Dot Gives Cosine and Cross Gives Sine: Pick by What the Question Tests
The dot product tests how parallel two vectors are, and the cross product tests how perpendicular they are. Many questions are solved just by knowing which product is zero in which situation.
Dot Product and Cross Product is the central idea in this question pattern. Identify the condition in the stem before applying the method below.
Dot product
it is a scalar and is zero when the vectors are perpendicular. A vector 5k is perpendicular to any vector in the xy-plane, such as 4i + 3j or 6i, but not to 7k.
Cross product
A × B has magnitude A B sinθ and direction given by the right-hand rule, perpendicular to both. With east, north and up as x, y and z, up × north points west. The magnitude |A × B| is the area of the parallelogram on the two vectors, and half of |d1 × d2| gives the area of a parallelogram with diagonals d1 and d2: for d1 = 3i + j - 2k and d2 = i - 3j + 4k, d1 × d2 = -2i - 14j - 10k with magnitude 10 sqrt(3), so the area is 5 sqrt(3).
Combining the two
If |A × B| = sqrt(3) A · B then tanθ = sqrt(3), so θ = 60° and |A - B|² = A² + B² - 2AB cos 60° = A² + B² - AB. Since sinθ and cosθ appear separately, divide the two magnitudes to obtain tanθ whenever both products are given.
OlympIQ exam lens
- ✓A · B = AB cosθ, a scalar; zero for perpendicular vectors
- ✓A × B = AB sinθ n̂, a vector, perpendicular to both; right-hand rule
- ✓Area of a parallelogram on A and B = |A × B|; with diagonals d1 and d2 it is (1/2)|d1 × d2|
- ✓Divide |A × B| by A · B to get tanθ
- ✓Orthogonality in components: set A · B = 0
Do not take the magnitude of the cross product as A B cosθ. The cross product uses sine, so it is zero for parallel vectors and largest for perpendicular ones, the reverse of the dot product.
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