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

Concept: Two lines in space are coplanar if the scalar triple product of the vector joining a point on each line and their direction vectors is zero. If \[ \vec{d_1}, \vec{d_2} \] are the direction vectors of the two lines and \[ \vec{P_1P_2} \] is the vector joining points on the two lines, then the coplanarity condition is \[ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 d_{1x} & d_{1y} & d_{1z} d_{2x} & d_{2y} & d_{2z} \end{vmatrix}=0 \] This determinant represents the scalar triple product.

Step 1: Identify points and direction ratios. From \[ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \] Point on line \(L_1\): \[ P_1(1,2,3) \] Direction ratios: \[ \vec{d_1}=(2,3,4) \] From \[ \frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{k} \] Point on line \(L_2\): \[ P_2(4,1,0) \] Direction ratios: \[ \vec{d_2}=(5,2,k) \]

Step 2: Find the vector joining the two points. \[ \vec{P_1P_2}=(4-1,\,1-2,\,0-3) \] \[ \vec{P_1P_2}=(3,-1,-3) \]

Step 3: Apply the coplanarity condition. \[ \begin{vmatrix} 3 & -1 & -3 2 & 3 & 4 5 & 2 & k \end{vmatrix}=0 \] Expanding the determinant: \[ 3(3k-8) -(-1)(2k-20) + (-3)(4-15)=0 \] \[ 9k-24 +2k-20 +33 =0 \] \[ 11k -11 =0 \] \[ k=1 \]