Theorem: A parallelogram is a rhombus if and only if the diagonals are perpendicular.
The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus, and determine the lengths of the sides and the angles between the sides.
So I found part A, the lengths of the sides, but I am having trouble finding part B, the angles between the sides.
For part A, the answer I have is [5(3)^0.5]/2, which is correct.
For part B, I think i have to use the formula
cos(theta)=(a.b)/(|a||b|), but when I plug in the given to the formula, I do not get the answer (same as back).
cos(theta)=(6-12+6)/(5(3)^0.5)
theta=90degrees
what did i do wrong?
The answers they have are 72 degrees and 108 degrees, and no i did not copy the question wrong.. I will re-type without checking above:
The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus and determine the lengths of the sides and the angles between the sides.
In my first answer, I had misread the question- the vectors you are given are the diagonals and they are clearly perpendicular. But you want the angles between the sides. You can get the lengths of the sides as the hypotenuses of right triangles where half the length of the diagonals are the lengths of the legs. Once you know the lengths of the sides, you can find the angles between the sides by using the "cosine law": where a and b are the lengths of the sides, c is the length of one of the diagonals, and C is the angle opposite that diagonal.