes x, y and z, meeting at the origin O(0, 0, 0). Use a right handed arrangement: rotation from the positive x direction towards the positive y direction follows the right hand rule towards positive z.

The symbols î, ĵ and k̂ denote unit vectors along the positive x, y and z axes respectively. Each has magnitude one. They provide fixed reference directions for expressing vectors in space.

For a point P with coordinates (x, y, z), its position vector r⃗ is O→P. The coordinates are its signed scalar components along the respective axes. Its component form and magnitude are:

r⃗ = xî + yĵ + zk̂; |r⃗| = √(x² + y² + z²).

Here √ denotes the non-negative square root, and a superscript ² denotes a square. The terms xî, yĵ and zk̂ are the vector components, whereas x, y and z are scalar components. These are different kinds of quantities.

How does component form test equality?

Write a⃗ = a₁î + a₂ĵ + a₃k̂ and b⃗ = b₁î + b₂ĵ + b₃k̂. The subscripts 1, 2 and 3 label the x, y and z components respectively. Equality requires a₁ = b₁, a₂ = b₂ and a₃ = b₃.

For a vector in the xy-plane, the z component is zero, so its component form reduces to xî + yĵ. The magnitude formula reduces correspondingly. Negative components locate a direction opposite to the positive direction of the relevant axis.

Worked example 1. Compare a⃗ = î + 2ĵ and b⃗ = 2î + ĵ. Do they have equal magnitudes, and are they equal vectors?

Answer: |a⃗| = √(1² + 2²) = √5 and |b⃗| = √(2² + 1²) = √5. Their magnitudes agree. Their corresponding components differ, so the vectors are not equal.

How are vectors added and subtracted?

The triangle law adds vectors by placing the initial point of the second at the terminal point of the first. The resultant, meaning their vector sum, joins the first initial point to the last terminal point.

For points A, B and C, this gives A→B + B→C = A→C. The arrows determine the order of travel. If the third side is instead taken from C back to A, the three directed sides form a closed route and their sum is 0⃗.

Part (ii) shows A→B labelled a⃗ and B→C labelled b⃗. The arrow A→C is labelled a⃗ + b⃗, joining the first starting point to the final endpoint.

See Fig. 10.8 in your NCERT textbook

The parallelogram law starts with two coinitial vectors represented by adjacent sides of a parallelogram. The diagonal through their common initial point represents the sum. This gives the same resultant as the triangle law.

Property: Addition is commutative and associative

For vectors a⃗, b⃗ and a third vector c⃗, a⃗ + b⃗ = b⃗ + a⃗ and (a⃗ + b⃗) + c⃗ = a⃗ + (b⃗ + c⃗). Commutative means the order can be exchanged; associative means the grouping can be changed.

The zero vector is the additive identity: a⃗ + 0⃗ = a⃗. The vector −a⃗ is its additive inverse, because a⃗ + (−a⃗) = 0⃗. Subtraction is defined by adding the negative: a⃗ − b⃗ = a⃗ + (−b⃗).

In component form, add or subtract corresponding coefficients. Thus a⃗ + b⃗ has components a₁ + b₁, a₂ + b₂ and a₃ + b₃. For a⃗ − b⃗, replace each of these additions by subtraction, preserving the order.

Worked example 2. Find the unit vector in the direction of the sum of a⃗ = 2î + 2ĵ − 5k̂ and b⃗ = 2î + ĵ + 3k̂.

Answer: Their sum is c⃗ = 4î + 3ĵ − 2k̂. Its magnitude is √(16 + 9 + 4) = √29. Dividing the sum by its magnitude gives the unit vector (4î + 3ĵ − 2k̂)/√29.

How does scalar multiplication produce unit vectors?

Let λ, read as lambda, be a real scalar. The product λa⃗ is a vector obtained by scaling a⃗. For a nonzero vector, positive λ preserves direction and negative λ reverses direction. Taking λ = 0 produces the zero vector.

|λa⃗| = |λ||a⃗|. Here |λ| means the absolute value of the scalar λ. This distinction matters: the sign controls direction, but the absolute value controls the change in magnitude.

Multiplication acts on every component. If a⃗ = a₁î + a₂ĵ + a₃k̂, then λa⃗ = (λa₁)î + (λa₂)ĵ + (λa₃)k̂. The coefficients change together, so a nonzero scalar multiple remains collinear with the original nonzero vector.

Result: Normalising a nonzero vector

Normalising means dividing a nonzero vector by its magnitude to obtain a unit vector. Write â for the unit vector in the direction of a⃗. Then â = a⃗/|a⃗|. Its magnitude is one because numerator and denominator have the same magnitude.

To obtain a specified positive magnitude while retaining a direction, first normalise the given vector and then multiply by the required magnitude. Multiplying the original vector directly by that number generally produces the wrong length.

Worked example 3. Find the unit vector in the direction of a⃗ = 2î + 3ĵ + k̂.

Answer: |a⃗| = √(2² + 3² + 1²) = √14. Therefore â = (2î + 3ĵ + k̂)/√14. Dividing each component by the same positive magnitude preserves direction.

Worked example 4. Find a vector of magnitude 7 units in the direction of î − 2ĵ.

Answer: The given vector has magnitude √5. Its unit vector is (î − 2ĵ)/√5. Multiplying this unit vector by 7 gives (7/√5)î − (14/√5)ĵ, with the required magnitude and direction.

What are direction angles, direction cosines and direction ratios?

For a nonzero vector, let α, β and γ, read as alpha, beta and gamma, be its angles with the positive x, y and z axes respectively. These are its direction angles. Their cosines are the direction cosines.

Write l = cos α, m = cos β and n = cos γ. Here cos means the cosine trigonometric function. If a⃗ has components a₁, a₂ and a₃ and magnitude r = |a⃗|, then l = a₁/r, m = a₂/r, n = a₃/r.

Result: The direction cosines satisfy a normalisation condition

l² + m² + n² = 1. Substituting the component expressions gives (a₁² + a₂² + a₃²)/r². The numerator equals r² by the magnitude formula, so the quotient is one.

Direction ratios are numbers proportional to the direction cosines. A vector's scalar components provide direction ratios. Their squares do not, in general, add to one. The unit vector in the given direction is lî + mĵ + nk̂.

Retain the signs of the components when finding direction cosines. The magnitude used as the denominator is positive. A negative component therefore produces a negative direction cosine and indicates an angle greater than 90° and at most 180° with that positive coordinate axis.

Worked example 5. Find the direction ratios and direction cosines of a⃗ = î + ĵ − 2k̂.

Answer: Its components give direction ratios 1, 1, −2. Its magnitude is √(1 + 1 + 4) = √6. The direction cosines are 1/√6, 1/√6 and −2/√6; their squares add to 1.

The direction cosines also describe projections on the coordinate axes after normalisation. They encode direction independently of the vector's positive magnitude, while the original components retain both its size and its direction.

How is the vector joining two points found?

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points. The subscript 1 identifies coordinates of P, and the subscript 2 identifies coordinates of Q. Their position vectors start at the same origin O.

The vector directed from P to Q is P→Q = O→Q − O→P. Thus the terminal position vector is taken first and the initial position vector is subtracted. This order determines the direction of the answer.

P→Q = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂.

How are length and direction obtained from the joining vector?

Its magnitude is √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²). This is the distance between P and Q. Reversing the endpoints changes every component's sign but leaves the squared components and the distance unchanged.

Once the joining vector is known, its unit vector follows by division by its magnitude, provided P and Q are distinct. Its scalar components supply direction ratios. Dividing those components by the magnitude supplies direction cosines.

Worked example 6. Find the vector from P(2, 3, 0) to Q(−1, −2, −4).

Answer: Subtract the coordinates of P from those of Q: P→Q = (−1 − 2)î + (−2 − 3)ĵ + (−4 − 0)k̂ = −3î − 5ĵ − 4k̂.

A position vector and a joining vector should not be confused. O→Q locates Q relative to the origin. P→Q describes the change from P to Q. They have the same terminal point but generally different initial points.

How do the internal and external section formulae work?

Let p⃗ and q⃗ be the position vectors of points P and Q. Let R divide the line through them in a ratio m:n, where m and n now denote positive ratio numbers. Let r⃗ be the position vector of R.

In internal division, R lies between P and Q, and the distance ratio is PR:RQ = m:n. The formula is r⃗ = (m q⃗ + n p⃗)/(m + n). Notice that m multiplies the position vector of Q.

In external division, R lies outside the segment PQ and PR:QR = m:n. The formula is r⃗ = (m q⃗ − n p⃗)/(m − n), with m ≠ n, meaning m is not equal to n.

How can the weights be kept in the correct order?

Match the ratio to the named distances before substitution. Keep the order of the endpoint vectors consistent with that ratio. For external division, the subtraction in the numerator must correspond to the subtraction in the denominator.

The midpoint is the internal division point with equal ratio numbers. Its position vector is (p⃗ + q⃗)/2. Equal ratio numbers cannot be substituted into the external formula because its denominator would be zero.

Worked example 7. Let a⃗ and b⃗ be vectors, with O→P = 3a⃗ − 2b⃗ and O→Q = a⃗ + b⃗. Find the position vector of R dividing PQ in the ratio 2:1, internally and externally.

Answer: Internally, O→R = [2(a⃗ + b⃗) + (3a⃗ − 2b⃗)]/3 = 5a⃗/3. Externally, O→R = [2(a⃗ + b⃗) − (3a⃗ − 2b⃗)]/(2 − 1) = 4b⃗ − a⃗.

These formulae apply directly to vectors and component by component to coordinates. The same ratio and denominator must be used for every component; changing the weights between coordinates would describe a different point.

How does the scalar product determine angles and perpendicularity?

For nonzero vectors a⃗ and b⃗, let θ, read as theta, be the angle between them, taken from 0 to π radians. Here π radians equals 180°. Their scalar product, also called the dot product, is a⃗ · b⃗ = |a⃗||b⃗| cos θ.

The dot symbol · denotes this product. Its answer is a scalar. If either vector is zero, the angle is undefined, but the dot product is defined to be zero. The angle formula must therefore be used only when both magnitudes are nonzero.

In components, a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃. Multiply corresponding components and add. Thus cos θ = (a⃗ · b⃗)/(|a⃗||b⃗|) for two nonzero vectors.

Property: Dot products are commutative and distributive

The identities are a⃗ · b⃗ = b⃗ · a⃗ and a⃗ · (b⃗ + c⃗) = a⃗ · b⃗ + a⃗ · c⃗. Distributive means that the product can be expanded across a sum. Also, (λa⃗) · b⃗ = λ(a⃗ · b⃗). Taking a vector's dot product with itself gives a⃗ · a⃗ = |a⃗|².

Two nonzero vectors are perpendicular precisely when their dot product is zero. For unit coordinate vectors, î · î = ĵ · ĵ = k̂ · k̂ = 1, whereas dot products of different coordinate unit vectors are zero.

Worked example 8. Find the angle between a⃗ = î + ĵ − k̂ and b⃗ = î − ĵ + k̂.

Answer: a⃗ · b⃗ = 1 − 1 − 1 = −1. Both magnitudes are √3, giving cos θ = −1/3. Thus θ = cos⁻¹(−1/3), where cos⁻¹ denotes inverse cosine with values from 0 to π.

Expanding dot products also gives |a⃗ + b⃗|² = |a⃗|² + 2(a⃗ · b⃗) + |b⃗|² and |a⃗ − b⃗|² = |a⃗|² − 2(a⃗ · b⃗) + |b⃗|². Keep the sign of the middle term consistent with the operation.

What is the difference between scalar and vector projection?

A projection describes the part of a vector along a chosen directed line. Let b⃗ be a nonzero vector defining that direction, and let b̂ = b⃗/|b⃗| be its unit vector. Let θ be the angle between a⃗ and b⃗ when a⃗ is nonzero.

The scalar projection of a⃗ on b⃗ is the signed number s = a⃗ · b̂ = (a⃗ · b⃗)/|b⃗|. Here s names the scalar projection. For nonzero a⃗, the same quantity is |a⃗| cos θ.

The vector projection, denoted here by p⃗, includes direction: p⃗ = s b̂ = [(a⃗ · b⃗)/|b⃗|²]b⃗. It lies along the line defined by b⃗. Its magnitude is |s|, the absolute value of the scalar projection.

How does the sign affect the projected direction?

When s is positive, the projection vector points in the direction of b⃗. When s is negative, it points oppositely. When s is zero, the projection vector is 0⃗. The scalar projection can therefore be negative while the projection vector's magnitude cannot.

Worked example 9. Find the scalar projection of a⃗ = 2î + 3ĵ + 2k̂ on b⃗ = î + 2ĵ + k̂.

Answer: a⃗ · b⃗ = 2 + 6 + 2 = 10 and |b⃗| = √6. The scalar projection is 10/√6 = 5√6/3. The divisor is the magnitude of the vector onto which the projection is taken.

For the same data, the vector projection is [10/6](î + 2ĵ + k̂) = (5/3)(î + 2ĵ + k̂). This follows by multiplying the scalar projection by b̂. It is a vector, whereas 10/√6 is a scalar.

The scalar components of a vector are its scalar projections along the positive coordinate axes. This connects projection with the earlier component form: the axes supply the reference directions, and the signs of the components preserve directional information.

How is the vector product calculated and interpreted?

The vector product, or cross product, of nonzero vectors a⃗ and b⃗ is a⃗ × b⃗ = |a⃗||b⃗| sin θ n̂. Here × denotes the cross product, sin denotes sine, and n̂ is a unit vector perpendicular to both vectors.

For nonparallel vectors, the right hand rule fixes n̂: curl the fingers from a⃗ towards b⃗ through their angle θ; the thumb indicates n̂. The angle lies from 0 to π. For parallel vectors the cross product is zero.

The vectors a⃗ and b⃗ share an initial point, with θ marked between them. The unit vector n̂ points perpendicular to their plane, and a dotted continuation is labelled −n̂.

See Fig. 10.23 in your NCERT textbook

If either vector is zero, its cross product with the other is defined as 0⃗. For two nonzero vectors, a⃗ × b⃗ = 0⃗ precisely when they are parallel or collinear. This tests parallelism rather than perpendicularity.

Property: Reversing the order reverses the cross product

b⃗ × a⃗ = −(a⃗ × b⃗). Also, a⃗ × (b⃗ + c⃗) = a⃗ × b⃗ + a⃗ × c⃗, and (λa⃗) × b⃗ = λ(a⃗ × b⃗). The product is distributive but is not commutative in general.

The coordinate rules are î × ĵ = k̂, ĵ × k̂ = î and k̂ × î = ĵ. Reversing any of these orders changes the sign. The cross product of each coordinate unit vector with itself is 0⃗.

In components, a⃗ × b⃗ = (a₂b₃ − a₃b₂)î − (a₁b₃ − a₃b₁)ĵ + (a₁b₂ − a₂b₁)k̂. Calculate the î, ĵ and k̂ coefficients separately, taking particular care with the minus sign before the ĵ bracket.

Worked example 10. Find the cross product and its magnitude for a⃗ = 2î + ĵ + 3k̂ and b⃗ = 3î + 5ĵ − 2k̂.

Answer: a⃗ × b⃗ = (−2 − 15)î − (−4 − 9)ĵ + (10 − 3)k̂ = −17î + 13ĵ + 7k̂. Its magnitude is √(289 + 169 + 49) = √507.

How do cross products give areas and perpendicular unit vectors?

The magnitude |a⃗ × b⃗| = |a⃗||b⃗| sin θ has a direct area interpretation. If a⃗ and b⃗ represent adjacent sides of a parallelogram, it equals the area of that parallelogram. The product combines a base length with the corresponding perpendicular height.

For two sides of a triangle drawn from the same vertex, area = ½|a⃗ × b⃗|, where ½ means one half. With vertices A, B and C, use A→B and A→C so that both vectors start from A.

How should a coordinate area calculation proceed?

1. Identify the two adjacent side vectors, subtracting endpoint coordinates in a consistent order.
2. Calculate their cross product, retaining the sign of every component.
3. Find its magnitude by taking the square root of the sum of the squared components.
4. Use that magnitude for a parallelogram, or half of it for a triangle, and report square units.

Worked example 11. Find the area of the triangle with A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1).

Answer: A→B = ĵ + 2k̂ and A→C = î + 2ĵ. Their cross product is −4î + 2ĵ − k̂, with magnitude √(16 + 4 + 1) = √21. The triangle's area is √21/2 square units.

Worked example 12. Find the area of a parallelogram with adjacent sides a⃗ = 3î + ĵ + 4k̂ and b⃗ = î − ĵ + k̂.

Answer: a⃗ × b⃗ = 5î + ĵ − 4k̂. Its magnitude is √(25 + 1 + 16) = √42. The parallelogram's area is √42 square units, without the factor one half.

For nonparallel vectors, divide their cross product by its magnitude to obtain a perpendicular unit vector. Both (a⃗ × b⃗)/|a⃗ × b⃗| and its negative are perpendicular to both given vectors. The chosen order determines which direction is obtained first.

If the cross product is zero, this normalisation cannot be performed. For distinct points, a zero cross product of two joining vectors also supplies a test of collinearity: the vectors lie along the same line, and their triangle has zero area.

Glossary

• Scalar — A quantity described by magnitude alone, without requiring an associated direction.
• Vector — A quantity possessing both magnitude and direction, represented by a directed line segment.
• Magnitude — The non-negative length of a vector, measured between its initial and terminal points.
• Position vector — The vector directed from a chosen origin to the point being described.
• Zero vector — A vector of zero magnitude whose initial and terminal points coincide.
• Unit vector — A vector with magnitude one, used to specify a direction.
• Coinitial vectors — Two or more vectors whose initial points are the same point.
• Collinear vectors — Vectors parallel to the same line, irrespective of their magnitudes and directions.
• Scalar components — Signed coefficients of the coordinate unit vectors in a vector's component form.
• Direction cosines — Cosines of the angles a nonzero vector makes with the positive coordinate axes.
• Direction ratios — Numbers proportional to direction cosines, supplied by a vector's scalar components.
• Dot product — A scalar obtained by multiplying vector magnitudes and the cosine of their included angle.
• Scalar projection — The signed scalar component of a vector along a specified directed line.
• Vector projection — The vector along a chosen line obtained by multiplying scalar projection by that line's unit vector.
• Cross product — A vector perpendicular to both nonparallel factors, with magnitude determined by their magnitudes and included angle.

Common errors and misconceptions

• Misconception: Equal magnitudes make two vectors equal. Correct: Equality requires the same magnitude and direction; corresponding components must all agree.
• Misconception: Every vector can be normalised. Correct: The zero vector cannot be divided by its magnitude, because its magnitude is zero.
• Misconception: The vector from P to Q is the position of P minus the position of Q. Correct: Subtract the initial position vector from the terminal position vector.
• Misconception: Direction ratios must have squared sum one. Correct: Direction cosines satisfy this condition; direction ratios do not, in general.
• Misconception: A zero dot product proves that one factor is zero. Correct: Two nonzero perpendicular vectors also have zero dot product.
• Misconception: Cross products can be reversed without changing the answer. Correct: Reversal negates the cross product; retain the stated order.
• Misconception: Scalar projection is necessarily a non-negative length. Correct: It is signed; its absolute value is the magnitude of the projection vector.
• Misconception: The cross-product magnitude gives the area of a triangle directly. Correct: It gives the parallelogram area; the triangle area is half that magnitude.

Exam-style questions with model answers

Q1. State the magnitude and direction properties of the zero vector and a unit vector. [2 marks]

1. The zero vector has magnitude zero and cannot be assigned a definite direction.
2. A unit vector has magnitude one and specifies a direction; it is distinct from the zero vector.

Q2. Find a unit vector in the direction of a⃗ = 2î + 3ĵ + k̂. [3 marks]

1. Use the component formula for magnitude: |a⃗| = √(2² + 3² + 1²). All three scalar components contribute through their squares.
2. This gives |a⃗| = √14, which is nonzero. The given vector can therefore be divided by its magnitude.
3. The required unit vector is â = a⃗/|a⃗| = (2î + 3ĵ + k̂)/√14. Division by a positive magnitude preserves its direction.

Q3. Find the direction ratios and direction cosines of a⃗ = î + ĵ − 2k̂, and verify the squared-sum condition. [4 marks]

1. Read the scalar components from the given vector. They give direction ratios 1, 1 and −2 in the x, y and z directions.
2. Calculate the magnitude: |a⃗| = √(1² + 1² + (−2)²) = √6.
3. Divide each component by this magnitude. The direction cosines are l = 1/√6, m = 1/√6 and n = −2/√6.
4. Check the condition: l² + m² + n² = 1/6 + 1/6 + 4/6 = 1, as required.

Q4. Points P and Q have position vectors 3a⃗ − 2b⃗ and a⃗ + b⃗ respectively, where a⃗ and b⃗ are given vectors. Find the position vectors dividing PQ internally and externally in the ratio 2:1. [4 marks]

1. For internal division, use the weighted sum of the endpoint position vectors, with the weight 2 on Q and the weight 1 on P.
2. The internal position vector is [2(a⃗ + b⃗) + (3a⃗ − 2b⃗)]/(2 + 1) = 5a⃗/3.
3. For external division, use the corresponding difference in both numerator and denominator: [2(a⃗ + b⃗) − (3a⃗ − 2b⃗)]/(2 − 1).
4. The numerator simplifies to −a⃗ + 4b⃗ and the denominator is 1. The external position vector is therefore 4b⃗ − a⃗.

Q5. For a⃗ = î + ĵ − k̂ and b⃗ = î − ĵ + k̂, calculate the angle between the vectors. [3 marks]

1. Calculate the scalar product using corresponding components: a⃗ · b⃗ = 1(1) + 1(−1) + (−1)(1) = −1.
2. Each vector has magnitude √(1 + 1 + 1) = √3. Both are nonzero, so the formula for the angle between them applies.
3. Therefore cos θ = −1/(√3 × √3) = −1/3 and θ = cos⁻¹(−1/3), taking the angle between 0 and π radians.

Q6. Find both the scalar and vector projections of a⃗ = 2î + 3ĵ + 2k̂ on b⃗ = î + 2ĵ + k̂. Explain the distinction between your answers. [5 marks]

1. Calculate the dot product of the supplied vectors: a⃗ · b⃗ = 2(1) + 3(2) + 2(1) = 10. This result is a scalar.
2. The vector defining the projection direction is b⃗. Its magnitude is √(1² + 2² + 1²) = √6, and its squared magnitude is 6.
3. The scalar projection is s = (a⃗ · b⃗)/|b⃗| = 10/√6 = 5√6/3. It is positive for these vectors.
4. The vector projection is [(a⃗ · b⃗)/|b⃗|²]b⃗ = (5/3)(î + 2ĵ + k̂), found by multiplying s by the unit vector along b⃗.
5. The scalar answer gives the signed component along b⃗. The vector answer also specifies direction, here the same direction as b⃗ because the scalar projection is positive.

Q7. Find the area of the triangle with vertices A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1), using a cross product. [5 marks]

1. Take the two side vectors from the common vertex A. Subtracting A's coordinates from B's gives A→B = 0î + ĵ + 2k̂.
2. Subtracting A's coordinates from C's gives A→C = î + 2ĵ + 0k̂. These are the adjacent side vectors needed for the area formula.
3. Their cross product is [1(0) − 2(2)]î − [0(0) − 2(1)]ĵ + [0(2) − 1(1)]k̂ = −4î + 2ĵ − k̂.
4. Its magnitude is √((−4)² + 2² + (−1)²) = √21. This is the area of the parallelogram determined by the two side vectors.
5. A triangle occupies half that parallelogram, so its area is ½|A→B × A→C| = √21/2 square units.

Q8. Find the area of a parallelogram whose adjacent sides are a⃗ = 3î + ĵ + 4k̂ and b⃗ = î − ĵ + k̂. [3 marks]

1. Use the cross product of the given adjacent sides: a⃗ × b⃗ = [1(1) − 4(−1)]î − [3(1) − 4(1)]ĵ + [3(−1) − 1(1)]k̂.
2. This simplifies to 5î + ĵ − 4k̂. Its magnitude is √(5² + 1² + (−4)²) = √42.
3. The parallelogram area equals the magnitude of the cross product, so the required area is √42 square units. No factor of one half is used.

Key takeaways

• A vector combines magnitude and direction; equality requires both, while equality of magnitudes compares lengths alone.
• Component form separates a vector into scalar coefficients and fixed coordinate unit vectors, making algebraic operations systematic.
• Normalise a nonzero vector by dividing by its magnitude, then scale the unit vector if another magnitude is required.
• Direction cosines have squared sum one; scalar components supply direction ratios, whose squared sum is not generally one.
• Find a joining vector by subtracting the initial position vector from the terminal position vector, preserving the requested direction.
• Use the dot product for angles and perpendicularity, retaining the nonzero conditions when applying the angle formula.
• Scalar projection is signed; vector projection includes direction and has magnitude equal to the absolute scalar projection.
• The cross product tests collinearity and gives area through its magnitude, with a factor of one half for triangles.

Test yourself

Why does |a⃗| = |b⃗| not establish a⃗ = b⃗?

It establishes equal lengths but does not establish equal directions. Equal vectors require both magnitude and direction to agree.

Can the zero vector have a unit vector obtained by normalisation?

No. Normalisation divides by the magnitude, which is zero for the zero vector, so that division is undefined.

What is the sum of a triangle's directed sides taken in order?

The sum is the zero vector because the directed route returns to its initial point.

What condition do direction cosines l, m and n satisfy?

They satisfy l² + m² + n² = 1, since they are the components of a unit vector.

What is the position vector of the midpoint of points with position vectors p⃗ and q⃗?

It is (p⃗ + q⃗)/2, obtained from internal division with equal ratio numbers.

For two nonzero vectors, what does a zero dot product establish?

It establishes that the vectors are perpendicular, because the cosine of their included angle is zero.

For two nonzero vectors, what does a zero cross product establish?

It establishes parallelism or collinearity; their directions may be the same or opposite.

Why is the magnitude of a scalar projection different from its signed value?

The signed value records agreement or opposition to the chosen direction. Its absolute value gives the non-negative projection length.