Find the equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes x + 2y + 3z = 4 and 2x - 3y + 4z = 5.
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Step-by-Step Solution
Step 1: Identify Normal Vectors of Given Planes
The normal vector to a plane given by the equation Ax+By+Cz=D is n=Ai^+Bj^+Ck^. For the given planes, we extract their normal vectors.
Step 2: Find Normal Vector of the Required Plane
If a plane is perpendicular to two other planes, its normal vector must be perpendicular to the normal vectors of both those planes. Therefore, the normal vector of the required plane, n, is parallel to the cross product of the normal vectors of the two given planes, n1 and n2.
Step 3: Calculate the Cross Product
We calculate the determinant to find the cross product. This gives us the normal vector to the plane we are looking for. The components of this vector are the coefficients A, B, and C for the plane's equation.
Step 4: Form the Equation of the Plane
The equation of a plane passing through a point (x1,y1,z1) with normal vector Ai^+Bj^+Ck^ is given by A(x−x1)+B(y−y1)+C(z−z1)=0. We substitute the point (1,2,−3) and the components of the normal vector found in the previous step.
Step 5: Simplify the Equation
We expand and simplify the equation to obtain the general form of the plane's equation.