Problem 75 - Advanced Engineering Mathematics

Question

In Problems find an equation of the plane that contains the given line and is orthogonal to the indicated plane. $$ x=4+3 t, y=-t, z=1+5 t ; x+y+z=7 $$

Step-by-Step Solution

Answer

Equation of the plane: \(-3x - 4y + 2z + 10 = 0\).

Step 1: Identify the directional vector of the line

The line is given in parametric form: \(x=4+3t, y=-t, z=1+5t\). The vector \(\mathbf{d}\) that represents the direction of the line is \(\langle 3, -1, 5 \rangle\).

Step 2: Find the normal vector of the given plane

The given plane is in the equation form \(x + y + z = 7\). The normal vector \(\mathbf{n_1}\) of this plane is \(\langle 1, 1, 1 \rangle\), which is the vector of the coefficients of \(x, y,\) and \(z\).

Step 3: Obtain the normal vector of the required plane

Since the desired plane contains the line, its normal vector \(\mathbf{n_2}\) must be orthogonal to \(\mathbf{d}\) and \(\mathbf{n_1}\). This can be found using the cross product: \(\mathbf{n_2} = \mathbf{d} \times \mathbf{n_1}\).

Step 4: Calculate the cross product

The cross product of \(\mathbf{d} = \langle 3, -1, 5 \rangle \) and \(\mathbf{n_1} = \langle 1, 1, 1 \rangle \) is calculated as follows:\[ \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & -1 & 5 \ 1 & 1 & 1 \end{vmatrix} = (-1)(1) - (5)(1) \mathbf{i} - (3)(1) - (5)(1) \mathbf{j} + (3)(1) - (-1)(1) \mathbf{k} \]\[ \mathbf{n_2} = (-6, -8, 4) \]Simplifying gives a normal vector of \(\langle -6, -8, 4 \rangle \).

Step 5: Formulate the equation of the desired plane

The equation of a plane can be given by \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\), where \(\langle a, b, c \rangle\) is the normal vector and \((x_0, y_0, z_0)\) is any point on the line. Using the initial vector of the line \((4, 0, 1)\) and the normal vector \(\langle -6, -8, 4 \rangle\), substituting gives:\[-6(x - 4) - 8(y - 0) + 4(z - 1) = 0\].

Step 6: Simplify the plane equation

Expanding the equation gives:\[-6x + 24 - 8y + 4z - 4 = 0\]\[-6x - 8y + 4z + 20 = 0\].Dividing throughout by 2 for simplicity:\[-3x - 4y + 2z + 10 = 0\].

Key Concepts

Parametric Form of a Line

In geometry, the parametric form of a line provides a way to express a line in terms of a parameter, usually denoted as \(t\).
This method is extremely useful for describing lines in three-dimensional space.
When a line is written in the parametric form, every point on it is expressed as:

• \(x = x_0 + at\)
• \(y = y_0 + bt\)
• \(z = z_0 + ct\)

Here, \((x_0, y_0, z_0)\) is a point on the line, and \(\langle a, b, c \rangle\) is the directional vector of the line.
This vector indicates the direction in which the line moves.
The parameter \(t\) varies over all real numbers and each value of \(t\) gives a unique point on the line.
This is what makes the parametric form so versatile and easy to work with.

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors.
It is especially useful in three dimensions where you need to find a vector normal to a plane.
To compute the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\), the result is defined as:

• \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\)

Here, \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the unit vectors along the x, y, and z axes.
The determinant of this matrix results in a new vector:

• \(\mathbf{c} = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k}\)

It's essential to note that the cross product is only defined in three dimensions.
It results in a vector that is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\).
This vector is crucial for defining the normal vector of a plane.

Normal Vector

A normal vector is a fundamental concept in plane geometry.
It is a vector that is perpendicular to a given plane or surface.
For any plane given by the equation \(ax + by + cz = d\), the vector \(\langle a, b, c \rangle\) is the normal vector.
This vector sets the orientation of the plane and plays a vital role in defining its characteristics.
A normal vector is useful because it helps define a plane uniquely.
It also assists in solving geometric problems, like finding angles between planes or checking if two planes are orthogonal when their normal vectors are cross-orthogonal.
In many applications, especially in physics and engineering, normal vectors are critical to express forces and other directional quantities.

Orthogonal Planes

In three-dimensional geometry, orthogonal planes are two planes that intersect at a right angle (90 degrees).
The concept of orthogonal planes is deeply tied to the idea of normal vectors.
Two planes are orthogonal if and only if their normal vectors are perpendicular to each other.
In mathematical terms, this condition translates to the dot product of the normal vectors being zero.
When dealing with problems that involve orthogonal planes, it's essential to find two vectors that are orthogonal by examining their cross product.
Once you have normal vectors of two planes and they satisfy this perpendicularity condition, you can be confident the planes are orthogonal.
Orthogonal planes have significant implications in practical applications like computer graphics where three-dimensional modeling is involved.