Question

Find a set of parametric equations for the line passing through the point (1,0,2) that is parallel to the plane given by \(x+y+z=5\) and perpendicular to the line \(x=t\) \(y=1+t, z=1+t\)

Short answer

The parametric equations of the line are \(x=1\), \(y=0\), \(z=2\).

Step-by-step solution

Find Normal Vector

First, find the normal vector of the plane. The normal vector of a plane given by an equation of the form \(ax+by+cz=d\) is \([a, b, c]\). Here, for the plane \(x+y+z=5\), the normal vector is \([1,1,1]\).

Find Vector of Given Line

The line given by \(x=t\), \(y=1+t\), \(z=1+t\) has the vector from \(t=0\) to \(t=1\) as its direction vector. That is, the line can be written in vector form as \(r(t) = t\vec{i} + (1+t)\vec{j} + (1+t)\vec{k}\). From \(t=0\) to \(t=1\), the vector is \(\vec{i}+\vec{j}+\vec{k} = [1,1,1]\).

Find Direction Vector

The required direction vector of the line is the cross product of the normal vector of the plane and the direction vector of the given line. So, it's \([1,1,1]\) cross \([1,1,1]\). It must be noted that the cross product of a vector with itself is always zero since they are not linearly independent. Hence, the direction vector of the line is \([0,0,0]\).

Formulate Parametric Equations

The parametric equations are formed using the coordinates of the given point and the components of the direction vector. Here, the direction vector is \([0,0,0]\), which coincides with the origin. So, this is a degenerate case and implies that every line through the point (1,0,2) that is parallel to the plane is also the point itself. Therefore, the parametric equations of the line are \(x=1\), \(y=0\), \(z=2\).

Key concepts

Cross Product

The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space. It has applications in physics, engineering, and mathematics, particularly in the context of determining a vector that is perpendicular to two given vectors.

For example, if we have two vectors \( \vec{A} = [A_1, A_2, A_3] \) and \( \vec{B} = [B_1, B_2, B_3] \), the cross product is given by \( \vec{A} \times \vec{B} = [A_2B_3-A_3B_2, A_3B_1-A_1B_3, A_1B_2-A_2B_1] \). The resulting vector is orthogonal to both \( \vec{A} \) and \( \vec{B} \) and the direction of the vector follows the right-hand rule.

In the case of our exercise, a cross product is used to find a direction vector for the line. However, an interesting note is that the cross product of a vector with itself, as we have with the direction vector of the line and the normal vector of the plane \( [1,1,1] \times [1,1,1] \), results in the zero vector \( [0,0,0] \). This suggests that finding a perpendicular direction in this specific scenario leads to a degenerate case.

Normal Vector

A normal vector is a vector that is perpendicular to a surface or a plane. In the context of our exercise, the normal vector of a plane represented by the equation \( ax + by + cz = d \) is straightforwardly the coefficients of the x, y, and z terms, succinctly written as the vector \( [a, b, c] \).

For the given plane \( x + y + z = 5 \), the normal vector is \( [1, 1, 1] \), which would generally be used to determine whether other vectors or planes are parallel or perpendicular to it. In our exercise scenario, the normal vector inadvertently highlighted the uniqueness of the line's direction vector being perpendicular to itself, something that is not usually encountered in typical geometric situations.

Direction Vector

The direction vector of a line in space is a vector that points in the direction of the line. It is obtained by taking any two points on the line and subtracting their coordinates to get the vector components.

In step 2 of our exercise, the direction vector for the given line defined by \( x = t \) and \( y = z = 1 + t \) is calculated. This line can be represented in vector form as \(\vec{r}(t) = t\vec{i} + (1+t)\vec{j} + (1+t)\vec{k}\), where the direction vector is derived from the difference between the coordinates at \( t = 1 \) and \( t = 0 \) resulting in \( \vec{v} = \vec{i} + \vec{j} + \vec{k} = [1, 1, 1] \). Usually, the direction vector is non-zero and unique, but as the solution demonstrates, this direction vector is an exception due to it being the same as the normal vector of the plane.

Vector Form

The vector form of a line in three-dimensional space gives a clear mathematical representation of the line in terms of a single variable, typically noted as \( t \). The generic vector equation is \( \vec{r}(t) = \vec{a} + t\vec{d} \), where \( \vec{a} \) is a point on the line (in vector form) and \( \vec{d} \) is the direction vector.

In our presented problem, the direction vector turned out to be zero, which means that we cannot use this direction to establish a path for the line. Therefore, the parametric equations do not describe a line but rather a fixed point in space, represented by the coordinates of the given point (1,0,2). This is a unique instance where the line that is supposed to be defined by the parametric equations collapses into a single point, showing that not all sets of parametric equations will result in a continuous line.