Calculus III Advanced (Course) (11.5) (Homework)

Section 11.5 Homework

From Calculus 10e by Larson and Edwards, p. 790. Exercises 10, 28, 29, 40, 42, 54, 82.

Exercise 11.5.10 Finding Parametric and Symmetric Equations

Find sets for (a) parametric equations and (b) symmetric equations for the line through the two points (if possible). (For each line, write the direction numbers as integers.)
(0,4,3), (-1,2,5)
Solution
(a). Let \(P(0,4,3)\) and \(Q(-1,2,5)\), then

\(\textbf{v} = \vec{PQ} = \langle 0-(-1), 4-2, 3-5 \rangle = \langle 1,2,-2 \rangle \).

This yields the direction numbers \(a=1\), \(b=2\), and \(c=-2\). The parametric equations are

\(x=1+0t\), \(y=4+2t\), and \(z=3-2t\).

(b) Symmetric equations are

$$\frac{x-1}{1}=\frac{y-4}{2}=\frac{z-3}{-2} $$

Exercise 11.5.28 Determining Parallel Lines

Determine whether the lines are parallel or identical.

$$L_{1}:\frac{x-3}{2}=\frac{y-2}{1}=\frac{z+2}{2} $$

$$L_{2}:\frac{x-1}{4}=\frac{y-1}{2}=\frac{z+3}{4} $$

$$L_{3}:\frac{x+2}{1}=\frac{y-1}{0.5}=\frac{z-3}{1} $$

$$L_{4}:\frac{x-3}{2}=\frac{y+1}{4}=\frac{z-2}{-1} $$

Solution
First step, extract the equations

\(L_{1}:P(3,2,-2)\:\langle 2,1,2 \rangle \)

\(L_{2}:P(1,1,-3)\:\langle 4,2,4 \rangle \)

\(L_{3}:P(-2,1,3)\:\langle 1,0.5,1 \rangle \)

\(L_{4}:P(3,-1,-2)\:\langle 2,4,-1 \rangle \)

The lines are parallel if each vector is a scalar multiple for another. The lines \(L_{1}\), \(L_{2}\), and \(L_{3}\) are parallel because

\( \langle 2,1,2 \rangle = 1/2 \langle 4,2,4 \rangle = 2\langle 1,0.5,1 \rangle \).

Identical lines are parallel, by definition, with a distance zero between the lines. This eliminates \(L_{4}\). Identical lines also have equivalent Symmetric equations. Let's find this by setting the \(x\) value equations equal for lines \(L_{1}\), \(L_{2}\), and \(L_{3}\) and see if they are equal.

$$L_{1}:\frac{x-3}{2} \ne L_{2}:\frac{x-1}{4} \text{ so they are parallel}$$

$$L_{2}:\frac{x-1}{4}=L_{3}:\frac{x+2}{1} \rightarrow -3=x \text{ so they are identical}$$

Exercise 11.5.29 Finding the Intersection Point

Determine if the lines intersect. If they intersect, find the intersection point and the cosine for the angle between the lines.

\(x=4t+2\), \(y=3\), and \(z=-t+1\)

\(x=2s+2\), \(y=2s+3\), and \(z=s+1\)

Solution
Rewrite as

\(x=2+4t\), \(y=3+0t\), and \(z=1-t\)

\(x=2+2s\), \(y=3+2s\), and \(z=1+s\)

Extract the points

\(P(2,3,1)\)

\(Q(2,3,1)\)

The lines do intersect and the point is (2,3,1).

Now for the angle
Extract the vectors

\(\textbf{n}= \langle 4, 0, -1 \rangle \)

\(\textbf{v}= \langle 2, 2, 1 \rangle \)

$$\cos \theta = \frac{ \textbf{u} \cdot \textbf{v} }{\| \textbf{u} \| \| \textbf{v} \|} = \frac{8+0-1}{\sqrt{17}\sqrt{9}} = \frac{7}{3\sqrt{17}}$$

Exercise 11.5.40 Find the Equation for a Plane

Find the equation for the plane passing through the point perpendicular to the given vector or line.
$$\text{Point }(3,2,2)$$ $$\text{Perpendicular to }\frac{x-1}{4}=y+2=\frac{z+3}{-3}$$ Solution
Rewrite as $$\text{Point }(3,2,2)$$ $$\text{Perpendicular to }\frac{x-1}{4}=\frac{y+2}{1}=\frac{z+3}{-3}$$ This produces the normal/perpendicular vector

\( \langle 4,1,-3 \rangle\)

which reduces to

\( (3,2,2) \cdot ( 4,1,-3 ) = 4x+y-3z = 8\)

Exercise 11.5.42 Find the Equation for a Plane with Three Points

The plane passes through (3,-1,2), (2,1,5), and (1,-2,-2).
Solution
Use the three points to produce two vectors.

\(\textbf{u} = \langle 2-3,1+1,5-2 \rangle = \langle -1,2,3 \rangle\)

\(\textbf{v} = \langle 1-3,-2-(-1),-2-2 \rangle = \langle -2,-1, -4 \rangle\)

the cross product is

\( \textbf{n} \)	\(=\textbf{u} \times \textbf{v} \)
	$$= \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ -1 & 2 & 3 \\ -2 & -1 & -4 \end{vmatrix} $$
	\(= -5\textbf{i} - 10\textbf{j} + 5 \textbf{k} \)
	\(= \langle a,b,c \rangle \)

Now for the plane equation

\(a(x-x_{1})+b(y-y_{1})+c(z-z_{1}) \)	\(=0\)
\(-5(x-3)-10(y-(-1))+5(z-2) \)	\(=0\)
\(-5x+15-10y+-10+5z-10\)	\(=0\)
\(-5x10y+5z-5 \)	\(=0.\)

Exercise 11.5.54 Find the Equation for a Plane with Two Points

Given the points, (1,0,2) and (2,0,1), find the equation for the plane.
Solution

\(x(1-2)+y(0-0)+z(2-1)=0 \)

\(-2x+z=0 \)

Exercise 11.5.82 Finding the Distance Between a Point and a Plane

Given (0,0,0) and \(5x+y-z=9\), find the distance between them. Solution
The vector \(\textbf{n}= \langle 5,1,-1 \rangle\) is normal to the plane. To find some arbitrary point \(P\) on the plane let \(y=0\) and \(z=0\) to produce \(P(5,0,0)\). Let \(Q(0,0,0)\) and

\vec{PQ} = \langle 0-5, 0-0,0-0 \rangle = \langle -5, 0,0 \rangle \)

the distance \(D\) is

$$D=\frac{| \vec{PQ} \cdot \textbf{n}| }{\| \textbf{n} \|}= \frac{| \langle 5,1,-1 \rangle \cdot \langle -5,0,0\rangle|}{\sqrt{25+1+1}}=\frac{|-25-0-0|}{\sqrt{27}}=\frac{25}{\sqrt{27}} \approx 4.811 $$

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Parent Article: Calculus III Advanced (Course)