<p>Electrical engineers are installing electricity cables on a building site. Points $(x,y,z)$ are defined relative to a main switching site at $(0,0,0)$, where units are metres. Cables are laid in straight lines and the widths of cables can be neglected.</p>
<p>An existing cable $C$ starts at the main switching site and goes in the direction $\left( \begin{matrix}3 \\1 \\-2 \\\end{matrix} \right)$. A new cable is installed which passes through points $P\,(1,2-1)$ and $Q\,(5,7,a)$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the value of $a$ for which $C$ and the new cable will meet.</div><p style="text-align: right;">[4]</p></li></ol>
<p>To ensure that the cables do not meet, the engineers use $a=-3$. The engineers wish to connect each of the points $P$ and $Q$ to a point $R$ on $C$.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">The engineers wish to reduce the length of cable required and believe in order to do this that angle $PRQ$ should be $90{}^\circ $. Show that this is not possible.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">The engineers discover that the ground between $P$ and $R$ is difficult to drill through and now decide to make the length of $PR$ as small as possible. Find the coordinates of $R$ in this case and the exact minimum length.</div><p style="text-align: right;">[5]</p></li></ol>

Electrical engineers are installing electricity cables on a building site. Points $(x,y,z)$ are defined relative to a main switching site at $(0,0,0)$, where units are metres. Cables are laid in straight lines and the widths of cables can be neglected.

An existing cable $C$ starts at the main switching site and goes in the direction $\left( \begin{matrix}3 \\1 \\-2 \\\end{matrix} \right)$. A new cable is installed which passes through points $P\,(1,2-1)$ and $Q\,(5,7,a)$.

Find the value of $a$ for which $C$ and the new cable will meet.

To ensure that the cables do not meet, the engineers use $a=-3$. The engineers wish to connect each of the points $P$ and $Q$ to a point $R$ on $C$.

The engineers wish to reduce the length of cable required and believe in order to do this that angle $PRQ$ should be $90{}^\circ $. Show that this is not possible.

The engineers discover that the ground between $P$ and $R$ is difficult to drill through and now decide to make the length of $PR$ as small as possible. Find the coordinates of $R$ in this case and the exact minimum length.

Vectors - General

Vectors

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2017 P1 Q10

$$l_c: \mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}, \quad \lambda \in \mathbb{R}$$

$$l_{PQ}: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 4 \\ 5 \\ a+1 \end{pmatrix}, \quad \mu \in \mathbb{R}$$

Lines will meet $\Rightarrow l_c = l_{PQ}$

$$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 4 \\ 5 \\ a+1 \end{pmatrix}$$

$$3\lambda = 1 + 4\mu \quad \Rightarrow \quad 3\lambda - 4\mu = 1 \quad --- (1)$$

$$\lambda = 2 + 5\mu \quad \Rightarrow \quad \lambda - 5\mu = 2 \quad --- (2)$$

From $(1)$ and $(2)$: $\lambda = \frac{3}{11}$, $\mu = -\frac{5}{11}$

$$-2\left(-\frac{3}{11}\right) = -1 - \frac{5}{11}(a+1)$$

Need to show that $\angle PRQ \neq 90° \Rightarrow \overrightarrow{PR} \cdot \overrightarrow{QR} \neq 0$

For some $\lambda \in \mathbb{R}$, $\overrightarrow{OR} = \lambda\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$

$$\overrightarrow{PR} = \lambda\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3\lambda - 1 \\ \lambda - 2 \\ -2\lambda + 1 \end{pmatrix}$$

$$\overrightarrow{QR} = \lambda\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 5 \\ 7 \\ -3 \end{pmatrix} = \begin{pmatrix} 3\lambda - 5 \\ \lambda - 7 \\ -2\lambda + 3 \end{pmatrix}$$

$$\overrightarrow{PR} \cdot \overrightarrow{QR} = \begin{pmatrix} 3\lambda - 1 \\ \lambda - 2 \\ -2\lambda + 1 \end{pmatrix} \cdot \begin{pmatrix} 3\lambda - 5 \\ \lambda - 7 \\ -2\lambda + 3 \end{pmatrix}$$

$$= 9\lambda^2 - 18\lambda + 5 + \lambda^2 - 9\lambda + 14 + 4\lambda^2 - 8\lambda + 3$$

$$= 14\left[\lambda^2 - \frac{35}{14}\lambda\right] + 22$$

$$= 14\left[\left(\lambda - \frac{35}{28}\right)^2 - \left(\frac{35}{28}\right)^2\right] + 22$$

$$= 14\left(\lambda - \frac{35}{28}\right)^2 + \frac{1}{8} > 0$$

Since $\overrightarrow{PR} \cdot \overrightarrow{QR} > 0$, hence $\angle PRQ$ cannot be $90°$.

Finding foot of perpendicular $P$ to $l_c$:

$$\overrightarrow{PR} = \begin{pmatrix} 3\lambda - 1 \\ \lambda - 2 \\ -2\lambda + 1 \end{pmatrix}$$

$$\overrightarrow{PR} \cdot \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} = 0$$

$$\begin{pmatrix} 3\lambda - 1 \\ \lambda - 2 \\ -2\lambda + 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} = 0$$

$$9\lambda - 3 + \lambda - 2 + 4\lambda - 2 = 0$$

$$14\lambda = 7 \quad \Rightarrow \quad \lambda = \frac{1}{2}$$

$$\overrightarrow{OR} = \frac{1}{2}\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$$

Coordinates of $R$: $\left(\frac{3}{2}, \frac{1}{2}, -1\right)$

$$|\overrightarrow{PR}| = \left| \begin{pmatrix} 3/2 - 1 \\ 1/2 - 2 \\ -1 + 1 \end{pmatrix} \right| = \left| \begin{pmatrix} 1/2 \\ -3/2 \\ 0 \end{pmatrix} \right|$$

$$= \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{3}{2}\right)^2} = \sqrt{\frac{5}{2}} \text{ units}$$

N2017 P1 Q10

N2017 P1 Q10

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