Question

Find the shortest distance from the origin to the line of intersection of the planes \(2 x+y-z=1\) and \(x-y+z=2\)

Short answer

The shortest distance is \( \frac{ \sqrt{22} }{22} \).

Step-by-step solution

Find a Direction Vector of the Line of Intersection

To find a direction vector of the line of intersection, we need to find the cross product of the normal vectors of the planes. The normal vector of the plane 1, \(2x+y-z=1\) is \(\mathbf{n_1} = [2, 1, -1]\). The normal vector of the plane 2, \(x-y+z=2\) is \(\mathbf{n_2} = [1, -1, 1]\). Now find the cross product \(\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}\).

Calculate the Cross Product

Finding the cross product of \(\mathbf{n_1} = [2, 1, -1]\) and \(\mathbf{n_2} = [1, -1, 1]\). Using the formula for the cross product, we have: \[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & -1 \ 1 & -1 & 1 \end{vmatrix} = [0, -1-1, -2-1] = [2, 3, -3] \]

Find a Point of Intersection

Find a common point that lies on both planes. Solve the system of equations given by the planes. One approach is to use substitution or elimination methods. By solving \(2x+y-z=1\) and \(x-y+z=2\), assume a value for \(z\), e.g., \(z = 0\), then solve for \(x\) and \(y\).

Solve for Specific Values

With \(z = 0\): The system becomes \(2x + y = 1\) and \(x - y = 2\). Solving these, we get \(x=1\) and \(y=-1\). Hence, a point on the line is (1, -1, 0).

Find the Distance from Origin to the Line

To find the shortest distance from the origin to the line, use the distance formula: \[ D = \frac{ \left| \mathbf{d} \cdot \mathbf{PQ} \right| }{ \| \mathbf{d} \| } \] where \(\mathbf{PQ} = (1-0, -1-0, 0-0) = [1, -1, 0] \), and \(\mathbf{d} = [2, 3, -3] \). Calculate the absolute value of the dot product and norm of the direction vector.

Calculate Dot Product and Norm

Compute the dot product \( \mathbf{d} \cdot \mathbf{PQ} = 2 \cdot 1 + 3 \cdot (-1) + (-3) \cdot 0 = -1 \). The norm of \(\mathbf{d}\) is \( \sqrt{2^2 + 3^2 + (-3)^2} = \sqrt{22} \).

Compute the Distance

Now compute the shortest distance using the formula: \[ D = \frac{ -1 }{ \sqrt{22} } = \frac{ 1 }{ \sqrt{22} } = \frac{ \sqrt{22} }{22} \].

Key concepts

Cross Product

The cross product is a vector operation that yields a vector perpendicular to two given vectors. For vectors \( \mathbf{A} \) and \( \mathbf{B} \), the cross product \( \mathbf{A} \times \mathbf{B} \) is found using a determinant:

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A1 & A2 & A3 \ B1 & B2 & B3 \end{vmatrix} \]

In our example, to find the direction vector of the line of intersection of two planes, we cross the normal vectors \( \mathbf{n_1} = [2, 1, -1] \) and \( \mathbf{n_2} = [1, -1, 1] \). The resulting direction vector \( \mathbf{d} = [2, 3, -3] \) ensures the vector lies on the intersection line and is perpendicular to both normal vectors.

Normal Vector

A normal vector is a perpendicular vector to a given surface, like a plane. It's derived directly from the coefficients of the planes' equations. For instance, the plane equation \( 2x + y - z = 1 \) has a normal vector \( \mathbf{n_1} = [2, 1, -1] \).

Similarly, the plane \( x - y + z = 2 \) has a normal vector \( \mathbf{n_2} = [1, -1, 1] \).
The cross product of these normal vectors helps us find a direction vector that lies on the intersection line of these planes. The perpendicularity property of the direction vector to both normal vectors confirms correctness.

Distance Formula

The distance from a point to a line in 3D can be calculated using the formula:

\[ D = \frac{ | \mathbf{d} \cdot \mathbf{PQ} | }{ \| \mathbf{d} \| } \]
Here, \( \mathbf{PQ} \) is a vector from the point to any point on the line, and \( \mathbf{d} \) is the direction vector of the line.

In our case, \( \mathbf{PQ} = [1, -1, 0] \) is from the origin (0,0,0) to the point (1, -1, 0) on the line. The direction vector \( \mathbf{d} = [2, 3, -3] \).

• The dot product \( \mathbf{d} \cdot \mathbf{PQ} = (2*1) + (3*-1) + (-3*0) = -1 \)
• The norm \( \| \mathbf{d} \| = \sqrt{2^2 + 3^2 + (-3)^2} = \sqrt{22} \)

This leads to the final distance formula calculation.

Norm Calculation

The norm of a vector, often referred to as its magnitude or length, is calculated using the Euclidean formula:

\[ \| \mathbf{A} \| = \sqrt{A1^2 + A2^2 + A3^2} \]
It's a scalar value representing the length of the vector in space. For direction vector \( \mathbf{d} = [2, 3, -3] \), we calculate the norm:

• \[ \| \mathbf{d} \| = \sqrt{2^2 + 3^2 + (-3)^2} = \sqrt{4 + 9 + 9} = \sqrt{22} \]

All these steps come together to give us the shortest distance from the origin to the line of intersection, which we found to be \( \frac{ \sqrt{22} }{22} \).