Exercise practice: Vector Algebra

5.1 Exercise practice: the chapter's exercise sections

The chapter has four numbered exercise sections and one miscellaneous exercise. Below is a worked set from each, chosen to cover every question type the board repeats. State the formula first, then substitute, then simplify.

5.1.1 Exercise 10.1

E1.1 Classify as scalar or vector: 10 kg, 2 metres north-west, 40 degrees, 40 watt, 20 m/s squared. Mass, angle, power and acceleration magnitude are scalars; the displacement of 2 metres north-west is a vector. The point of the exercise is that a unit alone does not decide the type; direction does.

E1.2 Which vectors in a square ABCD are equal, coinitial and collinear but not equal? The sides AB and DC are equal and parallel, and so are AD and BC. Vectors with a common initial point, such as AB and AD at A, are coinitial. Vectors along the same side line, such as AB and BA, are collinear but not equal because their directions are opposite.

E1.3 Mark each claim true or false. A vector and its reverse are collinear (true). Equal magnitude is guaranteed for any two parallel vectors (false). Matching lengths force two vectors to be parallel (false). Parallel vectors with matching lengths are equal (false, because direction can still differ).

5.1.2 Exercise 10.2

E2.1 Find the magnitude of a = 2 i - 7 j - 3 k. The length is the square root of 4 + 49 + 9, which is root 62.

E2.2 Give a unit vector along a = 2 i + 3 j + k. Since |a| is root 14, divide the vector by root 14 to get the answer.

E2.3 Write down the direction cosines of 2 i + 3 j + 6 k. Its length is 7, so the three cosines are 2/7, 3/7 and 6/7.

E2.4 Show that A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear. The vector AB is i + 4 j - 4 k and BC is i + 4 j - 4 k. The two are equal, and B is common, so the three points lie on one line.

E2.5 Find the position vector of the midpoint of the segment joining P(2, 3, 4) and Q(4, 1, -2). The midpoint is the average of the position vectors, which gives 3 i + 2 j + k.

E2.6 Prove that 2 i - 3 j + 4 k and -4 i + 6 j - 8 k are collinear. The second is minus two times the first, so the two are scalar multiples of each other. That relation is the collinearity test, and the claim follows.

5.1.3 Exercise 10.3

E3.1 Find the angle between a = 2 i - j + k and b = 3 i - 2 j + k. The dot product is 6 + 2 + 1 = 9, and the lengths are root 6 and root 14, so the cosine is 9 divided by root 84.

E3.2 Find the projection of a = i - j on b = i + j. The dot product is 1 - 1 = 0, so the projection is zero and the vectors are perpendicular.

E3.3 Find λ if a + λb is perpendicular to c, for the vectors given in the question. Put the dot product (a + λb) dot c equal to zero, expand as a dot c plus λ times b dot c, and solve the linear equation for lambda.

E3.4 If |a| = 3, |b| = 4 and a dot b = 6, find |a - 2b|. Use the expansion: |a - 2b| squared equals 9 plus 4 times 16 minus 4 times 6, that is 9 plus 64 minus 24, which is 49, so the length is 7.

5.1.4 Exercise 10.4

E4.1 Find a cross b for a = i - j + 2 k and b = 2 i + 3 j - k. The determinant gives i(1 - 6) minus j(-1 - 4) plus k(3 + 2), which is -5 i + 5 j + 5 k.

E4.2 Find a unit vector perpendicular to both a = 2 i + j + k and b = i + 2 j + k. The cross product is -i - j + 3 k, with magnitude root 11, so the unit vector is that vector divided by root 11.

E4.3 A triangle has vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). Find its area. Take AB as i + 2 j + 3 k and AC as 4 j + 3 k. Their cross product is 6 i - 3 j + 4 k with magnitude root 61, so the area is half of root 61.

E4.4 Find the area of the parallelogram with adjacent sides 2 i + 3 j + 4 k and 3 i + j + 2 k. The cross product is 2 i + 8 j - 7 k, so the area is the square root of 117, which is 3 root 13.

5.1.5 Miscellaneous exercise

M1 Find a unit vector in the XY-plane making an angle of 30 degrees with the positive x-axis. In the plane the unit vector is cos theta i plus sin theta j with theta equal to 30 degrees, so it is (root 3 / 2) i + (1 / 2) j.

M2 The points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) lie on one line. Prove it and state the ratio in which B cuts AC. AB is 4 i + 2 j + 6 k and BC is 6 i + 3 j + 9 k, so BC is three halves of AB; the points are collinear and B cuts AC internally in the ratio 2 : 3.

M3 The sides 2 i - 4 j + 5 k and i - 2 j - 3 k are adjacent sides of a parallelogram. Give a unit vector along one diagonal and the area. One diagonal is the sum, 3 i - 6 j + 2 k, whose magnitude is 7, so the unit vector along it is that vector over 7. The other diagonal is the difference. The area is the magnitude of the cross product of the two adjacent sides.

M4 The three corners of a triangle have position vectors (1, -1, 2), (2, 1, -1) and (3, -1, 2). Find its area. The sides from the first vertex are i + 2 j - 3 k and 2 i + 0 j + 0 k, whose cross product is 6 j + 4 k with magnitude root 52, so the area is root 13.