Calculating for unknown vector.

[-Dragoon-]

Homework Statement

The vertices of a triangle are given by A(-3, 1, 2), B(1, -3, -1), and C(3, -1, -1). What are the coordinates of D(x, y, z) such that ABCD is a rectangle?

Homework Equations

\vec{AB} = \vec{CD}
\vec{AB} = \vec{OB} - \vec{OA}
\vec{CD} = \vec{OD} - \vec{OC}

The Attempt at a Solution

Since ABCD is a triangle, and two sides are equal:
\vec{AB} = \vec{CD} \vec{BC} = \vec{AD}
I decide to go with \vec{AB} = \vec{CD} to find vector D
First I find \vec{AB}:
\vec{AB} = \vec{OB} - \vec{OA}
\vec{AB} = (1 - (-3), -3 - 1, -1 - 2)
\vec{AB} = (4, -4, -3)

So, now that I have found vector AB, I can now find vector CD:
\vec{CD} = \vec{OD} - \vec{OC}
\vec{CD} = (x - 3, y - (-1), z - (-1)) = (4, -4, -3)
So, by using the values of vector AB to find the unknowns of vector D, I yield:
x = 7, y = -5, z = -3 (7, -5, -3). But, my book says the solution is x = -1, y = 3, z= 2 (-1, 3, 2). Instead of calculating vector AB = vector CD, they went with vector BC = vector AD. But, since it was a rectangle, shouldn't I have come to same answer as the textbook? Where did I go wrong? Thanks in advance.

 

[jbunniii]

What you have constructed is a parallelogram but not a rectangle. You have AB parallel to CD, and AC parallel to BD, but you do not have AC and BD perpendicular to AB and CD. (Simple check using dot product.)

Your problem is with setting AB = CD. If you sketch the three points A, B, C, you will see from the geometry that you need AB = DC if you are going to form a rectangle.

 

[-Dragoon-]

What you have constructed is a parallelogram but not a rectangle. You have AB parallel to CD, and AC parallel to BD, but you do not have AC and BD perpendicular to AB and CD. (Simple check using dot product.)

Your problem is with setting AB = CD. If you sketch the three points A, B, C, you will see from the geometry that you need AB = DC if you are going to form a rectangle.

Ah, I see. I forgot the law that vectors of equal magnitudes and same direction are equal. But, the same does not apply for a parallelogram? Because if I were to draw a parallelogram, wouldn't AB and CD have different directions?

 

[jbunniii]

Ah, I see. I forgot the law that vectors of equal magnitudes and same direction are equal. But, the same does not apply for a parallelogram? Because if I were to draw a parallelogram, wouldn't AB and CD have different directions?

You started with three points: A, B, and C. If you make AB one of the sides of the rectangle, then you have two choices for where to put D.

One choice is to make CD = AB, the other is to make DC = AB. Both choices will give you a vector parallel to AB and the same magnitude as AB.

In the first case, CD = AB, the other two sides will be AC and BD.

In the second case, DC = AB, the other two sides will be AD and BC.

Either way, you will get a parallelogram.

But a rectangle is a parallelogram with an additional constraint: the angles have to be right angles, or equivalently, two of the vectors must be perpendicular to the other two.

There's no general rule regarding which vector you should form (i.e. should AB equal CD, or DC) in order to get a rectangle. In fact, for most triangles, neither choice will result in a rectangle. In order to do that, you need to start with a right triangle. (i.e. one of the angles has to be a right angle). So a good first step in this problem would be to identify which angle is the right angle, then it should be clear how to choose D.

P.S. The above is probably going to be confusing unless you draw a picture. Draw a triangle with vertices A, B, C, then draw a line through C that is parallel to AB. Then you you can see that you have two choices for where to put D: to the left of C, or to the right of C, by the distance |AB|. You can see that only one choice (at most) will give you a rectangle.

 

[-Dragoon-]

You started with three points: A, B, and C. If you make AB one of the sides of the rectangle, then you have two choices for where to put D.

Okay, I follow you so far.

One choice is to make CD = AB, the other is to make DC = AB. Both choices will give you a vector parallel to AB and the same magnitude as AB.

In the first case, CD = AB, the other two sides will be AC and BD.


In the second case, DC = AB, the other two sides will be AD and BC.

Either way, you will get a parallelogram.

But, won't vector AC and vector BD be diagonal and have completely different directions, and thus not parallel to CD and AB? Or is this a technique that can be done with parallelograms? Does the direction of a vector not matter but only the magnitudes to decide they are equal in a parallelogram?

But a rectangle is a parallelogram with an additional constraint: the angles have to be right angles, or equivalently, two of the vectors must be perpendicular to the other two.

Understood

There's no general rule regarding which vector you should form (i.e. should AB equal CD, or DC) in order to get a rectangle. In fact, for most triangles, neither choice will result in a rectangle. In order to do that, you need to start with a right triangle. (i.e. one of the angles has to be a right angle). So a good first step in this problem would be to identify which angle is the right angle, then it should be clear how to choose D.

So, in order to do that, I would have to know the measurements of the vectors and solve accordingly?