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Vectors in two dimensions

Further Mathematics · WAEC and JAMB · SS2 and SS3

Vectors carry a full question on most papers, usually magnitude, unit vector and the angle between two vectors. Everything rests on two formulas, so the marks are lost on arithmetic and on confusing position vectors with displacement vectors.

What you need to know

  • A vector has both magnitude and direction, while a scalar has magnitude only. Speed is a scalar, velocity is a vector, and a question that mentions direction or bearing is almost always a vector question.
  • In two dimensions a vector is written as ai + bj, where i points along the positive x-axis and j along the positive y-axis. The same vector may be written as a column with a on top and b below, and the two notations mean exactly the same thing.
  • The magnitude of a = ai + bj is |a| = sqrt(a^2 + b^2), which is just Pythagoras. So 3i + 4j has magnitude sqrt(9 + 16) = 5, and you should know the 3-4-5 and 5-12-13 triples on sight because examiners use them constantly.
  • A unit vector has magnitude 1 and is found by dividing a vector by its own magnitude. The unit vector along 3i + 4j is (3i + 4j)/5 = 0.6i + 0.8j, and you can check it because 0.6^2 + 0.8^2 = 1.
  • The position vector of a point P(x, y) is the vector from the origin, written OP = xi + yj. The displacement from P to Q is PQ = OQ - OP, which is always the far point minus the near point.
  • The scalar product of a = a1 i + a2 j and b = b1 i + b2 j is a.b = a1 b1 + a2 b2. The result is a number, not a vector, which is why it is also called the dot product.
  • The angle between two vectors comes from cos theta = (a.b)/(|a| |b|). Compute the dot product and the two magnitudes separately, write them down, then divide, rather than trying to do it in one line.
  • Two vectors are perpendicular exactly when their scalar product is zero. So to find k such that 2i + kj is perpendicular to 6i - 4j, set 12 - 4k = 0, giving k = 3.
  • Two vectors are parallel when one is a scalar multiple of the other, so a1/b1 = a2/b2. Parallel vectors give a dot product equal to plus or minus the product of the magnitudes, never zero.
  • The resultant of two or more vectors is their sum, found by adding the i components and adding the j components separately. For forces in equilibrium the resultant is the zero vector, so each component sum is zero.
  • The direction of a vector is usually given as the angle it makes with the positive x-axis, from tan theta = b/a. Take care with the quadrant: a negative i component with a positive j component puts the vector in the second quadrant, so add 180 degrees to the calculator value.
  • A bearing is measured clockwise from north and is always written with three figures, so due east is 090 degrees. Convert a bearing to an angle from the x-axis before using any vector formula, because the two are measured in opposite senses.
  • To show that three points are collinear, show that two of the displacement vectors between them are parallel and share a common point. For instance if AB = 2i + 3j and BC = 4i + 6j, then BC = 2 AB, so A, B and C lie on one straight line.

Key terms

Vector
A quantity having both magnitude and direction, such as displacement, velocity or force.
Position vector
The vector from the origin to a given point, so that point P(x, y) has position vector xi + yj.
Unit vector
A vector of magnitude one, obtained by dividing a vector by its own magnitude.
Scalar product
The number a1 b1 + a2 b2 formed from two vectors, equal to |a||b| cos theta.
Resultant
The single vector equivalent to the combined effect of two or more vectors added together.
Collinear points
Points lying on the same straight line, shown by proving that the displacement vectors joining them are parallel and share a point.

Formulae

  • |a| = sqrt(a1^2 + a2^2)
  • Unit vector of a = a/|a|
  • PQ = OQ - OP
  • a.b = a1 b1 + a2 b2
  • a.b = |a| |b| cos theta
  • cos theta = (a.b)/(|a| |b|)
  • Perpendicular vectors: a.b = 0
  • Parallel vectors: a1/b1 = a2/b2
  • Direction angle: tan theta = a2/a1
  • Resultant of a and b = (a1 + b1)i + (a2 + b2)j
  • Distance between P(x1, y1) and Q(x2, y2) = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Worked examples

Points P and Q have coordinates (2, -1) and (5, 3). Find the displacement vector PQ, its magnitude, the unit vector in the direction of PQ, and the angle PQ makes with the positive x-axis.

  1. PQ = OQ - OP = (5 - 2)i + (3 - (-1))j = 3i + 4j.
  2. Magnitude: |PQ| = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
  3. Unit vector: (3i + 4j)/5 = 0.6i + 0.8j. Check: 0.6^2 + 0.8^2 = 0.36 + 0.64 = 1.
  4. Direction: tan theta = 4/3, so theta = 53.13 degrees.
  5. Both components are positive, so the vector lies in the first quadrant and no adjustment to the angle is needed.

Find the angle between the vectors a = 3i + 4j and b = 5i + 12j, correct to one decimal place.

  1. Scalar product: a.b = (3)(5) + (4)(12) = 15 + 48 = 63.
  2. Magnitudes: |a| = sqrt(9 + 16) = 5 and |b| = sqrt(25 + 144) = sqrt(169) = 13.
  3. Apply cos theta = (a.b)/(|a||b|) = 63/(5 x 13) = 63/65.
  4. 63/65 = 0.96923, so theta = cos inverse of 0.96923.
  5. This gives theta = 14.25 degrees, which is 14.3 degrees to one decimal place.
  6. The answer is sensible because both vectors point up and to the right, so the angle between them must be small.

The mistake to avoid

Writing PQ as OP - OQ instead of OQ - OP, which reverses the vector and gives a direction angle 180 degrees out. The second standing error is taking the square root of the sum before squaring, so that |3i + 4j| is written as 3 + 4 = 7 rather than 5.

In the exam

Set out the dot product and the two magnitudes on three separate lines before combining them, because each one carries a mark. When the question asks for a direction, say clearly whether you are giving an angle from the x-axis or a bearing, and sketch the vector roughly so you can see which quadrant the answer must fall in before you trust your calculator.