Vector Question from Hong Kong 1990

This question comes from the HKCEE 1990 paper (that’s before I was born!). This was an interesting vectors question that one of my IB students in Mathematics SL dug up for me. There aren’t any vectors in the HSC syllabus yet, but they soon will be…


In the figure above, \(\vec{x}\) and \(\vec{y}\) are unit vectors, each of which makes an angle \(\theta\) with \( \vec{z}\), where \(0<\theta<\frac{\pi}{2}\).

  1. Show that \(\vec{x}\cdot\vec{z}=\vec{y}\cdot\vec{z}\).
  2. Let \(\vec{z}=m\vec{x}+n\vec{y}\). By expressing  \(\vec{x}\cdot\vec{z}\) and \(\vec{y}\cdot\vec{z}\) in terms of \(m\), \(n\) and \(\theta\), show that \(m=n\).


Part 1 Solution

Since \(\vec{x}\) and \(\vec{y}\) are unit vectors, then they have a length of 1. Using this, we observe that:

\[\begin{align*}\vec{x}\cdot\vec{z}&=|\vec{x}||\vec{z}|\cos \theta\\&=|\vec{z}|\cos\theta\end{align*}\]

\[\begin{align*}\vec{y}\cdot\vec{z}&=|\vec{y}||\vec{z}|\cos \theta\\&=|\vec{z}|\cos\theta\end{align*}\]

Therefore: \[\vec{x}\cdot{z} = \vec{y}\cdot\vec{z}\]

Part 2 Solution

Without loss of generality, we can let \(\vec{x}=\left(\begin{matrix}1\\0\end{matrix}\right)\), which is a unit vector. We can do this due to the fact that any coordinate system can be rotated and translated so that the vector \(\vec{x}\) is mapped to \(\left(\begin{matrix}1\\0\end{matrix}\right)\).

Let \(\vec{y}=\left(\begin{matrix}y_1\\y_2\end{matrix}\right)\) for some \(y_1\), \(y_2\in\mathbb{R}\).

By considering the horizontal and vertical vector components of \(\vec{y}\) as shown below:

We can see that \(y_1=|\vec{y}|\cos 2\theta=\cos 2\theta\), since \(\vec{y}\) is a unit vector.

Now as \(\vec{z}=m\vec{x}+n\vec{y}\):


\[\begin{align*}\vec{x}\cdot\vec{z}&=\left(\begin{matrix}1\\0\end{matrix}\right)\cdot\left(\begin{matrix}m+ny_1\\ny_2\end{matrix}\right)\\&=m+ny_1\\&=m+n\cos 2\theta\end{align*}\]

\[\begin{align*}\vec{y}\cdot\vec{z}&=\left(\begin{matrix}y_1\\y_2\end{matrix}\right)\cdot\left(\begin{matrix}m+ny_1\\ny_2\end{matrix}\right)\\&=my_1+ny_1^2+ny_2^2\\&=my_1+n(y_1^2+y_2^2)\\&=my_1+n|\vec{y}|^2\\&=my_1+n\\&=m\cos 2\theta + n\end{align*}\]

From part 1, since \(\vec{x}\cdot\vec{z}=\vec{y}\cdot\vec{z}\):

\[\begin{align*}m+n\cos 2 \theta &= m\cos 2\theta +n\\ m-m\cos 2\theta &= n-n\cos 2\theta \\ m(1-\cos 2\theta) & = n (1-\cos 2 \theta)\\ \mbox{therefore } m&=n\end{align*}\]


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