Question 8
An equation of the line passing through the points $\text{A}(2,4,5)$ and $\text{B}(3,-2,1)$ is
Not quite. This line goes through A, but its direction $3\hat{\boldsymbol{i}}-2\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}$ is the position vector of B. The direction of the line is $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(1,-6,-4)$, and $(3,-2,1)$ is not a multiple of it.
Not quite. The direction $(1,-6,-4)$ is right, but the point $(-3,2,-1)$ has every sign of B changed. That is what you get by reading the point straight off numerators like $x-3$ without changing the signs. The line misses both A and B.
Not quite. Here the point and the direction have swapped jobs. $(1,-6,-4)$ is $\overrightarrow{AB}$, which should be the direction, and $(2,4,5)$ is the position vector of A. B fits this line, because all three fractions equal $1$ there, but at A they give $\frac12,\ \frac{10}{4},\ \frac95$, which aren’t equal.
That’s it. Step through the hints below to see why.
Drag the picture or use the sliders to turn it. Choose an option to draw its line in black. The ring is the line’s point at $t=0$, and the parameter slider moves the orange point along the line.
This line starts at A, but its direction $(3,-2,1)$ is the position vector of B. The direction has to be $\overrightarrow{AB}=(1,-6,-4)$, so this line misses B.
The direction $(1,-6,-4)$ is $\overrightarrow{AB}$, but the point $(-3,2,-1)$ is B with every sign changed. This line is parallel to AB and misses both points.
This line passes through B but misses A. Its point $(1,-6,-4)$ is the vector $\overrightarrow{AB}$ and its direction $(2,4,5)$ is the position vector of A, so the two jobs are swapped.
This line passes through B at $t=0$ and A at $t=1$. Its direction $(-1,6,4)$ is $-\overrightarrow{AB}$, which is parallel to AB. This is option (D).
A line needs a point on it and a direction. With two points, use either point and the vector between them.
$$\begin{aligned}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\&=\begin{pmatrix}3\\-2\\1\end{pmatrix}-\begin{pmatrix}2\\4\\5\end{pmatrix}=\begin{pmatrix}1\\-6\\-4\end{pmatrix}\end{aligned}$$
Read the point and the direction from each option. In Cartesian form $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ the point is $(x_0,y_0,z_0)$, so its signs are the opposite of the signs in the numerators.
$$\begin{aligned}\text{(A)}&:\ (2,4,5),\ (3,-2,1)\\\text{(B)}&:\ (-3,2,-1),\ (1,-6,-4)\\\text{(C)}&:\ (1,-6,-4),\ (2,4,5)\\\text{(D)}&:\ (3,-2,1),\ (-1,6,4)\end{aligned}$$
The direction must be a multiple of $(1,-6,-4)$, which leaves (B) and (D). The point has to be on the line as well. In (D) the point is B and the direction is $-\overrightarrow{AB}$. Check that A fits (D):
$$\frac{2-3}{-1}=\frac{4+2}{6}=\frac{5-1}{4}=1$$
The direction of the line is
$$\overrightarrow{AB}=\begin{pmatrix}3-2\\-2-4\\1-5\end{pmatrix}=\begin{pmatrix}1\\-6\\-4\end{pmatrix}$$
Option (D) uses the point $\text{B}(3,-2,1)$ and the direction $(-1,6,4)=-\overrightarrow{AB}$, which is parallel to $\overrightarrow{AB}$. A is on it too:
$$\frac{2-3}{-1}=\frac{4+2}{6}=\frac{5-1}{4}=1$$
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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