Problem 38 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The parametric equation of the line of intersection is $\vec{r}(t) = \left( \frac{7}{2}t, \frac{-5}{2}t - 1, 3t \right)$.

Step 1: Find a Point on the Line of Intersection

To find a point on the line of intersection, we need to solve the system of two equations formed by the given planes Q and R: $$ Q: x-y-2z=1 \\ R: x+y+z=-1 $$ To solve this system, we can add the two equations to eliminate the y variable: $$ Q+R: x-y-2z+x+y+z = 1-1 \\ 2x-z = 0 $$ Now we can assign an arbitrary value for one of the variables left (let’s choose \(z=t\)) and find the corresponding values for the other variables: $$ x = \frac{1}{2}z \\ x = \frac{1}{2}t $$ Now we have the parametric representation of the variables x and z. This implies that for any value of \(t\), we can find the values of x and z that lie on the line of intersection. To find the corresponding y value: $$ y = z - x - 1 \\ y = t - \frac{1}{2}t - 1 \\ y = \frac{1}{2}t - 1 $$ So the parametric representation of the point on the line of intersection is: $$ P(t) = (\frac{1}{2}t, \frac{1}{2}t - 1, t) $$

Step 2: Find the Direction Vector of the Line of Intersection

To find the direction vector of the line of intersection, we need to find the cross product of the normal vectors of the two planes. The normal vectors of planes Q and R are given by the coefficients of the variables in the equations: $$ \vec{N_Q} = (1, -1, -2) \\ \vec{N_R} = (1, 1, 1) $$ Now we compute the cross product \(\vec{N_Q} \times \vec{N_R}\): $$ \vec{d} = \vec{N_Q} \times \vec{N_R} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -2 \\ 1 & 1 & 1 \end{vmatrix} \\ \vec{d} = (3\hat{i} - 3\hat{j} + 2\hat{k}) $$ So the direction vector of the line of intersection is \(\vec{d} = (3, -3, 2)\).

Step 3: Write the Equation of the Line of Intersection

Now, we have the parametric representation of a point P(t) on the line of intersection and the direction vector \(\vec{d} = (3, -3, 2)\). The parametric equation of the line can be written as: $$ \vec{r}(t) = P(t) + t \vec{d} \\ \vec{r}(t) = (\frac{1}{2}t, \frac{1}{2}t - 1, t) + t(3, -3, 2) \\ \vec{r}(t) = (\frac{1}{2}t + 3t, \frac{1}{2}t - 1 - 3t, t + 2t) \\ \vec{r}(t) = \left( \frac{7}{2}t, \frac{-5}{2}t - 1, 3t \right) $$ So the parametric equation of the line of intersection is: $$ \vec{r}(t) = \left( \frac{7}{2}t, \frac{-5}{2}t - 1, 3t \right) $$