Title: Default for 9.5 Lines and Planes
Course: HELM Workbooks 2023-2024

Exercise

Consider the vector \(\overrightarrow{OP}\), where \(P\) is the point with coordinates \([value]\).

1. Find the direction ratios \(l : m : n\) of \(\overrightarrow{OP}\).
\((l,m,n) = \)
2. Now find the cosines of the angles between the vector \(\overrightarrow{OP}\) and the \(x\),\(y\), and \(z\)-axes. Let these angles be \(\theta_x\), \(\theta_y\) and \(\theta_z\), respectively. Give exact answers, using square roots.
\(\cos \theta_x =\)
\(\cos \theta_y =\)
\(\cos \theta_z =\)
3. Lastly, find the angles between the vector \(\overrightarrow{OP}\) and the \(x\),\(y\), and \(z\)-axes. Give your answer in degrees, to three significant figures.
\( \theta_x =\) degrees.
\( \theta_y =\) degrees.
\( \theta_z =\) degrees.
Feedback: 1. We can just read off the direction rations \(l : m : n\) from the vector. \[(l,m,n) = \begin{bmatrix} [value] & [value] & [value] \end{bmatrix}\] 2. Next, we are asked for the cosines of the angles between \(\overrightarrow{OP}\) and the \(x\), \(y\) and \(z\)-axes. These are simply the direction cosines. In order to find them, we must first compute the magnitude of \(\overrightarrow{OP}\). \[\lvert\overrightarrow{OP}\rvert = [value]\] Then we have \[\begin{aligned} \cos \theta_x &= [value] \\ \cos \theta_y &= [value] \\ \cos \theta_z &= [value] \end{aligned}\] 3. Lastly, we are asked to find what each of these angles is. We can do this using the arccosine function in our calculator. Remember we must write our answers in degress and to three significant figures. \[\begin{aligned} \theta_x &= [value] \\ \theta_y &= [value] \\ \theta_z &= [value] \end{aligned}\]

Exercise

1. Find the Cartesian equation of the plane which passes through the point \(A\) and is perpendicular to the vector \([value]\). Here, \(A\) has position vector \([value] = [value]\), and \([value]= [value]\).
Give your answer as an equation (e.g 2*x + 3*y -9*z = -15).
The Cartesian equation of the plane passing through \(A\) and perpendicular to \([value]\) is:
2. What is the perpendicular distance from the origin to the plane? Give your answer exactly, involving square roots (remember the syntax, e.g. sqrt(17)).
The plane is a perpendicular distance of units from the origin.
Feedback: 1. Let \(R\) be a general point on our plane, with position vector \([value]= [value]\). Because \(A\) is a point in the plane, the vector \([value] - [value]\) is a vector lying within the plane. We know already that the plane must be perpendicular to the vector \([value]\); this must be true everywhere in the plane – and in particular, for our vector \([value]- [value]\). Recall from earlier that two vectors are perpendicular if and only if their scalar product (dot product) is equal to zero. Hence we have \(([value] - [value]) \cdot [value]= 0,\) or, upon rearrangement, \([value] \cdot [value] = [value] \cdot [value]\) as the vector equation of the plane. When we substitute in the particular vectors \([value]\) and \([value]\) for this question, we arrive at the Cartesian equation of the plane: \[[value] = [value]\] 2. The vector \([value]\) which is perpendicular to the plane may be normalised to give a unit vector \(\underline{\hat{n}}\) (recall from earlier how this is done). In this case, our unit vector perpendicular to the plane is \[\underline{\hat{n}} = \displaystyle[value].\] If we were to find the equation of the plane now, using the vector \(\underline{\hat{n}}\) instead of \([value]\), its vector equation would be \[[value] \cdot \underline{\hat{n}} =[value] \cdot \underline{\hat{n}}\] using the same idea as before. The quantity \(|\,[value] \cdot \underline{\hat{n}}\,| = d\) is understood to be the perpendicular distance from the origin to the plane. In this question, the plane is a perpendicular distance \(d = \displaystyle{[value]}\) units from the origin.

Exercise

Let the point \(A\) have position vector \([value] = [value]\), and the point \(B\) have position vector \([value] = [value]\).

1. Using this information, give the canonical vector equation of the line passing through \(A\) and \(B\). Input your answer, in column vector notation, in the boxes provided below. The line through \(A\) and \(B\) is given by the following equation, where \(t\) is a parameter which may take any real number as its value.
\(\underline{r} = \) \({}+t\ \)
2. Now give the equation of the line in its Cartesian form. Enter your answers as fractions, e.g. (k-5)/18.
The Cartesian equation of the line through \(A\) and \(B\) is:
\({}={}\) \({}={}\)
Feedback: 1. The canonical vector equation of the line passing through \(A\) and \(B\) is given by the formula \[\underline{r} = \underline{a} + t(\underline{b} - \underline{a}).\] This is the canonical form because we have \(\underline{r}(t = 0) = \underline{a}\) and \(\underline{r}(t = 1) = \underline{b}\). In our case, the values are \[\underline{r} = [value] + t\, [value].\] 2. Recall that the Cartesian equation of a line given by \(\underline{r} = \underline{a} + t(\underline{b} - \underline{a})\) is given by \[\frac{x - a_1}{b_1 - a_1} = \frac{y - a_2}{b_2 - a_2} = \frac{z - a_3}{b_3 - a_3},\] where \[\underline{a} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \quad \text{and} \quad \underline{b} = \begin{bmatrix} b_1 \\ b_2 \\b_3 \end{bmatrix}.\] Applying this formula in our case gives the equation: \[[value] = [value] = [value]\]

Exercise

In the following question, note that you can use the command sqrt(5) for the square root of \(5\), for example.

Points \(A\) and \(B\) have position vectors \([value]=[value]\), and \([value]= [value]\) respectively. Find

1. \(\overrightarrow{AB}=\) [value] \(+\) [value] \(+\) [value]
2. \(|\overrightarrow{AB}|=\)
3. The direction ratios of \(\overrightarrow{AB}\): : :
4. The direction cosines of \(\overrightarrow{AB}\), \((\ell,m,n)=\) \( \Big( \) , , \( \Big) \)
5. \(\ell^2+m^2+n^2=\) .
Feedback: Points \(A\) and \(B\) have position vectors \([value]=[value]\), and \([value]= [value]\) respectively. We have that 1. \(\overrightarrow{AB}= \underline{b} - \underline{a} = [value]\) 2. \(\lvert\overrightarrow{AB}\rvert=[value]\) 3. The direction ratios of \(\overrightarrow{AB}\) are [value] : [value] : [value] 4. The direction cosines of \(\overrightarrow{AB}\) are \[(\ell,m,n) = \left([value], [value], [value]\right).\] 5. We have \(\ell^2+m^2+n^2=1\). Can you see why this is true in general?

Exercise

What is the Cartesian equation of the plane that contains the point \(\left ([value],[value],[value] \right )\) and is perpendicular to the vector \(\underline{n} = [value] \, \underline{i}+[value] \,\underline{ j} [value]\,\underline{k}\)?

Hint: The Cartesian equation of a plane is an equation of the form
\[Ax + By + Cz = D,\] where \(A\), \(B\), \(C\) and \(D\) are constants.
Feedback: If \(\underline{r}= x\, \underline{i} + y\,\underline{ j} + z\, \underline{k}\) is the position vector of an arbitrary point in the plane and \(\displaystyle {\underline{r_0}}\) is the position vector of the point \(\left ( [value],[value],[value]\right )\) that is also in the plane, ie \(\underline{r_0} = [value]\,\underline{ i} + [value]\, \underline{ j} + [value]\, \underline{k}\), then the vector \[ \underline{r} -\underline{r_0}= \left ([value]\right )\, \underline{ i}+ \left ([value] \right ) \,\underline{j} +\left ([value] \right )\, \underline{k}\] lies in the plane and is perpendicular to \(\underline{n}\). Thus \( \left (\underline{r}-\underline{r_0} \right )\cdot \underline{n} = 0\), that is: \[ \left ([value] \right )\left ([value]\right ) + \left ([value] \right )\left ([value]\right )+ \left ([value] \right )\left ([value]\right ) =0.\] Expanding this out gives \[[value].\]

The direction ratio and direction cosines

Consider the point \(P(4,5)\) and its position vector \(4\underline{ i}+5\underline{j}\) shown in the following figure.

[Image: The vector (4,5). An arrow pointing 4 units to the right and 5 units upwards.]

The direction ratio of the vector \(\overrightarrow{OP}\) is defined to be 4:5. We can interpret this as stating that to move in the direction of the line \(OP\) we must move 4 units in the \(x\) direction for every 5 units in the \(y\) direction.

The direction cosines of the vector \(\overrightarrow{OP}\) are the cosines of the angles between the vector and each of the axes. Specifically, referring to Figure 1 these are \(\cos \alpha\) and \(\cos \beta\). Noting that the length of \(\overrightarrow{OP}\) is \(\sqrt{4^2+5^2}=\sqrt{41}\), we can write \(\cos\alpha=\dfrac{4}{\sqrt{41}},~ \cos\beta=\dfrac{5}{\sqrt{41}}.\) It is conventional to label the direction cosines as \(\ell\) and \(m\) so that \( \ell = \dfrac{4}{\sqrt{41}},~ m=\dfrac{5}{\sqrt{41}}.\) More generally we have the following result:

Key Point

For any vector \(\underline{r}=a\underline{i} +b\underline{j}\), its direction ratio is \(a:b\). Its direction cosines are

\(\qquad \ell =\dfrac{a}{\sqrt{a^2+b^2}}, \quad m=\dfrac{b}{\sqrt{a^2+b^2}}.\)

The final result in the previous Example is true in general:

Key Point

If \(\ell\) and \(m\) are the direction cosines of a vector lying in the \(xy\) plane, then \(\ell^2+m^2=1.\)

Exercise

Points \(P\) and \(Q\) have position vectors \([value] =[value]\), and \([value] = [value]\) respectively. Find

1. The direction ratio of \(\overrightarrow{PQ}\): \(:\)
2. The direction cosines of \(\overrightarrow{PQ}\), \((\ell,m)=\).
Feedback: The vector \(\overrightarrow{PQ}\), where \(P\) and \(Q\) have position vectors \(\underline{a} = [value]\) and \(\underline{b} = [value]\) respectively, is given by \[\overrightarrow{PQ} = \underline{b} - \underline{a} = [value].\] 1. The direction ratios of \(\overrightarrow{PQ}\) are \(\displaystyle[value]: \displaystyle[value]\). 2. To find the direction cosines, we must first find the magnitude of the vector. In this case, \[\lvert \overrightarrow{PQ} \rvert = \sqrt{([value])^2 + ([value])^2} = [value].\] Then, the direction cosines of \(\overrightarrow{PQ}\) are \[(\ell,m) = \displaystyle[value].\]

Direction ratios and cosines in three dimensions

The concepts of direction ratio and direction cosines extend naturally to three dimensions. Consider the following Figure.

[Image: A 3D vector diagram indicating the angles a vector makes with each of the x y and z axes.]

Given a vector \(\underline{r}=a\underline{i}+b\underline{j}+c\underline{ k}\) its direction ratios are \(a:b:c\). This means that to move in the direction of the vector we must must move \(a\) units in the \(x\) direction and \(b\) units in the \(y\) direction for every \(c\) units in the \(z\) direction.

The direction cosines are the cosines of the angles between the vector and each of the axes. It is conventional to label direction cosines as \(\ell\), \(m\) and \(n\) and they are given by \[\ell = \cos\alpha=\frac{a}{\sqrt{a^2+b^2+c^2}},\]\[m = \cos\beta=\frac{b}{\sqrt{a^2+b^2+c^2}},\]\[n = \cos\gamma=\frac{c}{\sqrt{a^2+b^2+c^2}}.\]

We have the following general result:

Key Point

For any vector \(\underline{r}=a{\underline i}+b\underline{j}+c\underline{k}\) its direction ratios are \(a:b:c\). Its direction cosines are \[\ell =\frac{a}{\sqrt{a^2+b^2+c^2}},\]\[m=\frac{b}{\sqrt{a^2+b^2+c^2}},\]\[n=\frac{c}{\sqrt{a^2+b^2+c^2}},\] where \(\ell^2+m^2+n^2 =1.\)

Example

Point \(A\) has coordinates \((3,5)\), and point \(B\) has coordinates \((7,8)\).

1. Write down the vector \(\overrightarrow{AB}\).
2. Find the direction ratio of the vector \(\overrightarrow{AB}\).
3. Find its direction cosines, \(\ell\) and \(m\).
4. Show that \(\ell^2+m^2=1.\)

Solution

1. \(\overrightarrow{AB}=\underline{b}-\underline{a} =4\underline{ i}+3\underline{ j}\).
2. The direction ratio of \(\overrightarrow{AB}\) is therefore 4:3.
3. The direction cosines are \(\ell= \dfrac{4}{\sqrt{4^2+3^2}} =\dfrac{4}{5},\: m = \dfrac{3}{\sqrt{4^2+3^2}} = \dfrac{3}{5}.\)
4. \(\ell^2+m^2=\left(\dfrac{4}{5}\right)^2 + \left(\dfrac{3}{5}\right)^2 = \dfrac{16}{25}+\dfrac{9}{25}=\dfrac{25}{25}=1.\)

Exercise

A line is inclined at \([value]^\circ\) to the \(x\)-axis and \([value]^\circ\) to the \(y\)-axis. What is the line's inclination to the \(z\)-axis? Any valid answer will suffice.

\(\phantom{.}^\circ\)

Give your answer in degrees.
Feedback: We know that the angle the vector makes with the \(x\)-axis is \(\theta_x=[value]^\circ\) and the angle the vector makes with the \(y\)-axis is \(\theta_y=[value]^\circ\). We see that \(\cos(\theta_x)=\displaystyle{[value]}\) and \(\cos( \theta_y)=\displaystyle{[value]}\). We can use the formula \(\cos^2( \theta_z)=1-\cos^2( \theta_x)-\cos^2( \theta_y)\) to calculate \(\theta_z\). In particular, we see that \(\cos( \theta_z)=\pm\displaystyle{[value]}\). Taking care to calculate \(\theta_z\) in degrees we see that \(\theta_z=[value]^\circ\) or \(\theta_z=[value]^\circ\).

The vector equation of a line

The answer to the previous task, \(\underline{ r}=\underline{ a} + t(\underline{ b}-\underline{ a})\), is the vector equation of the line through \(A\) and \(B\). It is a rule which gives the position vector \(\underline{r}\) of a general point on the line in terms of the given vectors \(\underline{a},\underline{b}\).

By varying the value of \(t\) we can move to any point on the line. For example, referring to Figure 3, when \(t=0,\) the equation gives

\(\underline{ r}=\underline{ a},\) which locates point \(A,\) when \(t=1,\) the equation gives \(\underline{ r}=\underline{ b},\) which locates point \(B.\)

If \(0<t<1\) the point \(P\) lies on the line between \(A\) and \(B\). If \(t>1\) the point \(P\) lies on the line beyond \(B\) (to the right in the figure). If \(t<0\) the point \(P\) lies on the line beyond \(A\) (to the left in the figure).

Key Point

The vector equation of the line through points \(A\) and \(B\) with position vectors \(\underline{a}\) and \(\underline{b}\) is \(\underline{r}=\underline{a}+ t(\underline{b}-\underline{a}).\)

Exercise

The vector equation of a line

Consider the straight line \(APB\) shown in the following figure. This is a line in three-dimensional space.

[Image: A 3D line passing through two points A and B, a point P is marked and the vector connecting the origin to point is given]

Points \(A\) and \(B\) are fixed and known points on the line, and have position vectors \(\underline{ a}\) and \(\underline{ b}\) respectively. Point \(P\) is any other arbitrary point on the line, and has position vector \(\underline{ r}\). Note that because \(\overrightarrow{AB}\) and \(\overrightarrow{AP}\) are parallel, \(\overrightarrow{AP}\) is simply a scalar multiple of \(\overrightarrow{AB}\), that is, \(\overrightarrow{AP}=t\overrightarrow{AB}\) where \(t\) is a number.

Referring to the figure, write down an expression for the vector \(\overrightarrow{AB}\) in terms of \(\underline{a}\) and \(\underline{b}\). Write \(\underline{a}\) and \(\underline{b}\) as a and b, respectively, ignoring the underline.

\(\overrightarrow{AB}=\)

Referring to the figure, use the triangle law for vector addition to find an expression for \(\underline{r}\) in terms of \(\underline{a}\), \(\underline{b}\) and \(t\), where \(\overrightarrow{AP}=t\overrightarrow{AB}\). If your answer involves multiplication, please indicate this by use of the * symbol.

\(\underline{r}=\)
Feedback: To find the vector \(\overrightarrow{AB}\) we simply subtract the position vector or \(A\) from the position vector of \(B\): \[\overrightarrow{AB} = \underline{b} - \underline{a}\] Next, we want to find the position vector \(\underline{r}\) of the point \(P\) on the line. We are given that \(\overrightarrow{AP} = t\overrightarrow{AB}\), so \[\underline{r} = \overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \underline{a} + t(\underline{b} - \underline{a}).\]

Exercise

Write down the vector equation of the line which passes through the points with position vectors \([value]=[value]\) and \([value]= [value]\).

Use the variable \(t\) in your answer.

\([value]=\) \([value]+{}\) \([value]\)

Also express the equation in column vector form.

\([value]=\) \({}+t\ \)
Feedback: The vector \(\overrightarrow{AB}\) is given by \[ [value]-[value] = [value] - ([value]) = [value].\] The equation of the line is then \[[value] = [value] + t([value]-[value]) = ([value]) + t([value]).\] Using column vector notation we could write \[[value] = [value]+t\,[value].\]

Exercise

Write down the vector equation of the line which passes through the points with position vectors \(\underline{ a}=[value]\) and \([value]= [value]\).

Use the variable \(t\) in your answer.

\([value]=\) \([value]+{}\) \([value]+{}\) \([value]\)

Also express the equation in column vector form.

\([value]=\) \({}+t\ \)
Feedback: The vector \(\overrightarrow{AB}\) is given by \[[value]-[value] = [value] - ([value]) = [value].\] The equation of the line is then \[\begin{aligned} [value] &= [value] + t([value]-[value]) \\ &= [value] + t ([value]). \end{aligned}\] Using column vector notation we could write \[[value] = [value]+t\,[value].\]

Cartesian form

On occasions it is useful to convert the vector form of the equation of a straight line into Cartesian form.

Suppose we write \(\underline{ a}=\begin{bmatrix}a_1 \\ a_2 \\ a_3 \end{bmatrix},\; \underline{ b}=\begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix},\; \underline{ r} = \begin{bmatrix}x \\ y \\ z \end{bmatrix}\) then \(\underline{ r} = \underline{ a} + t(\underline{ b}-\underline{ a})\) implies \(\; \begin{bmatrix}x \\ y \\ z \end{bmatrix} = \begin{bmatrix}a_1 \\ a_2 \\ a_3 \end{bmatrix} +t\begin{bmatrix} b_1-a_1 \\ b_2-a_2 \\ b_3-a_3\end{bmatrix} = \begin{bmatrix}a_1 + t(b_1-a_1) \\ a_2 + t(b_2-a_2) \\ a_3 + t(b_3-a_3) \end{bmatrix}\)

Equating the individual components we find

\(x = a_1 + t(b_1-a_1), \;\hbox{ or equivalently } t=\dfrac{x-a_1}{b_1-a_1}\)

\(y = a_2 + t(b_2-a_2), \;\hbox{ or equivalently } t=\dfrac{y-a_2}{b_2-a_2}\)

\(z = a_3 + t(b_3-a_3), \;\hbox{ or equivalently } t=\dfrac{z-a_3}{b_3-a_3}\)

Each expression on the right is equal to \(t\) and so we can write \(\dfrac{x-a_1}{b_1-a_1} = \dfrac{y-a_2}{b_2-a_2}=\dfrac{z-a_3}{b_3-a_3}.\) This gives the Cartesian form of the equations of the straight line which passes through the points with coordinates \((a_1,a_2,a_3)\) and \((b_1,b_2,b_3)\).

Key Point

The Cartesian form of the equation of the straight line which passes through the points with coordinates \((a_1,a_2,a_3)\) and \((b_1,b_2,b_3)\) is \[\frac{x-a_1}{b_1-a_1} = \frac{y-a_2}{b_2-a_2}=\frac{z-a_3}{b_3-a_3}.\]

Example

1. Write down the Cartesian form of the equation of the straight line which passes through the two points \((9,3,-2)\) and \((4,5,-1)\).
2. State the equivalent vector equation.

Solution

1. The Cartesian equation of the line is \(\dfrac{x-9}{4-9}=\dfrac{y-3}{5-3}= \dfrac{z-(-2)}{-1-(-2)},\) that is \(\dfrac{x-9}{-5}= \dfrac{y-3}{2} = \dfrac{z+2}{1}.\)
2. The vector equation of the line is \[ \underline{ r}=\underline{ a} +t(\underline{ b}-\underline{ a})\] \[= \begin{bmatrix} 9 \\ 3 \\ -2 \end{bmatrix} + t\left( \begin{bmatrix} 4 \\ 5 \\ -1 \end{bmatrix}- \begin{bmatrix} 9 \\ 3 \\ -2 \end{bmatrix}\right)\]\[=\begin{bmatrix} 9 \\ 3 \\ -2 \end{bmatrix} + t \begin{bmatrix} -5 \\ 2 \\1 \end{bmatrix}.\]

Exercise

Write down the vector equation of the line which passes through the points with position vectors \(\underline{ a}=[value]\) and \(\underline{ b}=[value]\).

Use the variable \(t\) in your answer.

\(\underline{r}=\) \(\underline{i}+{}\) \(\underline{j}+{}\) \(\underline{k}\)

Also express the equation in column vector form.

\(\underline{r}=\) \({}+t\ \)
Feedback: The vector \(\overrightarrow{AB}\) is given by \[[value]-[value] = [value] -([value]) = [value].\] The equation of the line is then \[\begin{aligned} \underline{r} &= \underline{a} + t(\underline{b}-\underline{a}) \\ &= [value] + t ([value]).\end{aligned}\] Using column vector notation we could write \[\underline{r} = [value]+t\,[value].\]

Exercise

Find the vector equation of the line passing through the point \(\underline{a}\) with position vector \(\underline{ a}=[value]\) and which is parallel to the vector \(\underline{d}= [value]\).

Express the equation in column vector form.

\(\underline{r} = \) \({}+t\ \)
Feedback: Since the line is parallel to \(\underline{d}\), we know that it must pass through the point \(\underline{a} + \underline{d}\). We are therefore looking for the equation of the line that passes through \(\underline{a}\) and \(\underline{a} + \underline{d}\). This is given by: \[\underline{r} = \underline{a} + t(\underline{a} + \underline{d} - \underline{a}) = \underline{a} + t\underline{d}\] In our particular case, this gives the equation \[\underline{r} = [value] + t\,[value].\]

The vector equation of a plane

Consider the plane shown in Figure 4.

[Image: A diagram of a 2D plane embedded in 3D vector space. Two points A and P on the plane are indicated and the vectors OA an]

Figure 4

Suppose that \(A\) is a fixed point in the plane and has position vector \(\underline{ a}\). Suppose that \(P\) is any other arbitrary point in the plane with position vector \(\underline{ r}\). Clearly the vector \(\overrightarrow{AP}\) lies in the plane.

Exercise

1. Referring to Figure 4, write the vector \(\overrightarrow{AP}\) in terms of \(\underline{a}\) and \(\underline{r}\).
Also shown in Figure 4 is a vector which is perpendicular to the plane and denoted by \(\underline{ n}\).

2. What relationship exists between \(\underline{n}\) and the vectors \(\underline{a}\) and \(\underline{r}\)?
Hint: think about the scalar product of \(\underline{n}\) with \(\overrightarrow{AP}\).
For technical reasons, the scalar product \(\underline{u}\cdot\underline{v}\) should be entered as "u*v".
\(=\)
Feedback: <h4 class="HELM_solution">Solution</h4><p>Because \(\overrightarrow{AP}\) and \(\underline{ n}\) are perpendicular their scalar product must equal zero, that is \(\qquad <br>(\underline{ r}-\underline{ a}).\underline{ n}=0\)so that \(\qquad <br>\underline{ r}.\underline{ n}=\underline{ a}.\underline{ n}\)<br></p>

The vector equation of a plane

The answer to the above question, \(\underline{ r} \cdot \underline{ n}=\underline{ a} \cdot \underline{ n},\) is the equation of a plane, written in vector form, passing through \(A\) and perpendicular to \(\underline{n}\).

In this formula it does not matter whether or not \(\underline{n}\) is a unit vector. If \(\underline{ \hat n}\) is a unit vector then \(\underline{ a} \cdot \underline{\hat n}\) represents the perpendicular distance from the origin to the plane which we usually denote by \(d\). Hence we can write \(\underline{ r} \cdot \underline{ \hat n}= d.\)

This is the equation of a plane, written in vector form, with unit normal \(\underline{\hat n}\) and which is a perpendicular distance \(d\) from \(O\).

Key Point

A plane with unit normal \(\underline{ \hat n}\), which is a perpendicular distance \(d\) from \(O\) is given by \[\underline{ r} \cdot \underline{ \hat n}=d.\]

Example

1. Find the vector equation of the plane which passes through the point with position vector \(3\underline{ i}+2\underline{ j}+5\underline{ k}\) and which is perpendicular to \(\underline{i}+\underline{k}.\)
2. Find the Cartesian equation of this plane.

Solution

1. Using the previous results we can write down the equation \(\qquad \underline{ r}\cdot(\underline{i}+\underline{k})= (3\underline{ i}+2\underline{ j}+5\underline{ k})\cdot(\underline{i}+\underline{k})=3+5=8.\)
2. Writing \(\underline{ r}\) as \(x\underline{ i}+y\underline{ j}+z\underline{ k}\) we have the Cartesian form: \(\qquad (x\underline{ i}+y\underline{ j}+z\underline{ k})\cdot (\underline{i}+\underline{k})=8\) so that \(\qquad x+z=8\).

Exercise

Find the equation of a plane which is normal to \([value]\) and which is a distance \(1\) from the origin. Give both vector and Cartesian forms.

The vector equation of the plane is:

\({}\cdot[value]=\)

The Cartesian equation of the plane is

\(=\)
Feedback: The unit normal to the plane is given by \[\underline{\hat{n}} = \frac{1}{[value]} [value] = [value].\] A plane with unit normal vector \(\underline{\hat{n}}\) which is of a distance \(1\) to the origin is given by the vector equation \[[value] \cdot \underline{r} = 1.\] Expanding out the vector expression we see the Cartesian equation of the plane is \[[value].\]

Exercise

What is the perpendicular distance from the origin to of the plane given by \(\underline{r}\cdot[value]=[value]\)?
Feedback: By dividing through by \(\lvert\underline{n}\rvert = [value]\), we can express the equation of the plane as \([value]\cdot \underline{\hat{n}} =c\) where \(\underline{\hat{n}}\) is a unit vector parallel to \(\underline{n}\) and \(c\) is a constant. Here \(\underline{\hat{n}} = \displaystyle{\frac{1}{[value]}[value]}\) and \(c=\displaystyle{\frac{[value]}{[value]}}\). The distance between the plane and the origin is thus given by \(\displaystyle{[value]}\).

Direction ratios and cosines in three dimensions

The vector equation of a line

The vector equation of a plane

Exercise

The vector equation of a plane

Consider the plane shown in the figure below.

[Image: A diagram of a 2D plane embedded in 3D vector space. Two points A and P on the plane are indicated and the vectors OA an]

Suppose that \(A\) is a fixed point in the plane and has position vector \(\underline{ a}\). Suppose that \(P\) is any other arbitrary point in the plane with position vector \(\underline{ r}\). Clearly the vector \(\overrightarrow{AP}\) lies in the plane.

1. Referring to the figure, write the vector \(\overrightarrow{AP}\) in terms of \(\underline{a}\) and \(\underline{r}\). Enter the vectors \(\underline{a}\) and \(\underline{r}\) as a and r, respectively, ignoring the underline.
\(\overrightarrow{AP} =\)
2. Also shown in the figure is a vector which is perpendicular to the plane and denoted by \(\underline{n}\).
What relationship exists between \(\underline{n}\) and the vectors \(\underline{a}\) and \(\underline{r}\)?
Hint: think about the scalar product of \(\underline{n}\) with \(\overrightarrow{AP}\). The scalar product \(\underline{u}\cdot\underline{v}\) should be entered as u*v.
\(=\)
Feedback: 1. An expression for \(\overrightarrow{AP}\) in terms of \(\underline{a}\) and \(\underline{r}\) is \(\overrightarrow{AP} = \underline{r} - \underline{a}\). 2. Because \(\overrightarrow{AP}\) and \(\underline{ n}\) are perpendicular their scalar product must equal zero, that is \((\underline{r}-\underline{a}) \cdot \underline{n}=0\) so that \(\underline{ r} \cdot \underline{n}=\underline{a} \cdot \underline{n}\).

Exercise

In all of the following questions give your answers exactly. Recall that \(\sqrt{2}\) can be entered by typing sqrt(2).

Points\(A\) and \(B\) have position vectors \(\underline{ a} =[value]\), and \(\underline{ b} = [value]\) respectively. Find:

1. \(\overrightarrow{AB}=\)
2. \(|\overrightarrow{AB}|=\)
3. The direction ratios of \(\overrightarrow{AB}\): :
4. The direction cosines of \(\overrightarrow{AB}\), \((\ell,m)=\)
5. \(\ell^2+m^2=\)
Feedback: We have that 1. (a) \(\overrightarrow{AB}=[value]\) 2. (b) \(|\overrightarrow{AB}|=[value]\) 3. (c) The direction ratios of \(\overrightarrow{AB}\): \([value]: [value]\) 4. (d) The direction cosines of \(\overrightarrow{AB}\), \((\ell,m)=\)[value] 5. (e) \(\ell^2+m^2=1\).