<p>The equation of a straight line $\ell $ is $r\,=\,\left( \begin{matrix}
1 \\
2 \\
3 \\
\end{matrix} \right)\,+\,t\left( \begin{matrix}
-1 \\
1 \\
1 \\
\end{matrix} \right)\,,\,\,\,t\,\in \mathbb{R}$.</p>
<p>The point $A\,$ on $\ell $ is given by $t\,=\,0\,$, and the origin of position vectors is $O$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Calculate the acute angle between $OA$ and $\ell $, giving your answer correct to the nearest degree.</div></li><li><div class="list-item-paragraph">Find the position vector of the point $P$ on $\ell $ such that $OP$ is perpendicular to $\ell $.</div></li><li><div class="list-item-paragraph">A point $Q$ on $\ell $ is such that the length of $OQ$ is $5$ units. Find the two possible position vectors of $Q$.</div></li><li><div class="list-item-paragraph">The points $R$ and $S$ on $\ell $ are given by $t\,=\,\lambda $ and $t\,=\,2\lambda $ respectively. Show that there is no value of $\lambda $ for which $OR$ and $OS$ are perpendicular.</div></li></ol>

The equation of a straight line $\ell $ is $r\,=\,\left( \begin{matrix}
1 \\
2 \\
3 \\
\end{matrix} \right)\,+\,t\left( \begin{matrix}
-1 \\
1 \\
1 \\
\end{matrix} \right)\,,\,\,\,t\,\in \mathbb{R}$.

The point $A\,$ on $\ell $ is given by $t\,=\,0\,$, and the origin of position vectors is $O$.

Calculate the acute angle between $OA$ and $\ell $, giving your answer correct to the nearest degree.

Find the position vector of the point $P$ on $\ell $ such that $OP$ is perpendicular to $\ell $.

A point $Q$ on $\ell $ is such that the length of $OQ$ is $5$ units. Find the two possible position vectors of $Q$.

The points $R$ and $S$ on $\ell $ are given by $t\,=\,\lambda $ and $t\,=\,2\lambda $ respectively. Show that there is no value of $\lambda $ for which $OR$ and $OS$ are perpendicular.

Vectors - General

Vectors

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N1993 P1 Q14

$t = 0$, $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$

$A\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$, $\mathbf{d}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

\[|\mathbf{a} \cdot \mathbf{d}| = |\mathbf{a}||\mathbf{d}|\cos\theta\]

\[\left|\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\right| = \sqrt{1^2 + 2^2 + 3^2}\sqrt{(-1)^2 + 1^2 + 1^2}\cos\theta\]

\[\cos\theta = \dfrac{|-1 + 2 + 3|}{\sqrt{14}\sqrt{3}} = \dfrac{4}{\sqrt{14}\sqrt{3}}\]

\[\theta = \cos^{-1}\left(\dfrac{4}{\sqrt{3}\sqrt{14}}\right) = 51.88^\circ\]

$\therefore$ Acute angle between $OA$ and $l = 52^\circ$ (nearest degree)

$P$ on $l \Rightarrow P$ satisfies $l$

$\Rightarrow$ for some $t \in \mathbb{R}$, $\overrightarrow{OP} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

Since $\overrightarrow{OP} \perp \mathbf{d} \Rightarrow \overrightarrow{OP} \cdot \mathbf{d} = 0$

\[\left[\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\right] \cdot \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = 0\]

\[\therefore \overrightarrow{OP} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \dfrac{4}{3}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 7/3 \\ 2/3 \\ 5/3 \end{pmatrix}\]

Position vector of $P = \dfrac{1}{3}\begin{pmatrix} 7 \\ 2 \\ 5 \end{pmatrix}$

For some $t \in \mathbb{R}$, $\overrightarrow{OQ} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

\[\left|\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\right| = 5\]

\[\left|\begin{pmatrix} 1 - t \\ 2 + t \\ 3 + t \end{pmatrix}\right| = 5\]

\[\sqrt{(1-t)^2 + (2+t)^2 + (3+t)^2} = 5\]

\[\sqrt{t^2 - 2t + 1 + t^2 + 4t + 4 + t^2 + 6t + 9} = 5\]

\[\therefore \overrightarrow{OQ} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + 1\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \text{ or } \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \dfrac{11}{3}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\]

\[= \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} \text{ or } \begin{pmatrix} 14/3 \\ -5/3 \\ -2/3 \end{pmatrix}\]

Possible position vectors of $Q$ are $\begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}$ or $\dfrac{1}{3}\begin{pmatrix} 14 \\ -5 \\ -2 \end{pmatrix}$

When $t = \lambda$, $\overrightarrow{OR} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

When $t = 2\lambda$, $\overrightarrow{OS} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + 2\lambda\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

\[\overrightarrow{OR} \cdot \overrightarrow{OS} = \left[\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\right] \cdot \left[\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + 2\lambda\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\right]\]

\[= \begin{pmatrix} 1 - \lambda \\ 2 + \lambda \\ 3 + \lambda \end{pmatrix} \cdot \begin{pmatrix} 1 - 2\lambda \\ 2 + 2\lambda \\ 3 + 2\lambda \end{pmatrix}\]

\[= (1 - \lambda)(1 - 2\lambda) + (2 + \lambda)(2 + 2\lambda) + (3 + \lambda)(3 + 2\lambda)\]

\[= 1 - 3\lambda + 2\lambda^2 + 4 + 6\lambda + 2\lambda^2 + 9 + 9\lambda + 2\lambda^2\]

Since $\overrightarrow{OR} \cdot \overrightarrow{OS} > 0$, $\therefore \overrightarrow{OR} \cdot \overrightarrow{OS} \neq 0$ for all $\lambda \in \mathbb{R}$.

Thus, there is no value of $\lambda$ for which $OS$ and $OR$ are perpendicular.

N1993 P1 Q14

N1993 P1 Q14

Video Solution: