Question

The shortest distance between the \(Z\)-axis and the line, \(x+y+2 z-3=0,2 x+3 y+4 z-4=0\) is (1) 1 (2) 2 (3) 3 (4) 0

Short answer

The shortest distance is 2.

Step-by-step solution

Identify the line direction vector

Find a direction vector for the line using the cross product of the normal vectors of the given planes. The normal vectors for the planes are \[ \mathbf{n_1} = (1, 1, 2) \ \mathbf{n_2} = (2, 3, 4) \ \]. Cross product: \[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 2 \ 2 & 3 & 4 \end{vmatrix} = (-2, 0, 1) \].

Find a point on the line

Solve the plane equations simultaneously to find a common point. Let \( z = 0 \): \[ x + y - 3 = 0 \rightarrow y = 3 - x \ 2x + 3y - 4 = 0 \rightarrow 2x + 3(3 - x) - 4 = 0 \rightarrow -x + 5 = 0 \rightarrow x = 5 \rightarrow y = -2 \]. The point is (5, -2, 0).

Set up the distance formula

Use the formula for the distance from a point to a line in 3D: \[ D = \frac{\| (\mathbf{r} - \mathbf{r_0}) \times \mathbf{d} \| }{\| \mathbf{d} \|} \], where \( \mathbf{r} = (0, 0, z) \) is any point on the Z-axis, \(\mathbf{r_0} = (5, -2, 0) \) the point on the line, and \( \mathbf{d} = (-2, 0, 1) \) the direction vector.

Calculate the distance

Plug in known values: \[ \mathbf{r - r_0} = (-5, 2, z) \ ( \mathbf{r - r_0}) \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -5 & 2 & z \ -2 & 0 & 1 \end{vmatrix} = (2, -z, 4) \ \|\mathbf{r - r_0} \times \mathbf{d}\| = \sqrt{2^2 + (-z)^2 + 4^2} = \sqrt{4 + z^2 + 16} = \sqrt{20 + z^2} \ \|d\| = \sqrt{(-2)^2 + 1^2} = \sqrt{5} \ D = \frac{\sqrt{20}}{\sqrt{5}} = 2 \ Hence, the distance is 2. \]

Key concepts

vector cross product

Understanding the vector cross product is essential for finding the shortest distance between a line and a point in 3D space. The cross product is a mathematical operation that takes two vectors and produces another vector that is perpendicular to both of the original vectors. To compute the cross product of vectors \( \textbf{a} = (a_1, a_2, a_3) \) and \( \textbf{b} = (b_1, b_2, b_3) \), we use the following formula: \( \textbf{a} \times \textbf{b} = (a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1) \).

In our solution, we find the direction vector of the line by taking the cross product of the normal vectors to the given planes. The normal vectors are \( (1, 1, 2) \) and \( (2, 3, 4) \). Performing the cross product, we get: \( \textbf{d} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 1 & 1 & 2 \ 2 & 3 & 4 \ \ \textbf{d} = (-2, 0, 1) \).

The resulting vector \( (-2, 0, 1) \) represents the direction of the line formed by the intersection of the planes.

distance formula in 3D

The distance formula in 3D is crucial for determining the shortest distance between a point and a line. The distance from a point to a line in 3D is found using the following formula: \( D = \frac{\| (\textbf{r - r_0}) \times \textbf{d} \|}{\| \textbf{d} \|} \), where \( \textbf{r} \) is the point in space, \( \textbf{r_0} \) is a point on the line, and \( \textbf{d} \) is the direction vector of the line.

First, we find a specific point on the line by solving the given plane equations. We chose \( z = 0 \) for simplicity, leading to the point (5, -2, 0). With this point and the direction vector \( (-2, 0, 1) \), we used the formula:

1. Find the vector between the point on the Z-axis and the point on the line: \( \textbf{r - r_0} = (0,0,z) - (5, -2, 0) = (-5, 2, z) \).

2. Compute the cross product of this vector with the direction vector \( \textbf{d} \): \( (\textbf{r - r_0}) \times \textbf{d} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ -5 & 2 & z \ -2 & 0 & 1 \ \ \ (2, -z, 4) \).
Compute its magnitude \( \|\textbf{r - r_0} \times \textbf{d}\| = \sqrt{2^2 + (-z)^2 + 4^2} = \sqrt{20 + z^2} \).

3. Finally, calculate \( \| \textbf{d} \| = \sqrt{(-2)^2 + 1^2} = \sqrt{5} \).
The distance is \( D = \frac{\sqrt{20}}{\sqrt{5}} = 2 \).

solving plane equations

Solving plane equations to find a common point is a critical part of finding the shortest distance in 3D space. The two given plane equations are: \( x + y + 2z - 3 = 0 \) and \( 2x + 3y + 4z - 4 = 0 \). To find a point that lies on the line of intersection of these planes, we can pick a value for one of the variables.

For simplicity, let \( z = 0 \). Substitute \( z = 0 \) in both equations:

1. From the first equation: \( x + y - 3 = 0 \Rightarrow y = 3 - x \).
2. Substitute \( y = 3 - x \) into the second equation: \( 2x + 3(3 - x) - 4 = 0 \).

Simplifying:
\( 2x + 9 - 3x - 4 = 0 \Rightarrow -x + 5 = 0 \Rightarrow x = 5 \).

With \( x = 5 \) and \( y = 3 - 5 = -2 \), we get the point (5, -2, 0). By solving these equations, we identify a point on the line that helps us use the distance formula effectively.