Find the equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes x + $2y + 3z = 4$ and $2x - 3y + 4z = 5$.

Step 1: Identify Normal Vectors of Given Planes

The normal vector to a plane given by the equation $Ax + By + Cz = D$ is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. 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, $\vec{n}$, is parallel to the cross product of the normal vectors of the two given planes, $\vec{n}_{1}$ and $\vec{n}_{2}$.

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 $(x_1, y_1, z_1)$ with normal vector $A\hat{i} + B\hat{j} + C\hat{k}$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$. 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.

More from this chapter

• Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square uni
• Let L_1: (x-1)/(1) = (y-2)/(1) = (z-1)/(2) and L_2: (x+1)/(1) = (y-2)/(2) = (z)/(4) be two lines. Let L_3 be a line passing through the point (α, β,) and be per
• The line L_1 is parallel to the vector a = -3 i + 2 j + 4 k and passes through the point (7, 6, 2) and the line L_2 is parallel to the vector b = 2 i + j + 3 k
• The distance of the line (x-2)/(2)=(y-6)/(3)=(z-3)/(4) from the point (1,4,0) along the line (x)/(4)=(y-2)/(2)=(z+3)/(3) is:
• Find the shortest distance between the skew lines: r⃗₁ = (i + j + k) + λ(2i + 3j + 4k) and r⃗₂ = (2i + 3j + 5k) + μ(i + 2j + 3k).
• Let the line passing through the points (-1, 2, 1) and parallel to the line (x-1)/(2) = (y+1)/(3) = (z)/(4) intersect the line (x+2)/(3) = (y-3)/(2) = (z-4)/(3)
• Let the shortest distance between the lines (x-3)/(3) = (y-α)/(-1) = (z-3)/(1) and (x+3)/(-3) = (y+7)/(2) = (z-β)/(4) be 3√30. Then the positive value of 5α + β
• Consider the lines L_1: x - 1 = y - 2 = z and L_2: x - 2 = y - 2 = z - 1. Let the feet of the perpendiculars from the point P(5,1,-3) on the lines L_1 and L_2 b

All Three Dimensional Geometry solutions →