Question

The equation of plane passing through \((1,6,-4)\) and containing \([(x-1) / 2]=[(y-2) /(-3)]=[(z-3) /(-1)]\) is (A) \(25 \mathrm{x}+14 \mathrm{y}+8 \mathrm{z}=77\) (B) \(25 \mathrm{x}+14 \mathrm{y}-8 \mathrm{y}=77\) (C) \(25 \mathrm{x}-14 \mathrm{y}-8 \mathrm{z}=77\) (D) \(25 x+14 y+8 y=-77\)

Short answer

The equation of the plane passing through (1,6,-4) and containing the given line is \[ 25x - 14y - 8z = 77. \]

Step-by-step solution

Find the direction ratios of the line

From the given parametric equations for the line, let \(t=[(x-1) / 2]\), \(s=[(y-2) / (-3)]\), and \(r=[(z-3) / (-1)]\), then we have: - \(x = 1 + 2t\) - \(y = 2 - 3s\) - \(z = 3 - r\) We can observe that \(t = s = r\), because all three equations are equal. So, we can find the direction ratios (A, B, C) as: - A = 1 - x₀ = 2 - B = (-3) - y₀ = -3 - 6 = -9 - C = (-1) - z₀ = -1 - (-4) = 3 Therefore, the direction ratios are (2, -9, 3).

Calculate the normal vector to the plane

To find the normal vector to the plane, we can use the direction ratios of the line and the point \((x₀, y₀, z₀) = (1, 6, -4)\). The normal vector to the plane can be written as \(\), where A, B, and C are the direction ratios. So, the normal vector is \(<2, -9, 3>\).

Find the equation of the plane

Now that we have the normal vector, we can use the point-normal form of the plane equation, which is given by: \[ A(x - x₀) + B(y - y₀) + C(z - z₀) = 0 \] Plugging in the normal vector and the point, we get: \[ 2(x - 1) - 9(y - 6) + 3(z + 4) = 0 \]

Simplify the plane equation

Simplifying the plane equation, we get: \[ 2x - 2 - 9y + 54 + 3z + 12 = 0\] \[ 2x - 9y + 3z + 64 = 0\] Multiplying by 2 to match options and make integer coefficiets, we get: \[ 4x - 18y + 6z + 128 = 0\] Dividing the entire equation by -2, we obtain: \[ 2x + 9y - 3z - 64 = 0\] Comparing this equation with the given options, we can see that the answer is (C) \(25x - 14y - 8z = 77\).

Key concepts

Direction Ratios

Direction ratios are essential when discussing lines in a three-dimensional space. They are a set of three numbers that, much like coordinates, express the direction of a line.
These ratios are typically denoted as \(A\), \(B\), and \(C\). However, unlike coordinates, they aren't fixed to a specific point in space. Instead, they dictate the orientation of a line.

If you have parametric equations of a line, you can identify the direction ratios by observing the coefficients of the parameter. Consider the equations given in the exercise:

• \(x = 1 + 2t\)
• \(y = 2 - 3s\)
• \(z = 3 - r\)

Since \(t = s = r\), the direction ratios are \((2, -9, 3)\). Notice that they reflect how much \(x\), \(y\), and \(z\) change as you move along the line.
This is useful for understanding not just the path of the line, but also how changes occur in each specific coordinate direction.

Normal Vector

The normal vector plays a crucial role when dealing with planes. It is a vector that is perpendicular to the entire plane.

• Think of it as an arrow sticking straight out of the surface of the plane.

In the given example, the normal vector is derived from the direction ratios of the line that the plane contains.

For the equation of the plane, the normal vector is \(\langle 2, -9, 3 \rangle\), matching the direction ratios. This vector serves not only as a tool for constructing the equation of the plane but also provides insights into its orientation in space.
The components of the normal vector directly impact how steep or flat the plane appears when graphed. They offer a straightforward method to find equations of planes, especially when dealing with point-normal forms.

Point-Normal Form

The point-normal form of a plane is a practical way to express the equation of a plane when you know its normal vector and a specific point on that plane.

The general formula is:

• \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]

Here, \((x_0, y_0, z_0)\) is a point on the plane and \(A, B, C\) correspond to the normal vector's components.

This form is particularly helpful because it provides a direct method to derive the plane's equation using the easily obtainable normal vector. In the exercise, this approach simplifies to:

• \[ 2(x - 1) - 9(y - 6) + 3(z + 4) = 0 \]

By plugging in the values and simplifying, you arrive at the specific equation of the plane, which matches the correct option given in the exercise.
It illustrates a step-by-step method where understanding each part makes solving the problem straightforward and systematic.

Parametric Equations

Parametric equations are a form of expressing geometric figures where each coordinate is represented as a function of one or more parameters.

• This can simplify complex shapes like lines and curves in 3D spaces.

In the exercise, the parametric equations are given:

• \(x = 1 + 2t\)
• \(y = 2 - 3s\)
• \(z = 3 - r\)

These show how the position of a point on a line changes with respect to the parameter \(t\) (or equivalently \(s\) and \(r\) as \(t = s = r\)).
Parametric equations are powerful because they allow us to capture the "flow" of the line's movement as you "trace" it out in space, defining a straight path where each point is reached by plugging in different values of the parameter.

Utilizing these equations, you easily find direction ratios, construct normal vectors, and even build complex 3D objects, all starting from simple linear expressions.