**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$.

I like how this is exactly what we were talking about in physics when we weren't paying attention. And 37 degrees isn't an ugly angle--it's a very useful angle.

ReplyDelete#1 and #2 should be easy.

ReplyDelete#3: Show that the vector from A to the orthocenter is perpendicular to the vector from B to C (i.e (A+B+C)-A is perpendicular to B-C). Easy by using the dot product.

#4: We know H = A+B+C and O is the origin. The centroid, otherwise known as the center of mass, if the average of the coordinates of the triangle (A+B+C)/3. We then have O, G, and H on the same line through the origin and OG = (A+B+C)/3, GH = (A+B+C)-(A+B+C)/3 = 2(A+B+C)/3. So OG:GH = 1:2.