# Quiz 3: Lines and Planes

## Question 1

Find the equation of the line joining points
$P\left(2,1,-1\right)$ and
$Q\left(0,3,1\right)$, in
vector form.

**Not correct. Choice (a) is false.**

**Not correct. Choice (b) is false.**

**Your answer is correct.**

A vector parallel to the line is the vector
$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right].$
Hence the line can be represented by the vector equation
$$x=\overrightarrow{OP}+t\overrightarrow{PQ}.$$
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:
$$x=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 3\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right].$$

**Not correct. Choice (d) is false.**

## Question 2

A line has parametric equations
$$x=2+3t,\phantom{\rule{1em}{0ex}}y=1+2t,\phantom{\rule{1em}{0ex}}z=-6-t.$$
A vector parallel to the line is:

**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.

**Your answers are correct**

**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.**False**.**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.**False**.

## Question 3

Given the parametric equations of a line,
$$x=3+5t,\phantom{\rule{1em}{0ex}}y=2-t,\phantom{\rule{1em}{0ex}}z=3+2t,$$
find the equation of the same line in vector form.

**Not correct. Choice (a) is false.**

**Not correct. Choice (b) is false.**

**Not correct. Choice (c) is false.**

**Your answer is correct.**

In vector form, the equation of a line is written
$x=p+td$,
where $p$
is the position vector (relative to the origin) of a point
$P$ on the
line and $d$
is a direction vector for the line. The components of the direction vector are simply the
coefficients of $t$
in the parametric equations.

## Question 4

Find the equation of the line $\ell $
through $\left(1,2,1\right)$
parallel to the line given by the parametric equations

$$x=2+3t,\phantom{\rule{1em}{0ex}}y=1+t,\phantom{\rule{1em}{0ex}}z=5+4t.$$

**Not correct. Choice (a) is false.**

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 $.

**Your answer 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 $x=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 2\hfill \\ \hfill 8\hfill \end{array}\right].$

**Not correct. Choice (c) is false.**

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!

**Not correct. Choice (d) is false.**

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.

## Question 5

Suppose that $P$
and $Q$ are
two distinct points in 3-dimensional space. How many planes are there which contain
both $P$
and $Q$?

**Not correct. Choice (a) is false.**

**Not correct. Choice (b) is false.**

**Not correct. Choice (c) is false.**

**Your answer 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.

## Question 6

Find the general equation of the plane which goes through the point
$\left(3,1,0\right)$ and is perpendicular
to the vector $\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right].$

**Not correct. Choice (a) is false.**

**Not correct. Choice (b) is false.**

**Your answer 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, $x-y+2z=2.$

**Not correct. Choice (d) is false.**

**Not correct. Choice (e) is false.**

## Question 7

Find the equation of the unique plane through the three points
$A=\left(3,-2,1\right),\phantom{\rule{0.3em}{0ex}}B=\left(1,1,5\right),\phantom{\rule{0.3em}{0ex}}C=\left(-2,4,0\right).$

**Not correct. Choice (a) is false.**

First find a vector perpendicular (normal) to the plane by calculating
$\overrightarrow{AB}\times \overrightarrow{BC}$. 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.

**Not correct. Choice (b) is false.**

First find a vector perpendicular (normal) to the plane by calculating
$\overrightarrow{AB}\times \overrightarrow{BC}$. 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.

**Not correct. Choice (c) is false.**

First find a vector perpendicular (normal) to the plane by calculating
$\overrightarrow{AB}\times \overrightarrow{BC}$. 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.

**Your answer is correct.**

The vector
$\overrightarrow{AB}$ equals
$\left[-2,3,4\right]$ and the
vector $\overrightarrow{BC}$ 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
$\overrightarrow{AB}\times \overrightarrow{BC}=\left[-27,-22,3\right]$. 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.

## Question 8

Find a vector perpendicular to the two lines
$$x=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}x=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \\ \hfill 7\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 4\hfill \\ \hfill -2\hfill \\ \hfill 7\hfill \end{array}\right]$$

**Not correct. Choice (a) is false.**

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.

**Not correct. Choice (b) is false.**

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.

**Your answer is correct.**

**Not correct. Choice (d) is false.**

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.

**Not correct. Choice (e) is false.**

## Question 9

The vectors $u$
and $v$
are non-parallel. Which of the following vectors are perpendicular to
$u\times v$?

**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 $u\times v$
and is therefore a vector in the same or opposite direction, not perpendicular to
$u\times v$.

**There is at least one mistake.**

For example, choice (c) should be false.

This is the
negative of $u\times v$
and is therefore a vector in the opposite direction to
$u\times v$.

**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
$$\left(u+v\right)\times v=u\times v+v\times v=u\times v+0=u\times v.$$
Therefore this option is certainly not perpendicular to
$u\times v$ – it is
equal to it.

**Your answers are correct**

**True**.**False**. This is just a scalar multiple of $u\times v$ and is therefore a vector in the same or opposite direction, not perpendicular to $u\times v$.**False**. This is the negative of $u\times v$ and is therefore a vector in the opposite direction to $u\times v$.**True**.**True**.**False**. The properties of cross product give $$\left(u+v\right)\times v=u\times v+v\times v=u\times v+0=u\times v.$$ Therefore this option is certainly not perpendicular to $u\times v$ – it is equal to it.

## Question 10

Find the acute angle between the planes
$3x+y+z=0$ and
$x-2y+z=3$.

**Not correct. Choice (a) is false.**

Hint: Find the angle between the normals to the two planes.

**Your answer is correct.**

The two planes are not parallel, since the first plane is perpendicular to
$n=\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$and the second plane
is perpendicular to $m=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right]$,
and $n$ is not parallel
to $m$. 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
$n$ and
$m$. This
angle $\theta $
is given by the now well-known formula
$$cos\theta =\frac{n\cdot m}{\left|\right|n\left|\right|\phantom{\rule{1em}{0ex}}\left|\right|m\left|\right|}=\frac{2}{\sqrt{11}\sqrt{6}}.$$
So, the angle between the two planes is the (acute) angle
$\theta ={cos}^{-1}\frac{2}{\sqrt{66}}\approx 1.32$
radians.

**Not correct. Choice (c) is false.**

Hint: Find the angle between the normals to the two planes.

**Not correct. Choice (d) is false.**

Hint: Find the angle between the normals to the two planes.