Question

Find the value of \(k\) if the lines \[ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \] and \[ \frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{k} \] are coplanar.

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(2\)

Hint:

For two lines in 3D to be coplanar, use the scalar triple product condition: \[ [\vec{P_1P_2},\vec{d_1},\vec{d_2}] = 0 \] If the determinant formed by these three vectors equals zero, the lines lie in the same plane.

Answer 1

The Correct Option is B

To determine the value of \( k \) that makes the given lines coplanar, we use the condition that the lines are coplanar if the scalar triple product created by their direction vectors and a vector between any two points on the lines is zero.

First, we identify the direction vectors of the given lines:

• For the line \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), the direction vector is \(\mathbf{a} = \langle 2, 3, 4 \rangle\).
• For the line \( \frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{k} \), the direction vector is \(\mathbf{b} = \langle 5, 2, k \rangle\).

Next, we find a vector connecting a point on the first line to a point on the second line:

• Selecting point \( P(1, 2, 3) \) on the first line and point \( Q(4, 1, 0) \) on the second line, vector \(\mathbf{PQ}\) is computed as follows:
• \(\mathbf{PQ} = \langle 4 - 1, 1 - 2, 0 - 3 \rangle = \langle 3, -1, -3 \rangle\).

The condition for coplanarity is that the scalar triple product \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{PQ} = 0\).

First, calculate the cross product \(\mathbf{a} \times \mathbf{b}\):

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 5 & 2 & k \end{vmatrix} = \mathbf{i}(3k - 8) - \mathbf{j}(2k - 20) + \mathbf{k}(4 - 15) \]

\[ = \langle 3k - 8, - (2k - 20), -11 \rangle = \langle 3k - 8, 20 - 2k, -11 \rangle \]

Next, calculate the dot product \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{PQ}\):

\[ (3k - 8)\cdot 3 + (20 - 2k)\cdot (-1) + (-11)\cdot (-3) = 0 \]

Breaking it down, this equates to:

\[ 3(3k - 8) - (20 - 2k) + 33 = 0 \]

Expanding the terms, we get:

\[ 9k - 24 - 20 + 2k + 33 = 0 \]

Simplifying, we combine like terms:

\[ 11k - 11 = 0 \]

Solving for \( k \), we add 11 to both sides:

\[ 11k = 11 \]

Dividing both sides by 11:

\[ k = 1 \]

Therefore, the value of \( k \) that makes the lines coplanar is \( \boxed{1} \).