Skip to main content

MTH 254 Calculus IV Winter 2026

Richard Zhao

Print preview

Paper size

Worksheet Assigned Problems for Section 12.5

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.
🔗
The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.
🔗

12.5.3.

Write the equation of the plane with normal vector \(\v{n} = \la -1,2,1 \ra\) passing through the point \(\lp 4,1,5 \rp\) in the scalar form \(ax + by + cz = d\text{.}\)
🔗
Solution.
The vector equation is
\begin{equation*} \la-1,2,1\ra\cdot\la x_0,y_0,z_0\ra=\la-1,2,1\ra\cdot\la 4,1,5\ra=-4+2+5=3 \end{equation*}
To obtain the scalar form we compute the dot product on the left-hand side above.
\begin{equation*} -x+2y+z=3 \end{equation*}
or in the other scalar form.
\begin{equation*} -(x-4)+2(y-1)+(z-5)=0 \end{equation*}
🔗
🔗
🔗

12.5.15.

Find a vector normal to the plane \(3(x - 4) - 8(y - 1) + 11z = 0\text{.}\)
🔗
Solution.
Using the scalar form of the equation of the plane, \(3x-8y+11z=4\text{,}\) a vector normal to the plane is the coefficients vector
\begin{equation*} \v{n}=\la3,-8,11\ra \end{equation*}
🔗
🔗
🔗

12.5.23.

Find an equation of the plane passing through the three points \(P = (1,0,0)\text{,}\) \(Q = (0,1,1)\text{,}\) and \(R = (2,0,1)\text{.}\)
🔗
Solution.
We use the vector form of the equation of the plane.
\begin{equation*} \v{n}\cdot\la x,y,z\ra=d \end{equation*}
To find the normal vector to the plane, \(\v{n}\text{,}\) we first compute the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) that lie in the plane, and then find the cross product of these vectors. This gives
\begin{align*} \overrightarrow{PQ} \amp =\la0,1,1\ra-\la1,0,0\ra=\la-1,1,1\ra\\ \overrightarrow{PR} \amp =\la2,0,1\ra-\la1,0,0\ra=\la1,0,1\ra \end{align*}
\begin{align*} \v{n}=\overrightarrow{PQ}\times\overrightarrow{PR} \amp= \begin{vmatrix}\v{i}\amp \v{j}\amp \v{k}\\ -1\amp 1\amp 1\\ 1\amp 0\amp 1\end{vmatrix} \\ \amp=\begin{vmatrix}1\amp 1\\ 0\amp 1\end{vmatrix}\v{i}-\begin{vmatrix}-1\amp 1\\ 1\amp 1\end{vmatrix}\v{j}+\begin{vmatrix}-1\amp 1\\ 1\amp 0\end{vmatrix}\v{k} \\ \amp=\v{i}+2\v{j}-\v{k} \\ \amp=\la1,2,-1\ra \end{align*}
We now choose any one of the three points in the plane, say \(P=(1,0,0),\) and compute \(d\text{.}\)
\begin{equation*} d=\v{n}\cdot\overrightarrow{OP}=\la1,2,-1\ra\cdot\la1,0,0\ra=1\cdot1+2\cdot0+(-1)\cdot0=1 \end{equation*}
Putting it all together, the equation of the plane is
\begin{align*} \la1,2,-1\ra\cdot\la x,y,z\ra \amp= 1\\ x+2y-z \amp= 1 \end{align*}
🔗
🔗
🔗

12.5.27.

In each case, determine whether or not the lines have a single point of intersection. If they do, give an equation of a plane containing them.
  1. \(\v{r}_1(t) = \la t, 2t-1, t-3 \ra\) and \(\v{r}_2(t) = \la 4, 2t-1, -1 \ra\)
    🔗
    🔗
  2. \(\v{r}_1(t) = \la 3t, 2t+1, t-5 \ra\) and \(\v{r}_2(t) = \la 4t, 4t-3, -1 \ra\)
    🔗
    🔗
🔗
Solution.
  1. We look for an intersection point by setting the components equal. Be careful to use different parameters for the two lines, say \(t\) for \(\v{r}_1\) and \(s\) for \(\v{r}_2\text{.}\)
    \begin{align*} x: \quad t \amp= 4 \\ y: \quad 2t - 1 \amp= 2s - 1 \\ z: \quad t - 3 \amp= -1 \end{align*}
    From the first equation, \(t=4\text{.}\) However, from the third equation, \(t - 3 = -1 \implies t = 2\text{.}\) Since \(t\) cannot be both 4 and 2, there is no solution. The lines do not intersect (and their direction vectors \(\la 1,2,1 \ra\) and \(\la 0,2,0 \ra\) are not parallel). Thus, they are skew lines and there is no single plane containing both.
    🔗
    🔗
  2. Again, we set the components equal using parameters \(t\) and \(s\text{.}\)
    \begin{align*} x: \quad 3t \amp= 4s \\ y: \quad 2t + 1 \amp= 4s - 3 \\ z: \quad t - 5 \amp= -1 \end{align*}
    From the third equation, \(t = 4\text{.}\) Plugging \(t=4\) into the first equation: \(3(4) = 4s \implies 12 = 4s \implies s = 3\text{.}\) Now we check if these values satisfy the second equation (the y-coordinate):
    \begin{gather*} \text{LHS: } 2(4) + 1 = 9 \\ \text{RHS: } 4(3) - 3 = 9 \end{gather*}
    They match! The lines intersect at the point derived from \(t=4\text{:}\)
    \begin{equation*} P = (3(4), 2(4)+1, 4-5) = (12, 9, -1) \end{equation*}
    🔗
    Since the lines intersect, they define a plane. The normal vector \(\v{n}\) is the cross product of the direction vectors \(\v{v}_1 = \la 3, 2, 1 \ra\) and \(\v{v}_2 = \la 4, 4, 0 \ra\text{.}\)
    \begin{align*} \v{n} = \v{v}_1 \times \v{v}_2 \amp = \begin{vmatrix} \v{i} \amp \v{j} \amp \v{k} \\ 3 \amp 2 \amp 1 \\ 4 \amp 4 \amp 0 \end{vmatrix}\\ \amp = \v{i}(0-4) - \v{j}(0-4) + \v{k}(12-8) \\ \amp = \la -4, 4, 4 \ra \end{align*}
    We can simplify this normal vector by dividing by -4 to get \(\v{n} = \la 1, -1, -1 \ra\text{.}\) Using the intersection point \((12, 9, -1)\text{,}\) the equation of the plane is:
    \begin{align*} 1(x - 12) - 1(y - 9) - 1(z - (-1)) \amp= 0 \\ x - 12 - y + 9 - z - 1 \amp= 0 \\ x - y - z \amp= 4 \end{align*}
    🔗
    🔗
🔗
🔗
🔗

12.5.33.

Draw the plane given by the equation \(x + y + z = 4\text{.}\)
🔗
Solution.
To draw the plane, we find the intercepts where the plane crosses the coordinate axes.
  • x-intercept: Set \(y=0, z=0\text{.}\) \(x = 4\text{.}\) Point \((4,0,0)\text{.}\)
    🔗
  • y-intercept: Set \(x=0, z=0\text{.}\) \(y = 4\text{.}\) Point \((0,4,0)\text{.}\)
    🔗
  • z-intercept: Set \(x=0, y=0\text{.}\) \(z = 4\text{.}\) Point \((0,0,4)\text{.}\)
    🔗
The traces form a triangle connecting these three points.
🔗
Figure 12.5.18. Plane \(x + y + z = 4\) with traces
🔗
🔗
🔗

12.5.41.

Find the intersection of the line \(\v{r}(t) = \la 1,1,0 \ra + t \la 0,2,4 \ra\) and the plane \(x + y + z = 14\text{.}\)
🔗
Solution.
The line has parametric equations
\begin{equation*} x(t)=1, \qquad y(t)=1+2t, \qquad z(t)=4t \end{equation*}
To find a value of \(t\) for which \((x,y,z)\) lies on the plane, we substitute the parametric equations in the equation of the plane and solve for \(t\text{.}\)
\begin{align*} x+y+z \amp = 14\\ 1+(1+2t)+4t \amp = 14\\ 6t \amp = 12 \qquad \implies \qquad t=2 \end{align*}
The point \(P\) of intersection has coordinates
\begin{equation*} x(2)=1, \qquad y(2)=1+2\cdot2=5, \qquad z(2)=4\cdot2=8 \end{equation*}
That is, \(P=(1,5,8)\text{.}\)
🔗
🔗
🔗

12.5.59.

By definition, the angle between two planes is the angle between their normal vectors.
🔗
Figure 12.5.19. Angle between Two Planes
🔗
Compute the angle between the two planes \(2x + 3y + 7z = 2\) and \(4x - 2y + 2z = 4\text{.}\)
🔗
Solution.
The planes \(2x+3y+7z=2\) and \(4x-2y+2z=4\) have the normal vectors of \(\v{n}_1=\la2,3,7\ra\) and \(\v{n}_2=\la4,-2,2\ra\) respectively. The cosine of the angle between \(\v{n}_1\) and \(\v{n}_2\) is
\begin{align*} \cos \theta \amp= \frac{\v{n}_1\cdot \v{n}_2}{\|\v{n}_1\|\|\v{n}_2\|} \\ \amp=\frac{\la2,3,7\ra\cdot\la4,-2,2\ra}{\|\la2,3,7\ra\|\|\la4,-2,2\ra\|} \\ \amp=\frac{8-6+14}{\sqrt{2^2+3^2+7^2}\sqrt{4^2+(-2)^2+2^2}} \\ \amp=\frac{16}{\sqrt{62}\sqrt{24}} \\ \amp\approx 0.415 \end{align*}
The solution for \(0\le\theta\lt\pi\) is \(\theta=1.143\) rad or \(\theta=65.49^{\circ}\text{.}\)
🔗
🔗
🔗

12.5.65.

Find a plane that is perpendicular to the two planes \(x + y = 3\) and \(x + 2y - z = 4\text{.}\)
🔗
Solution.
The vector forms of the equations of the planes are \(\la 1,1,0 \ra \cdot \la x,y,z \ra = 3\) and \(\la 1,2,-1 \ra \cdot \la x,y,z \ra = 4\text{,}\) hence the vectors \(\v{n}_1=\la1,1,0\ra\) and \(\v{n}_2=\la1,2,-1\ra\) are normal to the planes. We denote the equation of the planes which are perpendicular to the two planes by
\begin{equation*} ax + by + cz = d \end{equation*}
Then, the normal \(\v{n}=\la a,b,c\ra\) to the planes is orthogonal to the normals \(\v{n}_1\) and \(\v{n}_2\) of the given planes. Therefore, \(\v{n}\cdot \v{n}_1=0\) and \(\v{n}\cdot \v{n}_2=0\) which gives us
\begin{align*} \la a,b,c \ra \cdot \la 1,1,0 \ra \amp = 0\\ \la a,b,c \ra \cdot \la 1,2,-1 \ra \amp = 0 \end{align*}
We obtain the following equations.
\begin{equation*} \begin{cases} a+b=0 \\ a+2b-c=0 \end{cases} \end{equation*}
The first equation implies that \(b=-a\text{.}\) Substituting in the second equation we get \(a-2a-c=0\text{,}\) and hence \(c=-a\text{.}\) Substituting them back to the equation of the perpendicular planes, we get
\begin{equation*} ax-ay-az=d \qquad \implies \qquad x-y-z=\frac{d}{a} \end{equation*}
\(\frac{d}{a}\) is an arbitrary constant which we denote by \(f\text{.}\) Therefore, the planes which are perpendicular to the given planes are
\begin{equation*} x-y-z=f \end{equation*}
🔗
🔗
🔗

12.5.67.

Let \(\c{L}\) denote the line of intersection of the planes \(x - y - z = 1\) and \(2x + 3y + z = 2\text{.}\) Find parametric equations for the line \(\c{L}\text{.}\)
🔗
Hint.
To find a point on \(\c{L}\text{,}\) substitute an arbitrary value for \(z\) (say, \(z = 2\)) and then solve the resulting pair of equations for \(x\) and \(y\text{.}\)
🔗
🔗
Solution.
A direction vector for \(L\) is the cross product \(\v{v}=\v{n}_1\times \v{n}_2\text{.}\)
\begin{equation*} \v{v}=\v{n}_1\times \v{n}_2=\begin{vmatrix}\v{i}\amp \v{j}\amp \v{k}\\ 1\amp -1\amp -1\\ 2\amp 3\amp 1\end{vmatrix}=2\v{i}-3\v{j}+5\v{k}=\la2,-3,5\ra \end{equation*}
We now need to find a point on \(L\text{.}\) We choose \(z=2\) substitute in the equations of the planes and solve the resulting equations for \(x\) and \(y\text{.}\) This gives
\begin{align*} x-y-2=1 \amp \qquad \implies \qquad x-y=3\\ 2x+3y+2=2 \amp \qquad \implies \qquad 2x+3y=0 \end{align*}
The 1st equation implies that \(y=x-3\text{.}\) Substituting in the 2nd equation and solving for \(x\) gives
\begin{equation*} 2x+3(x-3)=0 \end{equation*}
\begin{equation*} 5x=9 \qquad \implies \qquad x=\frac{9}{5} \end{equation*}
This also implies that \(y=\dfrac{9}{5}-3=-\dfrac{6}{5}\text{.}\)
🔗
We conclude that the point \((\frac{9}{5},-\frac{6}{5},2)\) is on \(L\text{.}\) We now use the vector parametrization of a line to obtain the following parametrization for \(L\text{:}\)
\begin{equation*} \v{r}(t)=\la\frac{9}{5},-\frac{6}{5},2\ra+t\la2,-3,5\ra \end{equation*}
This yields the parametric equations
\begin{equation*} x(t) = \frac{9}{5}+2t, \qquad y(t) = -\frac{6}{5}-3t, \qquad z(t)=2+5t \end{equation*}
🔗
🔗
🔗
🔗
PrevTopNext