## MATH1002 Quizzes

Quiz 4: Lines and planes
Question 1 Questions
Find the equation of the line joining $P\left(2,1,-1\right)$ and $Q\left(0,3,1\right)$ in parametric vector form. Exactly one option must be correct)
 a) $\mathbf{r}=2\mathbf{i}+\mathbf{j}-\mathbf{k}+t\left(3\mathbf{j}+\mathbf{k}\right)$ b) $\mathbf{r}=3\mathbf{j}+\mathbf{k}+t\left(2\mathbf{i}+\mathbf{j}-\mathbf{k}\right)$ c) $\mathbf{r}=2\mathbf{i}+\mathbf{j}-\mathbf{k}+t\left(-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}\right)$ d) $\mathbf{r}=-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}+t\left(-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}\right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
A vector parallel to the line is the vector $\stackrel{⃗}{PQ}=\stackrel{⃗}{OQ}-\stackrel{⃗}{OP}=-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}$.
Hence the line can be represented by the parametric vector equation $\mathbf{r}=2\mathbf{i}+\mathbf{j}-\mathbf{k}+t\left(-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}\right).$ Note that there are many other ways of giving a vector parametric equation for this line. For example, instead of using the position vector of the known point $P$ in the equation, we could have used the position vector of $Q$, to give another version of the equation, $\mathbf{r}=3\mathbf{j}+\mathbf{k}+t\left(-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}\right).$
Choice (d) is incorrect
A line has Cartesian equations $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+6}{-1}.$ A vector parallel to the line is: Exactly one option must be correct)
 a) $3\mathbf{i}+2\mathbf{j}-\mathbf{k}$ b) $2\mathbf{i}-\mathbf{j}+6\mathbf{k}$ c) $\frac{2}{3}\mathbf{i}+\frac{1}{2}\mathbf{j}+6\mathbf{k}$ d) $\mathbf{i}+\mathbf{j}-7\mathbf{k}$

Choice (a) is correct!
When the cartesian equations of a line are given in the form $\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c},$ we can identify the coordinates of one point on the line (namely $\left(\alpha ,\beta ,\gamma \right)\right)$ and a vector parallel to the line, namely $a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$. In this case, $a=3,\phantom{\rule{0.3em}{0ex}}b=2,\phantom{\rule{0.3em}{0ex}}c=-1.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Given the parametric scalar equations of a line, $x=3+5t,\phantom{\rule{1em}{0ex}}y=2-t,\phantom{\rule{1em}{0ex}}z=3+2t,$ find the cartesian equations of the same line. Exactly one option must be correct)
 a) $\frac{x+3}{5}=\frac{y+2}{-1}=\frac{z+3}{2}$ b) $\frac{x-5}{3}=\frac{y+1}{2}=\frac{z-2}{3}$ c) $\frac{x}{15}=\frac{y}{-2}=\frac{z}{6}$ d) $\frac{x-3}{5}=\frac{y-2}{-1}=\frac{z-3}{2}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
In the three parametric scalar equations, the coefficients $5,\phantom{\rule{0.3em}{0ex}}-1,\phantom{\rule{0.3em}{0ex}}2$ of $t$ are the components of a vector parallel to the line. (Therefore the line has direction parallel to $5\mathbf{i}-\mathbf{j}+2\mathbf{k}$.) The numbers $5,\phantom{\rule{0.3em}{0ex}}-1,\phantom{\rule{0.3em}{0ex}}2$ become the denominators in the cartesian equations of the line. The constants $3,\phantom{\rule{0.3em}{0ex}}2,\phantom{\rule{0.3em}{0ex}}3$ in the scalar parametric equations are (in that order) the coordinates of a particular point $\left(3,2,3\right)$ on the line. This information allows us to form the numerators in the cartesian form, $x-3,\phantom{\rule{0.3em}{0ex}}y-2,\phantom{\rule{0.3em}{0ex}}z-3$.
Find the equation of the line through $\left(1,2,1\right)$ parallel to the line
$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-5}{4}.$
Exactly one option must be correct)
 a) $\mathbf{r}=\mathbf{i}+2\mathbf{j}+\mathbf{k}+t\left(2\mathbf{i}+\mathbf{j}+5\mathbf{k}\right)$ b) $\mathbf{r}=\mathbf{i}+2\mathbf{j}+\mathbf{k}+t\left(6\mathbf{i}+2\mathbf{j}+8\mathbf{k}\right)$ c) $\mathbf{r}=2\mathbf{i}+\mathbf{j}+5\mathbf{k}+t\left(3\mathbf{i}+\mathbf{j}+4\mathbf{k}\right)$ d) $\mathbf{r}=\mathbf{i}+2\mathbf{j}+\mathbf{k}+t\left(3\mathbf{i}-\mathbf{j}+4\mathbf{k}\right)$

Choice (a) is incorrect
Choice (b) is correct!
A vector parallel to the line is $3\mathbf{i}+\mathbf{j}+4\mathbf{k}$ (from the denominators in the cartesian form). So the line is also parallel to $6\mathbf{i}+2\mathbf{j}+8\mathbf{k}$. Therefore since the line passes through the point $\left(1,2,1\right)$, it has vector parametric equation $\mathbf{r}=\mathbf{i}+2\mathbf{j}+\mathbf{k}+t\left(6\mathbf{i}+2\mathbf{j}+8\mathbf{k}\right)$.
Choice (c) is incorrect
Choice (d) is incorrect
Find the parametric vector equation of the line through $\left(1,1,1\right)$ which is perpendicular to the two lines (1) and (2) below.
$\left(1\right)\phantom{\rule{1em}{0ex}}\mathbf{r}=\mathbf{i}-\mathbf{j}+t\mathbf{k}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left(2\right)\phantom{\rule{1em}{0ex}}\frac{x}{4}=y+1=\frac{z-2}{3}$
Exactly one option must be correct)
 a) $\mathbf{r}=\mathbf{i}+\mathbf{j}+\mathbf{k}+t\mathbf{i}$ b) $\mathbf{r}=\mathbf{i}+\mathbf{j}+\mathbf{k}+t\left(-\mathbf{i}+4\mathbf{j}\right)$ c) $\mathbf{r}=\mathbf{i}+\mathbf{j}+\mathbf{k}+t\left(-3\mathbf{i}+4\mathbf{k}\right)$ d) $\mathbf{r}=\mathbf{i}+\mathbf{j}+\mathbf{k}+t\left(\mathbf{i}+4\mathbf{j}\right)$

Choice (a) is incorrect
Choice (b) is correct!
Line (1) is parallel to $\mathbf{k}$ and line (2) is parallel to $4\mathbf{i}+\mathbf{j}+3\mathbf{k}$. Hence the required line, which is perpendicular to both, must be parallel to the cross product of these two vectors, $\mathbf{k}×\left(4\mathbf{i}+\mathbf{j}+3\mathbf{k}\right)=-\mathbf{i}+4\mathbf{j}.$ Its parametric vector equation is therefore $\mathbf{r}=\mathbf{i}+\mathbf{j}+\mathbf{k}+t\left(-\mathbf{i}+4\mathbf{j}\right).$
Choice (c) is incorrect
Choice (d) is incorrect
Suppose that $P$ and $Q$ are two distinct points in 3-dimensional space. How many planes are there which contain both $P$ and $Q$? Exactly one option must be correct)
 a) None. b) One only. c) Two only. d) None of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
There are infinitely many planes containing any two given points. To see this, visualise the line joining the points as the spine of a book, and the infinitely many planes as pages of the book.
Find the cartesian equation of the plane which goes through the point $\left(3,1,0\right)$ and is perpendicular to the vector $\mathbf{i}-\mathbf{j}+2\mathbf{k}$. Exactly one option must be correct)
 a) $x-y+2z=0$ b) $x-y+2z=4$ c) $x-y+2z=2$ d) $x-3y=0$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The equation of the plane through the point $\left(p,q,r\right)$ perpendicular to $a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ is $a\left(x-p\right)+b\left(y-q\right)+c\left(z-r\right)=0$. Hence in this case, the cartesian equation is $1\left(x-3\right)-1\left(y-1\right)+2\left(z-0\right)=0$, that is, $x-y+2z=2.$
Choice (d) is incorrect
Find the cartesian equation of the plane which contains the three points $A=\left(1,3,-1\right)$, $B=\left(1,1,5\right)$ and $C=\left(-2,1,0\right)$. Exactly one option must be correct)
 a) $10x+2y-z=17$ b) $10x+21y-6z=1$ c) $x+21y-6z=19$ d) $10x-18y-6z=-38$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
In order to know the equation of a plane, we must know a point on the plane and a vector perpendicular to the plane. We know three points on the plane, so we’re OK there. We can find a vector perpendicular to the plane by using the vector cross product. Notice that the vectors $\stackrel{⃗}{AB}$ and $\stackrel{⃗}{AC}$ are both parallel to the plane, so their cross product is normal to the plane. Since $\stackrel{⃗}{AB}=-2\mathbf{j}+6\mathbf{k}$ and $\stackrel{⃗}{AC}=-3\mathbf{i}-2\mathbf{j}+\mathbf{k}$, we calculate $\stackrel{⃗}{AB}×\stackrel{⃗}{AC}=10\mathbf{i}-18\mathbf{j}-6\mathbf{k}$. This is a normal vector to the plane. The cartesian equation is therefore $10\left(x-1\right)-18\left(y-3\right)-6\left(z+1\right)=0$, or $10x-18y-6z=-38.$
Find the cartesian equation of the plane which contains the point $\left(1,1,1\right)$ and is parallel to the two vectors $\mathbf{i}+\mathbf{j}-\mathbf{k}$ and $2\mathbf{i}-5\mathbf{k}$. Exactly one option must be correct)
 a) $3x+y-6z=2$ b) $x-y-4z=-4$ c) $5x+3y+2z=10$ d) $5x-3y+2z=4$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
A vector perpendicular to the plane is
$\left(\mathbf{i}+\mathbf{j}-\mathbf{k}\right)×\left(2\mathbf{i}-5\mathbf{k}\right)=\left|\begin{array}{ccc}\hfill \mathbf{i}\hfill & \hfill \mathbf{j}\hfill & \hfill \mathbf{k}\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -5\hfill \end{array}\right|=-5\mathbf{i}+3\mathbf{j}-2\mathbf{k}.$
So the cartesian equation of the plane is $-5\left(x-1\right)+3\left(y-1\right)-2\left(z-1\right)=0$ i.e $5x-3y+2z=4$.
Find the acute angle between the planes $3x+y+z=0$ and $x-2y+z=3$. Exactly one option must be correct)
The two planes are not parallel, since the first plane has normal vector $\mathbf{n}=3\mathbf{i}+\mathbf{j}+\mathbf{k}$ and the second plane has normal $\mathbf{m}=\mathbf{i}-2\mathbf{j}+\mathbf{k}$; so $\mathbf{n}$ is not parallel to $\mathbf{m}$. Denote the two planes by $ABCD$ and $EBCF$, respectively, so that $BC$ is the line of intersection.
The diagram above shows that the angle between two planes is the same as the angle between the normals to the planes, so we need to find the angle between the vectors $\mathbf{n}$ and $\mathbf{m}$. This angle $\theta$ is given by the now well-known formula $cos\theta =\frac{\mathbf{n}\cdot \mathbf{m}}{|\mathbf{n}||\mathbf{m}|}=\frac{2}{\sqrt{11}\sqrt{6}}.$ So, the angle between the two planes is $\theta \approx 1.32$ radians.