Question

The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar if \((1) k=0\) or \(-1\) \((2) \mathrm{k}=0\) or 1 (3) \(\mathrm{k}=0\) or \(-3\) (4) \(\mathrm{k}=3\) or \(-3\)

Short answer

The lines are coplanar if \( k = -1 \) or \( k = -3 \).

Step-by-step solution

Understanding the parametric form of the lines

Rewrite both equations in parametric form. The first line is given by \[ x = 2 + t, \ y = 3 + t, \ z = 4 - kt. \] The second line is given by \[ x = 1 + ks, \ y = 4 + 2s, \ z = 5 + s. \]

Finding points and direction vectors

From the parametric forms, identify points on the lines and their direction vectors. For Line 1: a point is (2, 3, 4) and the direction vector is (1, 1, -k). For Line 2: a point is (1, 4, 5) and the direction vector is (k, 2, 1).

Constructing vectors

Construct the vector connecting a point on the first line to a point on the second line, \( \ \overrightarrow{A_1B_1} = (1, 4, 5) - (2, 3, 4) = (-1, 1, 1). \)

Using the coplanarity condition

The lines are coplanar if the scalar triple product of the direction vectors and the constructed vector is zero. Calculate it as \[ \ (\overrightarrow{A_1B_1} \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2})) = -(1+k)(1+k)-1+1 = 0. \]

Solving the equation

Solve the equation formed: \[ -(1+k)(1+k) = 0. \]This simplifies to \[ k = -1 \].

Conclusion

Thus, lines are coplanar when \( k = 0 \) or \( k = -3 \) because substituting these values satisfies the coplanar condition.

Key concepts

Parametric Equations

Parametric equations allow us to represent lines using a parameter, typically denoted as 't' or 's'.
This method is very useful in 3D geometry. Let's see how it works:
The line equations from the exercise were transformed into their parametric forms as:
For Line 1: \(x = 2 + t, \ y = 3 + t, \ z = 4 - kt \).
Here, 't' is the parameter.
For Line 2: \(x = 1 + ks, \ y = 4 + 2s, \ z = 5 + s \).
Here, 's' is the parameter.
The variable changes (t and s) allow you to traverse or 'walk' along each line in 3D space.
In these equations, 't' and 's' can take any real number values.

Direction Vectors

Direction vectors give us the direction a line is pointing in 3D space.
From parametric equations, you can extract these vectors.
For Line 1: The direction vector is (1, 1, -k) since x increases by 1, y by 1, and z decreases by k for each unit increase in 't'.
For Line 2: The direction vector is (k, 2, 1) since x increases by k, y increases by 2, and z increases by 1 for each unit increase in 's'.
These vectors are essential for understanding trends of lines and further calculations like finding if lines are coplanar.

Scalar Triple Product

The scalar triple product is used to determine the volume of the parallelepiped formed by three vectors.
For vectors \( \textbf{a}, \textbf{b} \), and \( \textbf{c} \), scalar triple product is \( \textbf{a} \cdot ( \textbf{b} \times \$c) \).
If this product is zero, the vectors (and hence the lines they represent) are coplanar.
In this exercise, we constructed a vector \( \overrightarrow{A_1B_1} \) connecting points from both lines:
\( \overrightarrow{A_1B_1} = (-1, 1, 1) \).
This vector, along with our direction vectors, helps us confirm coplanarity.
Calculating the scalar triple product: \( -1(1*1 - (-k)2) + 1(-k*1 - 1*k) + 1(k*2 -1*1) = 0\)
Simplifying and solving it leads to the condition: \(k = -1 \).

Coplanarity Condition

For two lines to be coplanar, their direction vectors and the vector connecting points on each line must satisfy certain conditions.
Specifically, the scalar triple product must equal zero. This means:
Vectors are positioned in such a way that the volume of the parallelepiped they form is zero.
Lines with direction vectors \( \textbf{d}_1 = (1, 1, -k)\ \textbf{d}_2 = (k, 2, 1)\) are coplanar if:
\(( \overrightarrow{A_1B_1} \cdot (\textbf{d}_1 \times \textbf{d}_2)) = 0 \).
This calculation involves forming an equation that we solve to find the permissible values of 'k'.
In this exercise, solving the scalar triple product condition results in: \(k = -1 \).