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I've got two quite basic questions about vectors. I'm sorry if it isn't right to put two questions at the same thread. I'm quite confused about the technique of solving such problems.

  1. Let $\vec v=(3,-4)$, $\vec u=(1,2)$. Find two vectors $\vec w_1, \vec w_2$ so that:

(i) $\vec u=\vec w_1+\vec w_2$.

(ii) $\vec v || \vec w_1$

(iii) $\vec v \bot \vec w_2$

And the second question:

  1. Show that for all four points $A,B,C,D \in \mathbb R^n$ from some $n$.:

$\vec {AB} \cdot \vec {CD} + \vec {AC} \cdot \vec {DB} + \vec {AD} \cdot \vec {BC}=0$

Thanks in advance for any help!

richard
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  • In general, it is preferred that separate questions be asked in separate posts. Also, have you given all the context for your second question? It seems like there may be something missing. – Cameron Buie Nov 03 '13 at 13:29
  • I should have added that the points are at $\mathbb R^n$ for some n. Thank you! – richard Nov 03 '13 at 13:34

3 Answers3

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For your first, $\vec w_1$ will be of the form $(3a,-4a)$ and $\vec w_2$ will be of the form $(4b,3b)$ for real numbers $a$ and $b$, so setting their sum equal to $\vec u$ you need to solve: $$3a+4b=1$$ $$-4a+3b=2$$

Added: For your second, you could start by showing that $$\sum_i (b_i-a_i)(d_i-c_i) + \sum_i (c_i-a_i)(b_i-d_i) + \sum_i (d_i-a_i)(c_i-b_i)=0$$ by looking at each $i$ individually

Henry
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Let $\mathbf{w}_1$ be the orthogonal projection of $\mathbf{u}$ on $\mathbf{v}$, and let $\mathbf{w}_2$ be the difference between $\mathbf{u}$ and $\mathbf{w}_1$. By the projection formula,

$$\mathbf{w}_1=\operatorname{proj}_\mathbf{v}\mathbf{u}=\frac{\mathbf{u}\cdot\mathbf{v}}{||\mathbf{v}||^2}\mathbf{v},$$

and

$$\mathbf{w}_2=\mathbf{u}-\mathbf{w}_1.$$

Librecoin
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The technique involved in the first question is known as orthogonal projection. The rough idea is that we "cast a shadow" of $\vec u$ onto the line generated by $\vec v,$ and call that shadow vector $\vec w_1,$ then let $\vec w_2=\vec u-\vec w_1,$ and the desired properties will be satisfied.

To do this, we find the portion of $\vec u$ that is parallel to $\vec v$. First, we take the dot product $$\vec u\cdot \vec v=3-8=-5.$$ This is simply $\lVert \vec u\rVert\lVert\vec v\rVert\cos\theta,$ where $\theta$ is the angle between the directions of the two vectors. All we really want is a vector of magnitude $\lVert \vec u\rVert|\cos\theta|$ in a direction along the line generated by $\vec v,$ though. To compensate for this, we find $\vec v\cdot\vec v=9+16=25,$ and let $$\vec w_1=\frac{\vec u\cdot\vec v}{\vec v\cdot\vec v}\vec v=\frac{-5}{25}\langle 3,-4\rangle=\left\langle-\frac35,\frac45\right\rangle.$$ This is readily confirmed to be parallel to $\vec v$, and as desired, we have $$\lVert\vec w_1\rVert=\left\lVert\frac{\lVert \vec u\rVert\lVert \vec v\rVert\cos\theta}{\lVert \vec v\rVert\lVert \vec v\rVert}\vec v\right\rVert=\frac{\lVert \vec u\rVert|\cos\theta|}{\lVert \vec v\rVert}\lVert \vec v\rVert=\lVert \vec u\rVert|\cos\theta|.$$ Then, letting $\vec w_2=\vec u-\vec w_1,$ we have $\vec u=\vec w_1+\vec w_2$ by definition, and $$\begin{align}\vec w_2\cdot\vec v &= \vec u\cdot\vec v-\vec w_1\cdot\vec v\\ &= \vec u\cdot\vec v-\frac{\vec u\cdot\vec v}{\vec v\cdot\vec v}\vec v\cdot\vec v\\ &= \vec u\cdot\vec v-\vec u\cdot\vec v\\ &= 0,\end{align}$$ so $\vec v\perp\vec w_2.$

For your second question, try writing $$A=(a_1,...,a_n)\\B=(b_1,...,b_n)\\C=(c_1,...,c_n)\\D=(d_1,...,d_n)$$

Then, for example, we have $$\vec{AB}=\langle b_1-a_1,...,b_n-a_n\rangle\\\vec{CD}=\langle d_1-c_1,...,d_n-c_n\rangle$$ See if you can use this with the definition of dot product to prove the identity.

Cameron Buie
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