Problem 3 - Active Calculus Multivariable

Question

Consider the line which passes through the point \(\mathrm{P}(3,-5,-1),\) and which is parallel to the line \(x=1+6 t, y=2+2 t, z=3+6 t\) Find the point of intersection of this new line with each of the coordinate planes: xy-plane:_________,_________,_________ xz-plane:_________,_________,_________ yz-plane:_________,_________,_________

Step-by-Step Solution

Answer

xy-plane: \( (4, -\frac{23}{6}, 0) \) xz-plane: \( (18, 0, 14) \) yz-plane: \( (0, -6, -4) \)

Step 1: Equation of the New Line

Since the new line has to pass through the point P(3, -5, -1) and be parallel to the given line, we know that their direction vectors must be the same. The line given has the parametric form \(x=1+6t\), \(y=2+2t\), \(z=3+6t\). Thus, the direction of this line is given by the vector \( <6, 2, 6> \). Now we can write the equation of a line parallel to this, passing through the point P(3,-5,-1) using the general form for the equation of a line in parametric form as: \(x=3+6s\), \(y=-5+2s\), \(z=-1+6s\)

Step 2: Intersection with the xy-plane

If a point lies on the xy-plane, its z-coordinate is 0. To find the point of intersection, we set the z-coordinate of the line we found in Step 1 equal to 0 and solve for s: \(-1 + 6s = 0\) Solving for s, we get: \(s = \frac{1}{6}\) Substitute this value of s into the equations in step 1: \(x = 3 + 6\left(\frac{1}{6}\right) = 4\) \(y = -5 + 2\left(\frac{1}{6}\right) = -\frac{23}{6}\) So, the point of intersection with the xy-plane is \((4, -\frac{23}{6}, 0)\).

Step 3: Intersection with the xz-plane

If a point lies on the xz-plane, its y-coordinate is 0. To find the point of intersection, we set the y-coordinate of the line we found in Step 1 equal to 0 and solve for s: \(-5 + 2s = 0\) Solving for s, we get: \(s = \frac{5}{2}\) Substitute this value of s into the equations in step 1: \(x = 3 + 6 \left(\frac{5}{2}\right) = 18\) \(z = -1 + 6 \left(\frac{5}{2}\right) = 14\) So, the point of intersection with the xz-plane is \((18, 0,14)\)

Step 4: Intersection with the yz-plane

If a point lies on the yz-plane, its x-coordinate is 0. To find the point of intersection, we set the x-coordinate of the line we found in Step 1 equal to 0 and solve for s: \(3 + 6s = 0\) Solving for s, we get: \(s = -\frac{1}{2}\) Substitute this value of s into the equations in step 1: \(y = -5 + 2 \left(-\frac{1}{2}\right) = -6\) \(z = -1 + 6 \left(-\frac{1}{2}\right) = -4\) So, the point of intersection with the yz-plane is \((0, -6, -4)\). Now we have the points of intersection with the coordinate planes: xy-plane: \( (4, -\frac{23}{6}, 0) \) xz-plane: \( (18, 0, 14) \) yz-plane: \( (0, -6, -4) \)

Key Concepts

Coordinate Planes

In three-dimensional space, coordinate planes are the two-dimensional surfaces formed by any two of the three axes: the x-axis, y-axis, and z-axis. These planes divide the space and serve as references for determining positions of points and paths of lines in 3D.
The three main coordinate planes are:

• **xy-plane**: This plane is defined by the x and y axes, meaning all points on this plane have a z-coordinate of zero.
• **xz-plane**: This plane is defined by the x and z axes, meaning all points on this plane have a y-coordinate of zero.
• **yz-plane**: This plane is defined by the y and z axes, meaning all points on this plane have an x-coordinate of zero.

Understanding the concept of coordinate planes allows us to find where lines and shapes intersect or exist within a 3D space. Recognizing which plane a line intersects can help in analyzing and solving geometrical problems.

Intersection Points

Intersection points refer to the precise locations where lines or surfaces meet or cross each other. When calculating the intersection of a line with each of the coordinate planes, we aim to find the exact point where the line meets the plane.
To determine the intersection point with:

• **xy-plane**: Set the z-component of the line to zero.
• **xz-plane**: Set the y-component of the line to zero.
• **yz-plane**: Set the x-component of the line to zero.

By substituting the values accordingly into the line's parametric equations, you can find the value of the parameter that gives the exact intersection points on each plane. These calculations are crucial for applications in physics, engineering, and computer graphics.

Direction Vectors

A direction vector indicates the direction in which a line extends in space. It is a key component in line equation formation and analysis. Direction vectors are helpful in identifying the parallel nature between two lines, meaning if two lines are parallel, their direction vectors are scalar multiples of each other.
For a line with a parametric form, the direction vector is derived from the coefficients of the parameter in the parametric equations.
For example, if a line is defined as:

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

Then the direction vector is \.
These vectors ensure that lines can be appropriately translated or extended across a plane while maintaining their orientation. This becomes important when you need the line to maintain parallelism to another line, such as when solving exercises that involve parallel geometry.

Parametric Form of a Line

The parametric form of a line provides a way to express a line in terms of parameters, mainly used in 3D geometry. It uses a parameter, often denoted as \(t\) or \(s\), to represent how the position on the line changes as we move along it.
The general parametric equations for a line through point \(P(x_0, y_0, z_0)\) with direction vector \ are:

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

Using these parametric equations, one can easily determine the location of any point on the line by choosing specific values for the parameter. This representation simplifies many calculations, including finding intersection points and verifying geometric properties, like line parallelism or orthogonality.