## MATH1014 Quizzes

Quiz 3: Lines and Planes
Question 1 Questions
Find the equation of the line joining points $P\left(2,1,-1\right)$ and $Q\left(0,3,1\right)$, in vector form. Exactly one option must be correct)
 a) $\mathbf{x}=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 3\hfill \\ \hfill 1\hfill \end{array}\right]$ b) $\mathbf{x}=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 4\hfill \\ \hfill 0\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$ c) $\mathbf{x}=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right]$ d) $\mathbf{x}=\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
A vector parallel to the line is the vector $\stackrel{\to }{PQ}=\stackrel{\to }{OQ}-\stackrel{\to }{OP}=\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right].$ Hence the line can be represented by the vector equation $\mathbf{x}=\stackrel{\to }{OP}+t\stackrel{\to }{PQ}.$ Note that there are many other ways of giving a vector 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: $\mathbf{x}=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 3\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right].$
Choice (d) is incorrect
A line has parametric equations A vector parallel to the line is: (Zero or more options can be correct)
 a) $\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 4\hfill \\ \hfill -2\hfill \end{array}\right]$ b) $\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill -6\hfill \end{array}\right]$ c) $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$ d) $\left[\begin{array}{c}\hfill 5\hfill \\ \hfill 3\hfill \\ \hfill -7\hfill \end{array}\right]$

There is at least one mistake.
For example, choice (a) should be True.
When the equations of a line are given in parametric form we can identify the coordinates of two points on the line by letting the parameter take two different values, such as $t=0$ and $t=1$. Hence $\left(2,1,-6\right)$ is on the line, and so is $\left(5,3,-7\right)$. The vector from the first to the second point, namely $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$, is therefore parallel to the line. The vector $\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 4\hfill \\ \hfill -2\hfill \end{array}\right]$ is a scalar multiple of $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$ and is also parallel to the line.
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be True.
The point $\left(2,1,-6\right)$ is on the line (corresponding to $t=0$), and so is $\left(5,3,-7\right)$ (corresponding to $t=1$). The vector from the first to the second point, namely $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$, is therefore parallel to the line.
There is at least one mistake.
For example, choice (d) should be False.
Correct!
1. True When the equations of a line are given in parametric form we can identify the coordinates of two points on the line by letting the parameter take two different values, such as $t=0$ and $t=1$. Hence $\left(2,1,-6\right)$ is on the line, and so is $\left(5,3,-7\right)$. The vector from the first to the second point, namely $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$, is therefore parallel to the line. The vector $\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 4\hfill \\ \hfill -2\hfill \end{array}\right]$ is a scalar multiple of $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$ and is also parallel to the line.
2. False
3. True The point $\left(2,1,-6\right)$ is on the line (corresponding to $t=0$), and so is $\left(5,3,-7\right)$ (corresponding to $t=1$). The vector from the first to the second point, namely $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$, is therefore parallel to the line.
4. False
Given the parametric equations of a line, Find the equation of the same line in vector form. Exactly one option must be correct)
 a) $\mathbf{x}=\left[\begin{array}{c}\hfill 5\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill 3\hfill \end{array}\right]$ b) $\mathbf{x}=\left[\begin{array}{c}\hfill 5\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill -3\hfill \\ \hfill -2\hfill \\ \hfill -3\hfill \end{array}\right]$ c) $\mathbf{x}=\left[\begin{array}{c}\hfill 8\hfill \\ \hfill 1\hfill \\ \hfill 5\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 5\hfill \\ \hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$ d) $\mathbf{x}=\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill 3\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 5\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right]$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
In vector form, the equation of a line is written $\mathbf{x}=\mathbf{p}+t\mathbf{d}$, where $\mathbf{p}$ is the position vector (relative to the origin) of a point $P$ on the line and $\mathbf{d}$ is a direction vector for the line. The components of the direction vector are simply the coefficients of $t$ in the parametric equations.
Find the equation of the line $\ell$ through $\left(1,2,1\right)$ parallel to the line given by the parametric equations
Exactly one option must be correct)
 a) $\mathbf{x}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill 5\hfill \end{array}\right]$ b) $\mathbf{x}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 2\hfill \\ \hfill 8\hfill \end{array}\right]$ c) $\mathbf{x}=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill 5\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right]$ d) $\mathbf{x}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 4\hfill \end{array}\right]$

Choice (a) is incorrect
Line $\ell$ must be parallel to $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right]$. The line given in this option is parallel to $\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill 5\hfill \end{array}\right]$, which is not a direction vector for $\ell$.
Choice (b) is correct!
A vector parallel to the line is $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right]$ (from the coefficients of $t$ in the parametric equations). So the line is also parallel to $\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 2\hfill \\ \hfill 8\hfill \end{array}\right]$. Therefore since the line passes through the point $\left(1,2,1\right)$, it has vector equation $\mathbf{x}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 2\hfill \\ \hfill 8\hfill \end{array}\right].$
Choice (c) is incorrect
This line is parallel to the vector $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right]$, but does not contain the point $\left(1,2,1\right)$. If it did, there would be a value of the parameter $t$ satisfying all three of the following equations simultaneously: $1=2+3t,\phantom{\rule{2em}{0ex}}2=1+t,\phantom{\rule{2em}{0ex}}1=5+4t.$ It is easy to see that no such $t$ exists!
Choice (d) is incorrect
The line given in this option is parallel to the vector $\left[\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 4\hfill \end{array}\right]$, which is not parallel to the line given in the question.
Suppose that $P$ and $Q$ are two distinct points in 3-dimensional space. How many planes are there that 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 two distinct points. To see this, visualise the line joining the two points as the spine of a book, and the infinitely many planes as pages of the book.
Find the general equation of the plane which goes through the point $\left(3,1,0\right)$ and is perpendicular to the vector $\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right].$ Exactly one option must be correct)
 a) $x-y+2z=0$ b) $x-y+2z=4$ c) $x-y+2z=2$ d) $x-3y=0$ e) $x-3y=z$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The general equation of the plane through the point $\left(p,q,r\right)$ perpendicular to the vector $\left[\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right]$ is $a\left(x-p\right)+b\left(y-q\right)+c\left(z-r\right)=0$. In this particular case, the equation becomes $1\left(x-3\right)-1\left(y-1\right)+2\left(z-0\right)=0$, that is, $x-y+2z=2.$
Choice (d) is incorrect
Choice (e) is incorrect
Find the equation of the unique plane through the three points $A=\left(3,-2,1\right),\phantom{\rule{0.3em}{0ex}}B=\left(1,1,5\right),\phantom{\rule{0.3em}{0ex}}C=\left(-2,4,0\right).$ Exactly one option must be correct)
 a) $20x+22y+5=21$ b) $-27x+15y+8z=28$ c) $-20x+18y+z=-21$ d) $27x+22y-3z=34$

Choice (a) is incorrect
First find a vector perpendicular (normal) to the plane by calculating $\stackrel{\to }{AB}×\stackrel{\to }{BC}$. Then remember that the formula $ax+by+cz=d$ for the equation of a plane gives us the information that the vector $\left[a,b,c\right]$ is normal to the plane. So the only unknown constant left to find is the constant $d$. This can be evaluated by substituting the coordinates of any of the points $A,\phantom{\rule{0.3em}{0ex}}B,\phantom{\rule{0.3em}{0ex}}C$ into the equation.
Choice (b) is incorrect
First find a vector perpendicular (normal) to the plane by calculating $\stackrel{\to }{AB}×\stackrel{\to }{BC}$. Then remember that the formula $ax+by+cz=d$ for the equation of a plane gives us the information that the vector $\left[a,b,c\right]$ is normal to the plane. So the only unknown constant left to find is the constant $d$. This can be evaluated by substituting the coordinates of any of the points $A,\phantom{\rule{0.3em}{0ex}}B,\phantom{\rule{0.3em}{0ex}}C$ into the equation.
Choice (c) is incorrect
First find a vector perpendicular (normal) to the plane by calculating $\stackrel{\to }{AB}×\stackrel{\to }{BC}$. Then remember that the formula $ax+by+cz=d$ for the equation of a plane gives us the information that the vector $\left[a,b,c\right]$ is normal to the plane. So the only unknown constant left to find is the constant $d$. This can be evaluated by substituting the coordinates of any of the points $A,\phantom{\rule{0.3em}{0ex}}B,\phantom{\rule{0.3em}{0ex}}C$ into the equation.
Choice (d) is correct!
The vector $\stackrel{\to }{AB}$ equals $\left[-2,3,4\right]$ and the vector $\stackrel{\to }{BC}$ equals $\left[-3,3,-5\right]$. These two vectors are parallel to the plane and so their cross product is perpendicular to the plane. We find that $\stackrel{\to }{AB}×\stackrel{\to }{BC}=\left[-27,-22,3\right]$. The equation of the plane has the form $-27x-22y+3z=d$ where we can find $d$ by substituting the coordinates of any of the three original points. This gives $d=-34$ and the answer follows.
Find a vector perpendicular to the two lines $\mathbf{x}=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\mathbf{x}=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \\ \hfill 7\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 4\hfill \\ \hfill -2\hfill \\ \hfill 7\hfill \end{array}\right]$ Exactly one option must be correct)
 a) $\left[-16,-14,4\right]$ b) $\left[3,12,14\right]$ c) $\left[6,5,-2\right]$ d) $\left[6,15,2\right]$ e) $\left[6,-5,-2\right]$

Choice (a) is incorrect
The first line is parallel to $\left[1,0,3\right]$ and the second is parallel to $\left[4,-2,7\right]$. A vector perpendicular to both lines is therefore the cross product of these two vectors.
Choice (b) is incorrect
The first line is parallel to $\left[1,0,3\right]$ and the second is parallel to $\left[4,-2,7\right]$. A vector perpendicular to both lines is therefore the cross product of these two vectors.
Choice (c) is correct!
Choice (d) is incorrect
The first line is parallel to $\left[1,0,3\right]$ and the second is parallel to $\left[4,-2,7\right]$. A vector perpendicular to both lines is therefore the cross product of these two vectors.
Choice (e) is incorrect
The first line is parallel to $\left[1,0,3\right]$ and the second is parallel to $\left[4,-2,7\right]$. A vector perpendicular to both lines is therefore the cross product of these two vectors.
The vectors $\mathbf{u}$ and $\mathbf{v}$ are non-parallel. Which of the following vectors are perpendicular to $\mathbf{u}×\mathbf{v}$? (Zero or more options can be correct)
 a) $2\mathbf{u}$ b) $\left(\mathbf{u}\cdot \mathbf{v}\right)\left(\mathbf{u}×\mathbf{v}\right)$ c) $\mathbf{v}×\mathbf{u}$ d) $2\mathbf{u}+4\mathbf{v}$ e) $\left(\mathbf{u}×\mathbf{v}\right)×\mathbf{u}$ f) $\left(\mathbf{u}+\mathbf{v}\right)×\mathbf{v}$

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be False.
This is just a scalar multiple of $\mathbf{u}×\mathbf{v}$ and is therefore a vector in the same or opposite direction, not perpendicular to $\mathbf{u}×\mathbf{v}$.
There is at least one mistake.
For example, choice (c) should be False.
This is the negative of $\mathbf{u}×\mathbf{v}$ and is therefore a vector in the opposite direction to $\mathbf{u}×\mathbf{v}$.
There is at least one mistake.
For example, choice (d) should be True.
There is at least one mistake.
For example, choice (e) should be True.
There is at least one mistake.
For example, choice (f) should be False.
The properties of cross product give $\left(\mathbf{u}+\mathbf{v}\right)×\mathbf{v}=\mathbf{u}×\mathbf{v}+\mathbf{v}×\mathbf{v}=\mathbf{u}×\mathbf{v}+\mathbf{0}=\mathbf{u}×\mathbf{v}.$ Therefore this option is certainly not perpendicular to $\mathbf{u}×\mathbf{v}$ – it is equal to it.
Correct!
1. True
2. False This is just a scalar multiple of $\mathbf{u}×\mathbf{v}$ and is therefore a vector in the same or opposite direction, not perpendicular to $\mathbf{u}×\mathbf{v}$.
3. False This is the negative of $\mathbf{u}×\mathbf{v}$ and is therefore a vector in the opposite direction to $\mathbf{u}×\mathbf{v}$.
4. True
5. True
6. False The properties of cross product give $\left(\mathbf{u}+\mathbf{v}\right)×\mathbf{v}=\mathbf{u}×\mathbf{v}+\mathbf{v}×\mathbf{v}=\mathbf{u}×\mathbf{v}+\mathbf{0}=\mathbf{u}×\mathbf{v}.$ Therefore this option is certainly not perpendicular to $\mathbf{u}×\mathbf{v}$ – it is equal to it.
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 is perpendicular to $\mathbf{n}=\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$and the second plane is perpendicular to $\mathbf{m}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right]$, and $\mathbf{n}$ 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 $cos\theta =\frac{\mathbf{n}\cdot \mathbf{m}}{||\mathbf{n}||\phantom{\rule{1em}{0ex}}||\mathbf{m}||}=\frac{2}{\sqrt{11}\sqrt{6}}.$ So, the angle between the two planes is the (acute) angle $\theta ={cos}^{-1}\frac{2}{\sqrt{66}}\approx 1.32$ radians.