# Vector Operations in 3D - Problem 1

Some more properties of vectors that are exactly the same as in two dimensions are; for vectors U and V as long they're not zero vectors, cosine of theta, the angle between them, is U.V divided by the length of U times the length of V. U is perpendicular to V if and only if, U.V equals 0.

U is parallel to V if and only if U is a scalar multiple of V. So if their sum K, for which U equals K times V. So let's use these results in a problem. It says that A equal the point 1, 3, 4, B is 3, -1, 0, and C is 3, 2, 6. Show that AB and AC are perpendicular.

First I have to get components for AB. AB is 3 minus 1, 2. -1 minus 3, -4 and 0 minus 4, -4. We need components for AC. AC is 3 minus 1, 2, 2 minus 3, -1 and 6 minus 4, 2. Now the test to see if two vectors are perpendicular you have to take their dot product. If the dot product is 0, then they are perpendicular. So let's take AB.AC. Remember the order you do this doesn't matter, because the dot product is commutative.

So you get 2 times 2, 4 plus -4 times -1, another 4, plus -4 times 2, -8. This is 0, therefore, AB is perpendicular to AC.

Now it says find the area of triangle ABC. Well what we just proved is that side AB is perpendicular to side AC in this triangle. So this is a right triangle, and that means that the area is 1/2 the length of the two legs, the two perpendicular legs; AB and AC. So it's going to be 1/2 AB times AC.

So let's find the lengths AB and AC. So we have the components up here. The length of AB first. It's the square root of 4 plus 16 plus 16, the squares of the components, 4 plus 16 plus 16. That's 32 plus 4, 36, and root 36 is 6. What about AC? Square root.

Here we go 2² is 4, plus -1² is 1, plus 2² is 4, so 4 plus 1, plus 4 that's 9, and the square root of 9 is 3. So the area is 1/2 AB which is 6, the length of AC 3, so I get 3 times 3, 9.

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