Problem 36 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the equation of the line of intersection between the planes Q and R with the given equations: Plane Q: \(x + 2y - z = 1\) Plane R: \(x + y + z = 1\) Answer: The equation of the line of intersection is given by the parametric equations: $$x = -\frac{5}{2} + 3t$$ $$y = 1 - 2t$$ $$z = -\frac{1}{2} - t$$

Step 1: Find the Normal Vectors of the Planes

For plane Q, the coefficients of x, y, and z are the components of the normal vector. Similarly, for the plane R, the coefficients of x, y, and z are the components of the normal vector. Write the normal vectors as: $$\vec{n_Q} = \langle1, 2, -1\rangle$$ $$\vec{n_R} = \langle1, 1, 1\rangle$$

Step 2: Compute the Cross Product of the Normal Vectors

In order to get a vector parallel to the line of intersection, compute the cross product of the normal vectors: $$\vec{v} = \vec{n_Q} \times \vec{n_R}$$ $$\vec{v} = \langle(2)(1) - (-1)(1), (-1)(1) - (1)(1), (1)(1) - (2)(1)\rangle$$ $$\vec{v} = \langle2 + 1, -1 - 1, 1 - 2\rangle$$ $$\vec{v} = \langle3, -2, -1\rangle$$ So, the vector parallel to the line of intersection is \(\langle3, -2, -1\rangle\).

Step 3: Find a Point on the Line of Intersection

To find a point on the line, we can solve the simultaneous linear equations formed by the planes: $$x + 2y - z = 1 \ \ (1)$$ $$x + y + z = 1 \ \ (2)$$ To solve for x, y, and z, subtract equation (1) from equation (2): $$y + 2z = 0$$ We arbitrarily set y = 1, so $$z=-\frac{1}{2}$$ Substitute the value of y and z in equation (1) to get: $$x + 2(1) + \frac{1}{2} = 1$$ $$x = -\frac{1}{2} - 2$$ $$x = -\frac{5}{2}$$ So, the point on the line of intersection is \((-\frac{5}{2}, 1, -\frac{1}{2})\).

Step 4: Write the Equation of the Line of Intersection

Now that we have both a point on the line and a vector parallel to the line, we can write the parametric equation of the line using the point \(P(-\frac{5}{2}, 1, -\frac{1}{2})\) and the vector \(\vec{v}=\langle3, -2, -1\rangle\). The parametric equation is given as: $$x = -\frac{5}{2} + 3t$$ $$y = 1 - 2t$$ $$z = -\frac{1}{2} - t$$ Now we have the equation for the line of intersection between the planes Q and R.