DETAILS PREVIOUS ANSWERS SCALCET9 12.5.040. ation of the plane. the plane that passes through the line of intersection of the planes x-z=2 and y+4z=1 and is perpendicular to the plane x+y-2z=3 +5z=-11

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DETAILS PREVIOUS ANSWERS SCALCET9 12.5.040. ation of the plane. the plane that passes through the line of intersection of the planes  x-z=2 and  y+4z=1 and is perpendicular to the plane  x+y-2z=3  +5z=-11

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Find Direction Vector: Find the direction vector of the line of intersection of the two given planes. The direction vector of the line of intersection is given by the cross product of the normal vectors of the two planes. Normal vector of plane 1 (x-z=2): n1 = <1, 0, -1> Normal vector of plane 2 (y+4z=1): n2 = <0, 1, 4> Cross product: n1 x n2 = |i j k | |1 0 -1 | |0 1 4 | Calculating the determinant, we get: i(0*4 - 1*(-1)) - j(1*4 - 0*(-1)) + k(1*1 - 0*0) = i(4) - j(4) + k(1) = <4, -4, 1>Find Point of Intersection: Find a point of intersection of the two given planes. To find a point on the line of intersection, we can set one of the variables to zero and solve the system of equations formed by the two planes. Let's set y = 0, then we have: x - z = 2 0 + 4z = 1 From the second equation, we get z = 1/4. Plugging this into the first equation, we get: x - 1/4 = 2 x = 2 + 1/4 x = 9/4 So, one point on the line of intersection is (9/4, 0, 1/4).Use Normal Vectors: Use the direction vector of the line of intersection and the normal vector of the third plane to find the normal vector of the required plane. The normal vector of the third plane (x+y-2z=3): n3 = <1, 1, -2> The required plane must be perpendicular to this plane, so its normal vector must be parallel to n3. Since the required plane also contains the line of intersection, its normal vector must be orthogonal to the direction vector found in Step 1. Therefore, the normal vector of the required plane is the cross product of the direction vector and n3. Direction vector: d = <4, -4, 1> Cross product: d x n3 = |i j k | |4 -4 1 | |1 1 -2 | Calculating the determinant, we get: i((-4)*(-2) - 1*1) - j(4*(-2) - 1*4) + k(4*1 - (-4)*1) = i(8 - 1) - j(-8 - 4) + k(4 + 4) = i(7) - j(-12) + k(8) = <7, 12, 8>Write Plane Equation: Write the equation of the plane using the normal vector and the point found in Step 2. The equation of a plane is given by: a(x - x0) + b(y - y0) + c(z - z0) = 0 where is the normal vector of the plane and (x0, y0, z0) is a point on the plane. Using the normal vector <7, 12, 8> and the point (9/4, 0, 1/4), we get: 7(x - 9/4) + 12(y - 0) + 8(z - 1/4) = 0 Simplifying, we get: 7x - 63/4 + 12y + 8z - 2 = 0 Multiplying through by 4 to clear the fractions, we get: 28x - 63 + 48y + 32z - 8 = 0 Combining like terms, we get: 28x + 48y + 32z - 71 = 0

Find the equation of the plane. $\newline$ Find the plane that passes through the line of intersection of the planes $\newline$ $x-z=2$ and $\newline$ $y+4z=1$ and is perpendicular to the plane $\newline$ $x+y-2z=3$ $\newline$ $x+5z=-11$

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Find the equation of the plane.\newlineFind the plane that passes through the line of intersection of the planes \newlinex−z=2x-z=2x−z=2 and \newliney+4z=1y+4z=1y+4z=1 and is perpendicular to the plane \newlinex+y−2z=3x+y-2z=3x+y−2z=3\newlinex+5z=−11x+5z=-11x+5z=−11

Find the equation of the plane. $\newline$ Find the plane that passes through the line of intersection of the planes $\newline$ $x-z=2$ and $\newline$ $y+4z=1$ and is perpendicular to the plane $\newline$ $x+y-2z=3$ $\newline$ $x+5z=-11$