Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane containing the line \(x=3 t, y=1+t, z=2 t\) and parallel to the intersection of the planes \(2 x-y+z=0\) and \(y+z+1=0\).

Step-by-Step Solution

Answer

The plane equation is \(x - y - z + 1 = 0\).

Step 1: Understand the Given Line

We are given a line parameterized by the equations \(x=3t\), \(y=1+t\), and \(z=2t\). This means any point on the line can be represented as \((3t, 1+t, 2t)\).

Step 2: Find Direction Vector of Given Line

The coefficients of \(t\) in the line equations give us the direction vector: \((3, 1, 2)\). This vector is parallel to the direction of the line.

Step 3: Find Intersection of Given Planes

The planes are defined by \(2x-y+z=0\) and \(y+z+1=0\). To find the line of intersection, solve these equations simultaneously. Let \(z = t\). Then the system becomes \(2x - y + t = 0\) and \(y + t + 1 = 0\).

Step 4: Solve for Variables in Terms of Parameter

From \(y + t + 1 = 0\), \(y = -t - 1\). Substitute into \(2x - y + t = 0\) to get \(2x - (-t - 1) + t = 0\), simplifying to \(2x + 1 = 0\), \(2x = -1\), thus \(x = -\frac{1}{2}\).

Step 5: Derive Direction Vector for Intersection Line

From the above, the line of intersection is parameterized by \(x = -\frac{1}{2}\), \(y = -t - 1\), \(z = t\). This leads to a direction vector of intersection \((0, -1, 1)\), since \(x\) is constant.

Step 6: Determine Normal Vector of Desired Plane

To find the desired plane, we need a normal vector perpendicular to both the line on the plane and the direction of the intersection line. Cross the direction vectors, \((3, 1, 2)\) and \((0, -1, 1)\), resulting in the vector \((3, -3, -3)\). Simplify this to \((1, -1, -1)\).

Step 7: Equation of Plane Using a Point and a Normal Vector

A point on the line is easily obtained by using \(t=0\): \((0, 1, 0)\). Using the normal vector \((1, -1, -1)\) and the point \((0, 1, 0)\), the equation of the plane is: \[ 1(x-0) -1(y-1) -1(z-0) = 0 \] simplifying to \(x - y - z + 1 = 0\).

Key Concepts

Intersection of Planes

When two planes intersect, they form a line. To find this line, you need to solve the equations of the planes simultaneously. In this case, the planes given by the equations \(2x-y+z=0\) and \(y+z+1=0\) are intersected.
To find their intersection, choose a parameter, often denoted by \(t\), to express one of the variables. Substituting into both equations can help break down the problem into simpler parts.

• For the first equation get one variable in terms of others (e.g., \(z=t\)).
• Substitute into the other equation and solve for the remaining variables.

This will give you a parameterized form of the line of intersection, which helps in identifying the direction vector of this line.

Direction Vector

The direction vector is crucial for understanding the orientation of a line in space. For a line given in parametric form like \(x=3t\), \(y=1+t\), and \(z=2t\), the direction vector consists of the coefficients of \(t\).
Here, the line's direction vector is \((3, 1, 2)\). These values directly tell you the line's direction in 3D space.

• Each component of a direction vector shows how much the position changes along that axis as \(t\) increases.
• Direction vectors can be compared or crossed with others to find perpendicular vectors or to examine co-parallelism.

Understanding direction vectors lays the groundwork for solving more complex geometrical problems.

Parameterized Line

A parameterized line uses a parameter (commonly \(t\)) to express the coordinates of any point on the line. This form gives you a neat mapping from the parameter to the 3D space.
For the line \(x=3t\), \(y=1+t\), and \(z=2t\), as \(t\) varies over all real numbers, you generate all possible points on the line.

• This form is incredibly useful when describing movement or paths, as it provides a dynamic representation of the line.
• The parameter \(t\) is often free to choose, facilitating different interpretations or simplifications such as finding specific points by setting \(t\) to known values.

Parameterized lines are central in vector calculus and are foundational in understanding vector-based geometry.

Normal Vector

The normal vector of a plane is vital as it indicates a plane's orientation in space. It's perpendicular to every direction vector that lies on the plane.
To find a plane's normal vector, particularly when given multiple lines or parameters, often involves using the cross product.

• By crossing two direction vectors, you derive a vector orthogonal to both, serving as the normal vector.
• For example, crossing the vectors \((3, 1, 2)\) and \((0, -1, 1)\) results in \((3, -3, -3)\), which simplifies to \((1, -1, -1)\).

Once you have a normal vector, crafting the equation of a plane becomes straightforward using any point from the desired plane.