Problem 37 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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: $$ 5y + 5z - 9 = 0 \\ \Rightarrow y + z - \frac{9}{5} = 0 \\ \Rightarrow z = \frac{9}{5} - y $$

Step 3: 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 \(\textbf{d}\): $$ \textbf{d} = \textbf{n}_Q \times \textbf{n}_R = \langle 2, -1, 3 \rangle \times \langle -1, 3, 1 \rangle = \langle 8, -7, 7 \rangle $$

Step 4: 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: \(P(-\frac{4}{5}, 0, \frac{9}{5})\).

Step 5: 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.