Vectors in space

NUST NET (Engineering) · Maths · Vectors. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

A vector in space is written a1i + a2j + a3k. Its magnitude comes from Pythagoras in three dimensions, and dividing a vector by its magnitude gives the unit vector in its direction. The components of that unit vector are the direction cosines.

There are two ways to multiply vectors. The scalar product (dot product) gives a number and is used for angles, perpendicularity, projections and work done. The vector product (cross product) gives a new vector at right angles to both and is used for areas.

Combining them gives the scalar triple product, a single determinant that measures volume and tests whether three vectors lie in one plane.

Rules to remember

Common mistake

Dropping the minus sign on the j component of a cross product. Check the result quickly: its dot product with each of the two original vectors must be 0.

Worked example

Find the area of the parallelogram with adjacent sides a = i + 2j + 2k and b = 2i + j − 2k.

  1. a × b = (2 × (−2) − 2 × 1)i − (1 × (−2) − 2 × 2)j + (1 × 1 − 2 × 2)k = −6i + 6j − 3k.
  2. Check: its dot product with a is −6 + 12 − 6 = 0.
  3. |a × b| = √(36 + 36 + 9) = √81 = 9.

Answer: 9 square units

Practice questions

Try each one, then open the answer.

1. The area of the triangle with adjacent sides a = i + j − k and b = i − j + k is (in square units)

  1. A
    2√2
  2. B
    √2
  3. C
    2
  4. D
    1
Show answer

Answer: B. a × b = −2j − 2k, whose magnitude is 2√2. The triangle is half the parallelogram, so its area is √2; 2√2 is the parallelogram.

2. If |a| = 2, |b| = 3 and the angle between a and b is 60°, then |a − b| equals

  1. A
    √19
  2. B
    √7
  3. C
    1
  4. D
    √13
Show answer

Answer: B. |a − b|2 = |a|2 + |b|2 − 2a·b = 4 + 9 − 2(2)(3)(1/2) = 7, so |a − b| = √7. √19 is |a + b|.

3. The unit vector in the direction of i + 2j − 2k is

  1. A
    (1/3)(i + 2j − 2k)
  2. B
    (1/9)(i + 2j − 2k)
  3. C
    (1/√5)(i + 2j − 2k)
  4. D
    (1/√3)(i + 2j − 2k)
Show answer

Answer: A. The magnitude is √(1 + 4 + 4) = 3, so the unit vector is the vector divided by 3. Dividing by 9 uses the square of the magnitude.

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