Question

If lines \([(\mathrm{x}-1) / 2]=[(\mathrm{y}-3) / 4]=\mathrm{z}\) and \([(x-4) / 3]=[(1-y) / 2]=[(z-1) / 1]\) are co-planer, then the equation of plane containing these two lines is (A) \(6 \mathrm{x}+\mathrm{y}+16 \mathrm{z}=9\) (B) \(6 \mathrm{x}+\mathrm{y}-16 \mathrm{z}=9\) (C) \(6 \mathrm{x}-\mathrm{y}-16 \mathrm{z}=9\) (D) \(6 \mathrm{x}-\mathrm{y}+16 \mathrm{z}=9\)

Short answer

The equation of the plane containing the given lines is \(6x + y - 16z = 9\), which corresponds to option (A).

Step-by-step solution

Find a point and direction vector for each line

For the first line, we have the symmetric equation: \[(\mathrm{x}-1) / 2 = (\mathrm{y}-3) / 4 = \mathrm{z}\] Let us assign a parameter \(t\) for the first line: \[t = (\mathrm{x}-1) / 2 = (\mathrm{y}-3) / 4 = \mathrm{z}\] Solving for x, y, and z, we get the parametric equation of the first line: \[\mathrm{x} = 2t + 1, \mathrm{y} = 4t + 3, \mathrm{z} = t\] Hence, a point on the first line, \(P_1\), can be obtained by taking \(t = 0\), and the direction vector , \(d_1\), by taking the derivative with respect to \(t\): \[P_1 = (1, 3, 0), d_1 = (2, 4, 1)\] Similarly, for the second line, we have the symmetric equation: \[((\mathrm{x}-4) / 3)=((1-\mathrm{y}) / 2)=((\mathrm{z}-1) / 1)\] Let us assign a parameter \(s\) for the second line: \[s = (\mathrm{x}-4) / 3 = (1-\mathrm{y}) / 2 = (\mathrm{z}-1) / 1\] Solving for x, y, and z, we get the parametric equation of the second line: \[\mathrm{x} = 3s + 4, \mathrm{y} = 1 - 2s, \mathrm{z} = s + 1\] Hence, a point on the second line, \(P_2\), can be obtained by taking \(s = 0\), and the direction vector , \(d_2\), by taking the derivative with respect to \(s\): \[P_2 = (4, 1, 1), d_2 = (3, -2, 1)\]

Find the normal vector to the plane

Since the lines are coplanar, their direction vectors, \(d_1\) and \(d_2\), should be parallel to the plane. We can find the normal vector of the plane, \(\mathrm{n}\), by taking the cross product of \(d_1\) and \(d_2\): \[\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 1 \\ 3 & -2 & 1 \end{vmatrix} = (6, 1, -16) \]

Find the equation of the plane

Now, to find the equation of the plane, we can use the point-normal equation, where \(\vec{n} = (A, B, C)\) is the normal vector and \(P_1 = (x_1, y_1, z_1)\) is the point on the plane: \[A(x - x_1) + B(y - y_1) + C(z - z_1) = 0\] Substituting the values for \(\vec{n}\) and \(P_1\), we get: \[6(x - 1) + (y - 3) -16(z - 0) = 0 \Rightarrow 6x + y - 16z = 9\] Hence, the equation of the plane containing the given lines is: \(6x + y - 16z = 9\) which corresponds to option (A).

Key concepts

Symmetric Equations of Lines

When dealing with 3D geometry, one useful way to represent a line is using its symmetric equations. Symmetric equations are derived from the coordinates

• x
• y
• z

and their respective direction ratios:

• x-direction: a
• y-direction: b
• z-direction: c

A line in 3D can be represented in the symmetric form as:\[\frac{(x-x_1)}{a} = \frac{(y-y_1)}{b} = \frac{(z-z_1)}{c}\] Here, \((x_1, y_1, z_1)\) is a point on the line, and \((a, b, c)\) is the direction vector.
This form of representing lines is quite efficient when it comes to solving geometric problems such as finding the plane that contains a line.
In the given exercise, the lines are defined using such symmetric equations. Recognizing this form allows us to extract direction vectors and any specific points directly.

Parametric Equations

A line in 3D space can also be represented using parametric equations. These are derived from the symmetric equations by introducing a parameter, such as \(t\) or \(s\). In the parametric form, the equations look like:

• \(x = a t + x_1\)
• \(y = b t + y_1\)
• \(z = c t + z_1\)

Here, \((x_1, y_1, z_1)\) is a point on the line, \((a, b, c)\) is the direction vector, and \(t\) is the parameter.
Parametric equations are versatile and especially helpful in finding specific points along a line by simply assigning different values to the parameter.
In our problem, we used parametric equations to find a particular point and the direction vector for each line, which were essential in determining the coplanarity of the lines and finding the plane equation.

Cross Product

A cross product is a vector operation in three-dimensional space that yields a vector perpendicular to two given vectors.
Suppose you have two vectors, \(\vec{A} = (A_x, A_y, A_z)\) and \(\vec{B} = (B_x, B_y, B_z)\). The cross product \(\vec{A} \times \vec{B}\) is given by:\[\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix}\]This determinant results in a new vector \((C_x, C_y, C_z)\), computed as:

• \(C_x = A_y B_z - A_z B_y\)
• \(C_y = A_z B_x - A_x B_z\)
• \(C_z = A_x B_y - A_y B_x\)

Using the cross product, we can find a normal vector to the plane containing the two lines.
This normal vector is crucial as it helps in writing the equation of the plane.

Direction Vectors

Direction vectors are critical components when working with lines in space. In the context of a line, a direction vector defines the line's orientation through space.
Every line has a direction vector \( \vec{d} = (a, b, c) \), which is a vector that represents the change in each coordinate direction.

• \(a\) is the change in the x-direction
• \(b\) is the change in the y-direction
• \(c\) is the change in the z-direction

In our problem, we extracted direction vectors \(d_1\) and \(d_2\) from the parametric equations of each line. These vectors are essential for determining the plane's normal vector via the cross product.
Understanding direction vectors and how they interact through operations like the cross product is the foundational step for many problems, including finding planes or showing coplanarity.