Question

Find the symmetric equations of the line of intersection of the given pair of planes. \(4 x+3 y-7 z=1,10 x+6 y-5 z=10\)

Short answer

The symmetric equations are \(\frac{x - 4}{57} = \frac{y + 5}{50}\) and \(z = 0\).

Step-by-step solution

Identify Normal Vectors of the Planes

The normal vector of a plane is given by the coefficients of \(x\), \(y\), and \(z\) in its equation. Here, the normal vector for the first plane, \(4x + 3y - 7z = 1\), is \(\mathbf{n}_1 = \langle 4, 3, -7 \rangle\). The normal vector for the second plane, \(10x + 6y - 5z = 10\), is \(\mathbf{n}_2 = \langle 10, 6, -5 \rangle\).

Find Direction Vector of Line

The cross product of the normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\) gives the direction vector \(\mathbf{d}\) of the line of intersection. Calculate the cross product: \[ \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 3 & -7 \ 10 & 6 & -5 \end{vmatrix} = \langle (-3)(-5) - (6)(-7), (4)(-5) - (10)(-7), (4)(6) - (10)(3) \rangle \]Simplify to find \(\mathbf{d} = \langle 57, 50, 0 \rangle\).

Find a Point on the Line of Intersection

Substitute \(z = 0\) into both plane equations and solve the resulting system of equations to find the \(x\) and \(y\) coordinates of a point on the line. This gives the system:\[\begin{cases} 4x + 3y = 1 \ 10x + 6y = 10 \end{cases} \]Multiply the first equation by 2 and subtract from the second to eliminate \(y\):\[ 10x + 6y - (8x + 6y) = 10 - 2 \Longrightarrow 2x = 8 \Rightarrow x = 4 \]Substitute \(x = 4\) back into \(4x + 3y = 1\):\[ 4(4) + 3y = 1 \Longrightarrow 16 + 3y = 1 \Rightarrow 3y = -15 \Rightarrow y = -5 \]So the point is \((4, -5, 0)\).

Write Symmetric Equations

The symmetric equations for the line are found by using the point \((4, -5, 0)\) and direction vector \(\langle 57, 50, 0 \rangle\):\[ \frac{x - 4}{57} = \frac{y + 5}{50} = \frac{z}{0} \]Since the direction vector component in the \(z\)-direction is zero and if \(z = 0\) is a valid substitute for both equations forming the planes, \(z\) remains 0 along the line of intersection, indicating the line lies in the plane \(z = 0\).

Key concepts

Planes

In mathematics, especially in three-dimensional geometry, a plane is akin to a flat, two-dimensional surface that extends infinitely in all directions. Planes are often described using equations in the format \(Ax + By + Cz = D\), where \(A\), \(B\), and \(C\) are coefficients that affect the plane's orientation, and \(D\) is a constant that dictates the plane's position relative to the origin.

The interaction of two planes can result in three scenarios:

• They might not intersect at all if they are parallel.
• They might intersect along a line, which is the focus of our problem.
• They might coincide fully, meaning they're essentially the same plane.

Understanding planes is crucial when solving problems involving intersections, as it allows you to predict the shape and orientation of intersections by analyzing the planes' equations.

Normal Vectors

Normal vectors are vectors that are perpendicular to a surface, in this case, a plane. For a plane described by the equation \(Ax + By + Cz = D\), the normal vector can be directly obtained from the coefficients \(A\), \(B\), and \(C\).

Consider the equation of the plane \(4x + 3y - 7z = 1\). Here, the normal vector is \(\mathbf{n}_1 = \langle 4, 3, -7 \rangle\). Similarly, for the plane \(10x + 6y - 5z = 10\), the normal vector is \(\mathbf{n}_2 = \langle 10, 6, -5 \rangle\). These vectors are fundamental because they give insight into the orientation of the planes in three-dimensional space.

To find the line of intersection of two planes, we need the direction vector for this line, which is derived from the relationship between these normal vectors.

Cross Product

The cross product is a mathematical operation that takes two vectors and produces another vector that is perpendicular to both of the original vectors. This is very useful when dealing with the intersection of planes, as it helps find the direction vector of the line along which the two planes intersect.

For vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), their cross product \(\mathbf{a} \times \mathbf{b}\) is given by:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix}\]

Applying this to the normal vectors of our planes, \(\mathbf{n}_1 = \langle 4, 3, -7 \rangle\) and \(\mathbf{n}_2 = \langle 10, 6, -5 \rangle\), the resulting cross product is \(\mathbf{d} = \langle 57, 50, 0 \rangle\). This vector \(\mathbf{d}\) represents the direction of the line where both planes intersect.

Direction Vector

The direction vector of a line indicates the line's orientation in space and is key to writing symmetric equations. When finding the line of intersection between planes, the direction vector is obtained through the cross product of the planes' normal vectors.

In our problem, the cross product gave us \(\mathbf{d} = \langle 57, 50, 0 \rangle\), showing that the line runs parallel to this vector. Importantly, because the third component is zero, it points out that there is no change in the \(z\)-coordinate along the line of intersection.

The symmetric equations of a line are written using a direction vector and a point on the line. Typically, they take the form:\[\frac{x - x_0}{d_x} = \frac{y - y_0}{d_y} = \frac{z - z_0}{d_z}\]Substituting our point \((4, -5, 0)\) and direction vector \(\langle 57, 50, 0 \rangle\) leads us to \(\frac{x - 4}{57} = \frac{y + 5}{50} = \frac{z}{0}\). However, since \(z\) remains constant (zero), it underscores the line's confinement in the plane where \(z = 0\).