### Number 204 of a series

Math time again. Some time ago I had the idea to clear out shelf space, so I dumped all the copies of Schaum’s Outline books that are available on Kindle. This is from Schaum’s *Vector Analysis*, Chapter 2.

2.11. Show that the vectors

A= −i+j,B= −i−j– 2k,C= 2j+ 2kform a right triangle.

**i**, **j**, and **k** are the unit vectors parallel to the 3D coordinate axes. To be clear, imagine each vector as a straight line oriented in space but movable. Can you move the three lines into position to form a right triangle? Give a mathematical proof.

The answer is in the book, but you know better than to look for the answer. Post your answer as a comment below.

## Update and solution

First it is needed to demonstrate the three vectors satisfy the Pythagorean Theorem. Square the lengths of all vectors and see if the sums of two add up to the third. So we have:

|A|² = 2

|B|² = 6

|C|² =8

That works.

To form a closed polygon (triangle) the sum of the vectors must be zero.

Sum = (-1 -1)**i** (1 -1 +2)**j** (-2 +2)**k**, which is not zero. But if you reverse **A**, then

Sum = (1 -1)**i** (-1 -1 +2)**j** (-2 +2)**k**, which is zero.

Also, note the dot product of **A** and **B** is zero. The two vectors are orthogonal.

So the three vectors form a right triangle.

A right triangle must satisfy a squared + b squared = c squared. C squared = 8.

My comment got lost?

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