Question

Give the equation of the described plane in standard and general forms. Contains the point (5,7,3) and the line $$ \ell(t)=\left\\{\begin{array}{l} x=t \\ y=t \\ z=t \end{array}\right. $$

Short answer

Plane equation: \(4x - 2y - 2z = 0\)

Step-by-step solution

Identify Direction Vectors

The line is given by the parametric equations \(x = t\), \(y = t\), and \(z = t\). Therefore, a direction vector for the line is \(\mathbf{d} = \langle 1, 1, 1 \rangle\). Also, note the point \( P(5, 7, 3) \) that lies on the plane. We will need another vector to express the plane.

Find Normal Vector using Two Points on the Line

We have a point on the plane \((5, 7, 3)\) and a direction vector \(\mathbf{d} = \langle 1, 1, 1 \rangle\), plus the point \(A(0, 0, 0)\) on the line when \(t = 0\). To form the plane, we can use the cross product of two vectors, \(\mathbf{AP} = \langle 5, 7, 3 \rangle\) and \(\mathbf{d} = \langle 1, 1, 1 \rangle\).

Compute the Cross Product

The normal vector \(\mathbf{n}\) can be found by computing \(\mathbf{AP} \times \mathbf{d}\). \[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 5 & 7 & 3 \ 1 & 1 & 1 \end{vmatrix} = \mathbf{i}(7-3) - \mathbf{j}(5-3) + \mathbf{k}(5-7) \]\(\mathbf{n} = \langle 4, -2, -2 \rangle\).

Formulate Plane Equation

Using the normal vector \(\mathbf{n} = \langle 4, -2, -2 \rangle\) and the point \(P(5, 7, 3)\), the plane equation in standard form is:\[ 4(x - 5) - 2(y - 7) - 2(z - 3) = 0 \]

Convert to General Form

Expanding the standard form equation gives us the general form. Start by expanding:\[ 4x - 20 - 2y + 14 - 2z + 6 = 0 \]Simplify to obtain the general form:\[ 4x - 2y - 2z = 0 \]This is the equation of the plane in general form.

Key concepts

Parametric Equations

Parametric equations are a way of expressing a three-dimensional line or curve by breaking it down into simpler one-dimensional components using a parameter, typically denoted as \( t \). For example, the line \( \ell(t) \) in the exercise is defined by a set of parametric equations: \( x = t \), \( y = t \), and \( z = t \).
This form allows us to describe each coordinate point on the line in terms of the parameter \( t \), which varies over the entire set of real numbers.

• Each component function specifies how each coordinate changes as \( t \) changes.
• Such representation is useful for describing linear paths in space with ease.

The advantage of using parametric equations lies in their ability to express motion along paths and make it easy to relate changes in one coordinate with simultaneous changes in others.

Direction Vector

A direction vector gives a sense of direction along a line or within a plane. It's like an arrow that points along a straight path in space.
For the line \( \ell(t) \) given by the parametric equations above, the direction vector is \( \mathbf{d} = \langle 1, 1, 1 \rangle \).
This vector indicates that for every unit increase in \( t \), each coordinate (\( x, y, \) and \( z \)) increases by 1.

• This means the line moves diagonally in a 1:1:1 ratio.
• Every direction vector has components that indicate how much the line moves along each axis for a unit change in parameter.

Thus, direction vectors are fundamental for identifying orientation in 3D space and are crucial for calculations involving lines and planes.

Cross Product

The cross product is an essential operation in vector algebra, particularly in computer graphics and physics. When calculating the normal vector to a plane, the cross product becomes extremely useful.
This operation assigns a vector orthogonal (or perpendicular) to two given vectors.
For instance, in our exercise, to find the normal vector of the plane, we took the cross product of vectors \( \mathbf{AP} = \langle 5, 7, 3 \rangle \) and \( \mathbf{d} = \langle 1, 1, 1 \rangle \).

• The cross product \( \mathbf{AP} \times \mathbf{d} \) is computed using a determinant, which results here in the vector \( \langle 4, -2, -2 \rangle \).
• The result is a vector perpendicular to both \( \mathbf{AP} \) and \( \mathbf{d} \).

Thus, the cross product aids in defining planes in three dimensions by providing a vector that serves as a potential normal vector to the plane.

Normal Vector

A normal vector to a plane is a vector that is perpendicular to that plane. It plays a crucial role in determining the plane's orientation in space.
In our context, once we computed the cross product, we found the normal vector \( \mathbf{n} = \langle 4, -2, -2 \rangle \).

• Using this normal vector with a known point on the plane \( P(5, 7, 3) \), we formulate the plane's equation in standard form: \[ 4(x - 5) - 2(y - 7) - 2(z - 3) = 0 \].
• The parameters of the normal vector directly become coefficients in the plane's equation.

A normal vector is instrumental in forming three-dimensional geometry and setting the plane equation's constraints, thereby figuring prominently in spatial calculations.