Problem 64 - Advanced Engineering Mathematics

Question

Determine which of the following planes are parallel to the line \((1-x) / 2=(y+2) / 4=z-5\) (a) \(x-y+3 z=1\) (b) \(6 x-3 y=1\) (c) \(x-2 y+5 z=0\) (d) \(-2 x+y-2 z=7\)

Step-by-Step Solution

Answer

None of the planes are parallel to the given line.

Step 1: Identify Direction Vector of the Line

First, we need to determine the direction vector of the line given by the symmetric equations \( \frac{1-x}{2} = \frac{y+2}{4} = z-5 \). To find the direction vector, write each part in terms of a single parameter \( t \). For the first part, \( 1-x = 2t \) gives \( x = 1 - 2t \). The second part, \( y+2 = 4t \), gives \( y = 4t - 2 \), and for the third part, \( z = t + 5 \). So, the direction vector of the line is \( \langle -2, 4, 1 \rangle \).

Step 2: Identify the Normal Vectors of Planes

Next, we must identify the normal vector of each given plane.(a) The plane \( x - y + 3z = 1 \) has a normal vector \( \langle 1, -1, 3 \rangle \).(b) The plane \( 6x - 3y = 1 \) has a normal vector \( \langle 6, -3, 0 \rangle \).(c) The plane \( x - 2y + 5z = 0 \) has a normal vector \( \langle 1, -2, 5 \rangle \).(d) The plane \( -2x + y - 2z = 7 \) has a normal vector \( \langle -2, 1, -2 \rangle \).

Step 3: Determine Parallelism Conditions

A plane is parallel to a line if the line's direction vector is orthogonal to the plane's normal vector. Two vectors \( \mathbf{v} \) and \( \mathbf{n} \) are orthogonal if their dot product is zero: \( \mathbf{v} \cdot \mathbf{n} = 0 \). Our line's direction vector is \( \mathbf{v} = \langle -2, 4, 1 \rangle \).

Step 4: Calculate Dot Products

Now, compute the dot products with each plane's normal vector:(a) \( \langle -2, 4, 1 \rangle \cdot \langle 1, -1, 3 \rangle = -2 \times 1 + 4 \times (-1) + 1 \times 3 = -2 - 4 + 3 = -3 \).(b) \( \langle -2, 4, 1 \rangle \cdot \langle 6, -3, 0 \rangle = -2 \times 6 + 4 \times (-3) + 1 \times 0 = -12 - 12 = -24 \).(c) \( \langle -2, 4, 1 \rangle \cdot \langle 1, -2, 5 \rangle = -2 \times 1 + 4 \times (-2) + 1 \times 5 = -2 - 8 + 5 = -5 \).(d) \( \langle -2, 4, 1 \rangle \cdot \langle -2, 1, -2 \rangle = -2 \times (-2) + 4 \times 1 + 1 \times (-2) = 4 + 4 - 2 = 6 \).

Step 5: Identify Parallel Planes

Since none of the dot products calculated in Step 4 equals zero, we can conclude that none of the planes are parallel to the line.

Key Concepts

Direction Vectors

A direction vector is crucial in understanding the orientation of a line in space. It tells us the path along which the line extends. Considering the symmetric equation of the line \( \frac{1-x}{2} = \frac{y+2}{4} = z-5 \), we transform it into a parametric form by introducing a parameter \( t \). This process involves writing each part of the equation in terms of \( t \):

• From \( 1-x = 2t \), we get \( x = 1 - 2t \)
• From \( y+2 = 4t \), we obtain \( y = 4t - 2 \)
• Directly, \( z = t + 5 \)

These equations define the line's path, giving the direction vector as \( \langle -2, 4, 1 \rangle \), which provides a clear understanding of how the line extends in space.

Normal Vectors

Normal vectors have a pivotal role in defining the orientation of planes. They are perpendicular to the plane's surface, acting as the plane's spine. For each plane equation given:

• (a) \( x - y + 3z = 1 \) has a normal vector \( \langle 1, -1, 3 \rangle \)
• (b) \( 6x - 3y = 1 \) yields the normal vector \( \langle 6, -3, 0 \rangle \)
• (c) \( x - 2y + 5z = 0 \) provides \( \langle 1, -2, 5 \rangle \)
• (d) \( -2x + y - 2z = 7 \) gives \( \langle -2, 1, -2 \rangle \)

Each of these vectors tells us the tilt or orientation of its respective plane in three-dimensional space.

Orthogonal Vectors

Orthogonality in vector terms means perpendicularity. Two vectors are orthogonal if they meet at a right angle, and importantly, their dot product equals zero. In this case, determining if the planes are parallel to the line involves checking orthogonality between the line's direction vector and each plane's normal vector. For our exercise, this means checking if:- The dot product between the direction vector \( \langle -2, 4, 1 \rangle \) and each normal vector is zero. If it is, the plane and the line are parallel. Since none of the calculated dot products in the exercise equals zero, none of the planes are parallel to the line.

Dot Product

The dot product is a fundamental calculation in vector algebra used to determine the angle between vectors or their orthogonality. This operation involves multiplying corresponding components of two vectors and adding those products:

• For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

In this exercise, the dot product helps verify the orthogonality of the line's direction vector with the planes' normal vectors:

• (a) \(-2 \times 1 + 4 \times (-1) + 1 \times 3 = -3\)
• (b) \(-2 \times 6 + 4 \times (-3) + 1 \times 0 = -24\)
• (c) \(-2 \times 1 + 4 \times (-2) + 1 \times 5 = -5\)
• (d) \(-2 \times (-2) + 4 \times 1 + 1 \times (-2) = 6\)

Each result confirms that none of the planes are parallel to the line, as none of the results are zero.