## Tuesday, November 22, 2005

### Vectors... wow! Topic: Vector Geometry. Level: AMC/AIME.

Problem #1: Find the area of the triangle formed by the heads of two vectors, $A = (5, 67^{\circ})$ and $B = (4, 37^{\circ})$, and the origin.

Solution: So, we have two vectors with pretty ugly angles, but we do notice that they differ by exactly $30^{\circ}$! Hmm, then we know two sides and the angle between them...

And consequently we remember the formula $[ABC] = \frac{1}{2}ab\sin{\angle C}$, which makes us happy.

Hence the area of the triangle is $\frac{1}{2}(5)(4)(\sin{30^{\circ}}) = 5$. QED.

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Comment: Notice that the formula $\frac{|A||B|\sin{\theta}}{2} = \frac{|A \times B|}{2}$, where $A \times B$ is the cross product of $A$ and $B$.

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Problem #2: Show that the vectors $A = (3, 4, 5)$ and $B = (2, -4, 2)$ are perpendicular.

Solution: Consider the dot product of $A$ and $B$, $A \cdot B$. The general definition of this is $A \cdot B = |A||B|\cos{\theta}$, where $\theta$ is the angle between the two vectors. Notice that if two vectors are perpendicular to each other, the angle $\theta = 90^{\circ} \Rightarrow \cos{\theta} = 0 \Rightarrow A \cdot B = 0$.

We can easily show that the dot product is distributive (left as an exercise to the reader), that is, $A \cdot (B+C) = A \cdot B + A \cdot C$.

Thus $A \cdot B = [(3, 0, 0)+(0, 4, 0)+(0, 0, 5)] \cdot [(2, 0, 0)+(0, -4, 0)+(0, 0, 2)]$. Distributing this, we see that only the terms that are parallel remain because all the axes are perpendicular to each other ($\theta = 90^{\circ}$ between any two axes).

So $A \cdot B = (3, 0, 0) \cdot (2, 0, 0)+(0, 4, 0) \cdot (0, -4, 0)+(0, 0, 5) \cdot (0, 0, 2) = 6 +(-16)+10 = 0$. So $A$ is perpendicular to $B$. QED.

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Comment: In effect, we have just shown that given any two vectors $X = (x_1, x_2, \ldots, x_n)$ and $Y = (y_1, y_2, \ldots, y_n)$, we have $X \cdot Y = x_1y_1+x_2y_2+\cdots+x_ny_n$, an interesting and useful result (these are all in $n$-spaces, where there actually exist $n$ dimensions).

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Practice Problem #1: Show that the dot product is distributive: $A \cdot (B+C) = A \cdot B+ A \cdot C$.

Practice Problem #2: Find the area of the triangle formed by the points $(5,2)$, $(7,8)$, and $(2,1)$.

Practice Problem #3: Show that, given three vectors of equal magnitude $A$, $B$, and $C$, the orthocenter of the triangle formed by the heads of each of the vectors is $A+B+C$.

Practice Problem #4: Using Practice Problem #3, show that the centroid ($G$), circumcenter ($O$), and orthocenter ($H$) of a triangle are collinear and that $OG:GH = 1:2$.