Problem 76 - Calculus Early Transcendentals

Question

Find an equation of the plane passing through (0,-2,4) that is orthogonal to the planes \(2 x+5 y-3 z=0\) and \(-x+5 y+2 z=8\)

Step-by-Step Solution

Answer

Question: Find the equation of the plane orthogonal to the planes with equations \(2x + 5y - 3z = 0\) and \(-x + 5y + 2z = 8\), and passing through the point P(0, -2, 4). Answer: The equation of the plane is \(19x + y + 15z = 66\).

Step 1: Find the Normal Vectors of the Given Planes

The equations of the given planes are: \(2x + 5y - 3z = 0\) and \(-x + 5y + 2z = 8\) The normal vector \(\vec{N_1}\) of the first plane can be found by looking at the coefficients of x, y, and z: \(\vec{N_1} = (2, 5, -3)\) Similarly, the normal vector \(\vec{N_2}\) of the second plane is \(\vec{N_2} = (-1, 5, 2)\).

Step 2: Find the Cross Product of the Normal Vectors

To find a vector that is orthogonal to both \(\vec{N_1}\) and \(\vec{N_2}\), we have to take the cross product: \(\vec{N_3} = \vec{N_1} \times \vec{N_2}\) Using the determinant method: \(\vec{N_3} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 2 & 5 & -3 \\ -1 & 5 & 2 \end{vmatrix} = \boldsymbol{i}(5(2) - 2(-3)) - \boldsymbol{j}(2(2) - (-1)(-3)) + \boldsymbol{k}(2(5) - 5(-1))\) So, \(\vec{N_3} = (19\boldsymbol{i} - (-1)\boldsymbol{j} + 15\boldsymbol{k}) = (19, 1, 15)\)

Step 3: Find the Equation of the Plane

Now, we have the normal vector \(\vec{N_3}\) of the required plane and the point P(0, -2, 4) that it passes through. We can use the point-normal form of the plane equation to find the equation of our plane: \((x - x_0, y - y_0, z - z_0) \cdot \vec{N_3} = 0\) Substituting the values of P and \(\vec{N_3}\): \((x - 0, y - (-2), z - 4) \cdot (19, 1, 15) = 0\) Expanding the dot product, we get: \(19(x - 0) + 1(y - (-2)) + 15(z - 4) = 0\) Simplifying the equation, we obtain the final equation of the plane: \(19x + y + 15z = 66\)