A line L1 passes through a point with position vector  a⃗1  =  2i^ + 3j^ + 4k^ and is parallel to  b⃗1  =  2i^ + 3j^ + 4k^. Another line L2 passes through a point with position vector  a⃗2  =  3i^ + 4j^ + k^ and is parallel to  b⃗2  =  4i^ + j^ + 3k^. Equation of plane equidistant from L1 and L2 is

To find the equation of the plane equidistant from lines L1 and L2, we can follow these steps:

Step 1: Find the direction vectors of lines L1 and L2. The direction vector of L1 is b⃗1 = 2i^ + 3j^ + 4k^. The direction vector of L2 is b⃗2 = 4i^ + j^ + 3k^.

Step 2: Find the normal vector of the plane. Since the plane is equidistant from L1 and L2, the normal vector of the plane will be perpendicular to both b⃗1 and b⃗2. We can find the normal vector by taking the cross product of b⃗1 and b⃗2. n⃗ = b⃗1 × b⃗2 = (2i^ + 3j^ + 4k^) × (4i^ + j^ + 3k^) = (3(4) - 4(3))i^ - (2(4) - 4(3))j^ + (2(1) - 3(3))k^ = 0i^ + 4j^ - 7k^ = 4j^ - 7k^

Step 3: Find a point on the plane. We are given the point with position vector a⃗1 = 2i^ + 3j^ + 4k^ on line L1. We can use this point as a point on the plane.

Step 4: Write the equation of the plane. The equation of a plane can be written as: Ax + By + Cz = D where (A, B, C) is the normal vector of the plane and (x, y, z) are the coordinates of a point on the plane.

Using the normal vector 4j^ - 7k^ and the point (2, 3, 4), we can substitute these values into the equation: 4y - 7z = D

To find the value of D, we substitute the coordinates of the point (2, 3, 4) into the equation: 4(3) - 7(4) = D 12 - 28 = D D = -16

Therefore, the equation of the plane equidistant from L1 and L2 is: 4y - 7z = -16.