Question

Let points A(6,−1,3),B(0,0,5),C(−2,3,2) and D(4,2,0) be the vertices of parallelogram ABCD. What is the angle (in radians) at vertex B in parallelogram ABCD?

Answer 1

To find the angle at vertex B in parallelogram ABCD, we need to find vectors AB and BC and then use the dot product formula to find the angle.

Step-by-step explanation:

In order to find the angle at vertex B in parallelogram ABCD, we need to find vectors AB and BC and then use the dot product formula to find the angle. Here's how:

1. Find vector AB by subtracting the coordinates of point A from the coordinates of point B: AB = (0-6, 0-(-1), 5-3) = (-6, 1, 2)
2. Find vector BC by subtracting the coordinates of point B from the coordinates of point C: BC = (-2-0, 3-0, 2-5) = (-2, 3, 3)
3. Calculate the dot product of vectors AB and BC: AB · BC = (-6)(-2) + (1)(3) + (2)(3) = 12 + 3 + 6 = 21
4. Find the magnitudes of vectors AB and BC: |AB| = √((-6)^2 + 1^2 + 2^2) = √(36 + 1 + 4) = √41 and |BC| = √((-2)^2 + 3^2 + 3^2) = √(4 + 9 + 9) = √22
5. Use the formula cosθ = (AB · BC) / (|AB| |BC|) to find the cosine of the angle θ: cosθ = 21 / (√41 √22)
6. Finally, calculate the angle θ in radians using the inverse cosine function: θ = arccos(cosθ)

Using a calculator, the angle at vertex B in parallelogram ABCD is approximately 0.3167 radians.