3D Vector Length and Spatial Distance Practice

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Designed for Grade 12 students. Calculate the component vector and its magnitude for each problem. Simplify all radicals where possible.

Find the length of the vector u = (0, 3, 4).
Find the magnitude of the vector v = (2, -3, 6).
Calculate the length of vector w = (1, 2, 2).
Find the displacement vector AB and its length for points A(1, 0, 3) and B(3, 4, 7).
Find the distance between point P(5, -1, 2) and point Q(2, 3, -10).
A box has one corner at the origin (0, 0, 0) and the opposite corner at (4, 7, 4). Find the length of the main interior diagonal.
Vector a has components (-2, 5, -4). Find its exact length.

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3D Vector Length and Spatial Distance Practice

Answer key

Find the length of the vector u = (0, 3, 4).
|u| = √(0² + 3² + 4²) = √(0 + 9 + 16) = √25 = 5.
Find the magnitude of the vector v = (2, -3, 6).
|v| = √(2² + (-3)² + 6²) = √(4 + 9 + 36) = √49 = 7.
Calculate the length of vector w = (1, 2, 2).
|w| = √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3.
Find the displacement vector AB and its length for points A(1, 0, 3) and B(3, 4, 7).
Vector AB = (3 - 1, 4 - 0, 7 - 3) = (2, 4, 4). Length |AB| = √(2² + 4² + 4²) = √(4 + 16 + 16) = √36 = 6.
Find the distance between point P(5, -1, 2) and point Q(2, 3, -10).
Vector PQ = (2 - 5, 3 - (-1), -10 - 2) = (-3, 4, -12). Distance = √((-3)² + 4² + (-12)²) = √(9 + 16 + 144) = √169 = 13.
A box has one corner at the origin (0, 0, 0) and the opposite corner at (4, 7, 4). Find the length of the main interior diagonal.
The diagonal vector is d = (4, 7, 4). Length |d| = √(4² + 7² + 4²) = √(16 + 49 + 16) = √81 = 9.
Vector a has components (-2, 5, -4). Find its exact length.
|a| = √((-2)² + 5² + (-4)²) = √(4 + 25 + 16) = √45 = 3√5.