Vector Algebra — Practice

Question 1

Find the angle between a⃗ = 2î − ĵ + k̂ and b⃗ = î + ĵ − 2k̂.

Accepted Answer

a⃗·b⃗ = (2)(1) + (−1)(1) + (1)(−2) = 2 − 1 − 2 = −1. |a⃗| = √(4+1+1) = √6. |b⃗| = √(1+1+4) = √6. cos θ = (a⃗·b⃗)/(|a⃗||b⃗|) = −1/(√6·√6) = −1/6. θ = cos⁻¹(−1/6). The angle between the vectors is cos⁻¹(−1/6).

Question 2

Find the area of a triangle with vertices A(1, 2, 3), B(2, −1, 4), C(4, 5, −1).

Accepted Answer

AB⃗ = B − A = (2−1)î + (−1−2)ĵ + (4−3)k̂ = î − 3ĵ + k̂. AC⃗ = C − A = (4−1)î + (5−2)ĵ + (−1−3)k̂ = 3î + 3ĵ − 4k̂. AB⃗ × AC⃗ = |î  ĵ  k̂; 1 −3 1; 3  3 −4|. = î[(−3)(−4)−(1)(3)] − ĵ[(1)(−4)−(1)(3)] + k̂[(1)(3)−(−3)(3)]. = î[12−3] − ĵ[−4−3] + k̂[3+9]. = 9î + 7ĵ + 12k̂. |AB⃗ × AC⃗| = √(81+49+144) = √274. Area of triangle = ½|AB⃗ × AC⃗| = ½√274 square units.

Question 3

Show that vectors a⃗ = î − 2ĵ + 3k̂, b⃗ = −2î + 3ĵ − 4k̂, c⃗ = î − 3ĵ + 5k̂ are coplanar.

Accepted Answer

For coplanarity: [a⃗ b⃗ c⃗] = a⃗·(b⃗ × c⃗) must equal 0. Compute using determinant: [a⃗ b⃗ c⃗] = |1  −2  3; −2  3  −4; 1  −3  5|. Expand along Row 1: = 1·|3 −4; −3 5| − (−2)·|−2 −4; 1 5| + 3·|−2 3; 1 −3|. = 1·(15−12) + 2·(−10+4) + 3·(6−3). = 1·(3) + 2·(−6) + 3·(3). = 3 − 12 + 9. = 0. Since [a⃗ b⃗ c⃗] = 0, the three vectors are COPLANAR. ✓

Practice — Vector Algebra

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