Question

Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5.

Statement I: The parametric equations of the line of intersection of given planes are x = 3 + 14t, y = 1 + 2t, and z = 15t.

Statement II: The vector 14i + 2j + 15k is parallel to the given planes. 

Answer 1

Let a, b, c be the direction ratios of the line of intersection of 3x – 6y – 2z = 15 and 2x + y – 2z = 5. 

Then 3a – 6b – 2c = 0

and 2a + b – 2c = 0

Thus, 

Clearly, the vector 14i + 2j + 15k is parrallel to the line of intersections of the plane.

Now, we shall find the equation of the line put z = 0 in the given planes.

So 3x – 6y = 15 and 2x + y = 5.

On solving, we get x = 3 and y = – 1

Hence, the equation of the line is

The parametric equations of the line are

x = 14t + 3, y = 2t – 1, z = 15t.