Magnitude, position vector and direction cosines
A football pass tells how hard and in which direction the ball goes. Speed alone is a scalar, but speed with direction is a vector. In vectors we write both, using the unit vectors i, j, k along the three axes.
A vector is written as a = xi + yj + zk. Its magnitude is |a| = √(x² + y² + z²). The position vector of the point P(x, y, z) is the vector from the origin to P. The direction cosines are l = x/|a|, m = y/|a|, n = z/|a|, and they satisfy l² + m² + n² = 1.
Formula: |a| = √(x² + y² + z²), unit vector = a / |a|
Rule 1: Magnitude and unit vector — Find the length, then divide the vector by it.
- Easy: |3i + 4j| = 5, so the unit vector is (3i + 4j)/5.
- Medium: |2i − 2j + k| = 3, so the unit vector is (2i − 2j + k)/3.
- Exam-type: A vector of magnitude 10 along 2i − 2j + k is (10/3)(2i − 2j + k).
Rule 2: Direction cosines — Divide each component by the magnitude.
- Easy: For 2i + 3j + 6k the magnitude is 7, so the direction cosines are 2/7, 3/7, 6/7.
- Medium: For i + j + k the direction cosines are each 1/√3.
- Exam-type: The position vector of P(1, 2, 3) has magnitude √14.
A unit vector has magnitude 1, and l² + m² + n² = 1 always.
Forgetting the square root, or dividing by the sum of the components instead of the magnitude.
Divide by the length to make it a unit vector.
Practice questions
- 1Find the magnitude of 3i + 4j.
- 2Find the magnitude of 2i − 2j + k.
- 3Find the unit vector in the direction of 3i + 4j.
- 4Find the magnitude of the position vector of the point P(1, 2, 3).
- 5Find the direction cosines of 2i + 3j + 6k.
- 6Find the vector of magnitude 10 in the direction of 2i − 2j + k.
- 7For what value of λ is λi + 2λj + 2λk a unit vector?
- 8Check that the direction cosines of i + j + k satisfy l² + m² + n² = 1.
Show answers (try them yourself first)
- 15
- 23
- 3(3i + 4j)/5
- 4√(1 + 4 + 9) = √14
- 52/7, 3/7, 6/7
- 6(10/3)(2i − 2j + k) = (20/3)i − (20/3)j + (10/3)k
- 73|λ| = 1, so λ = 1/3 or −1/3
- 81/3 + 1/3 + 1/3 = 1
Types of vectors
Several people may walk in the same direction at different speeds, or in opposite directions. Describing them by magnitude and direction gives the standard types of vectors.
A zero vector has magnitude 0. A unit vector has magnitude 1. Equal vectors have the same magnitude and direction. A negative of a has the same magnitude and opposite direction. Collinear vectors are parallel to the same line. Coinitial vectors start at the same point.
Rule 1: Equal and negative — Compare components.
- Easy: a = 2i + 3j equals b = 2i + 3j, and −a = −2i − 3j.
- Medium: 3i − 2j + k has the negative −3i + 2j − k.
- Exam-type: If xi + 3j = 2i + yj, then x = 2 and y = 3.
Rule 2: Collinear vectors — One is a scalar multiple of the other.
- Easy: 4i + 6j = 2(2i + 3j), so these are collinear.
- Medium: 2i − 4j and −i + 2j are collinear, because the second is −½ times the first.
- Exam-type: 2i + 3j and 4i + xj are collinear if x = 6.
Collinear vectors have proportional components. Equal vectors have equal components.
Calling vectors equal because their magnitudes are equal. The direction must be equal too.
Same size and same way means equal, same line means collinear.
Practice questions
- 1Is i/√2 + j/√2 a unit vector?
- 2Is i + j a unit vector?
- 3Write the negative of 3i − 2j + k.
- 4Are 2i − 4j and −i + 2j collinear?
- 5Are 2i + 3j and 4i + 5j collinear?
- 6Find x if 2i + 3j and 4i + xj are collinear.
- 7Find x and y if xi + 3j = 2i + yj.
- 8What is the magnitude of the zero vector?
Show answers (try them yourself first)
- 1Yes, since √(1/2 + 1/2) = 1
- 2No, since its magnitude is √2
- 3−3i + 2j − k
- 4Yes, since the second is −½ times the first
- 5No, since 2/4 is not equal to 3/5
- 6x = 6
- 7x = 2 and y = 3
- 80