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Chapter 13: Problem 37

Find an equation of the line of intersection of the planes \(Q\) and \(R\). $$Q: 2 x-y+3 z-1=0 ; R:-x+3 y+z-4=0$$

Short Answer

Expert verified
Answer: The vector equation of the line where the planes Q and R intersect is given by: $\textbf{r}(t) = \langle -\frac{4}{5}, 0, \frac{9}{5} \rangle + t \langle 8, -7, 7 \rangle$.

Step by step solution

01

Solving given equations simultaneously for x and z

Let's write the given equations: $$ Q: 2x - y + 3z - 1 = 0 \\ R: -x + 3y + z - 4 = 0 $$ Now we will solve the given plane equations for x and z. To do so, we can multiply the second equation (R) by 2 and then add it to the first equation (Q): $$ 2(-x + 3y + z - 4) = -2x + 6y + 2z - 8 \\ 2x - y + 3z - 1 + (-2x + 6y + 2z - 8) = 0 \\ \Rightarrow 5y + 5z - 9 = 0 $$ We now have one equation with only two variables: y and z.
02

Simplifying and expressing z in terms of y

From the equation obtained in step 1, we can simplify it and express z in terms of y: $$ 5y + 5z - 9 = 0 \\ \Rightarrow y + z - \frac{9}{5} = 0 \\ \Rightarrow z = \frac{9}{5} - y $$
03

Finding the direction vector of the line

We know that the direction vector of the intersection line can be obtained by taking the cross product of the normal vectors of the given planes. Q has normal vector \(\textbf{n}_Q = \langle 2, -1, 3 \rangle\) and R has normal vector \(\textbf{n}_R = \langle -1, 3, 1 \rangle\). We'll compute \(\textbf{n}_Q \times \textbf{n}_R\) as the direction vector \(\textbf{d}\): $$ \textbf{d} = \textbf{n}_Q \times \textbf{n}_R = \langle 2, -1, 3 \rangle \times \langle -1, 3, 1 \rangle = \langle 8, -7, 7 \rangle $$
04

Finding a point on the line of intersection

We'll use the simplified equation from Step 2 to find a point that lies on both planes. We can choose any value for y, so let's choose \(y = 0\). Then, \(z = \frac{9}{5} - 0 = \frac{9}{5}\). Now, we'll substitute these values into any of the original plane equations to find x (we'll use equation Q): $$ 2x - 0 + 3(\frac{9}{5}) - 1 = 0 \\ \Rightarrow 2x = -\frac{18}{5} + 1 \\ \Rightarrow x = -\frac{4}{5} $$ Now we have a point on the intersection line: \(P(-\frac{4}{5}, 0, \frac{9}{5})\).
05

Writing the vector equation of the line

We have a point on the intersection line, \(P\), and the direction vector, \(\textbf{d}\). We can now write the vector equation of the line of intersection as: $$ \textbf{r}(t) = P + t\textbf{d} = \langle -\frac{4}{5}, 0, \frac{9}{5} \rangle + t \langle 8, -7, 7 \rangle $$ This is the vector equation of the line where the planes Q and R intersect.

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Most popular questions from this chapter

Rectangular boxes with a volume of \(10 \mathrm{m}^{3}\) are made of two materials. The material for the top and bottom of the box costs \(\$ 10 / \mathrm{m}^{2}\) and the material for the sides of the box costs \(\$ 1 / \mathrm{m}^{2}\). What are the dimensions of the box that minimize the cost of the box?

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(0,2)}(2 x y)^{x y}$$

a. Show that the point in the plane \(a x+b y+c z=d\) nearest the origin is \(P\left(a d / D^{2}, b d / D^{2}, c d / D^{2}\right),\) where \(D^{2}=a^{2}+b^{2}+c^{2} .\) Conclude that the least distance from the plane to the origin is \(|d| / D\). (Hint: The least distance is along a normal to the plane.) b. Show that the least distance from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) to the plane \(a x+b y+c z=d\) is \(\left|a x_{0}+b y_{0}+c z_{0}-d\right| / D\).

Let $$f(x, y)=\left\\{\begin{array}{ll}\frac{\sin \left(x^{2}+y^{2}-1\right)}{x^{2}+y^{2}-1} & \text { if } x^{2}+y^{2} \neq 1 \\\b & \text { if } x^{2}+y^{2}=1\end{array}\right.$$ Find the value of \(b\) for which \(f\) is continuous at all points in \(\mathbb{R}^{2}\).

Find the domains of the following functions. Specify the domain mathematically and then describe it in words or with a sketch. $$h(x, y, z)=\sqrt[4]{z^{2}-x z+y z-x y}.$$
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