Question

Let L₁ be the line joining (0, 0, 0) and (1, 2, 3) and L₂ be the line joining (2, 3, 4) and (3, 4, 5). The point of intersection of L₁ and L₂ is

• A. (0,0,0)
• B. (1,2,3)
• C. (2,3,4)
• D. (3,4,5)
• E. (1,1,1)

Answer 1

The Correct Option is B

We are given two lines \( L_1 \) and \( L_2 \). The parametric equation of a line joining two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) is given by: \[ (x, y, z) = (x_1, y_1, z_1) + t \left( (x_2 - x_1), (y_2 - y_1), (z_2 - z_1) \right), \] where \( t \) is a parameter. For the line \( L_1 \) joining \( (0, 0, 0) \) and \( (1, 2, 3) \), the parametric equations are: \[ x = 0 + t(1 - 0) = t, \quad y = 0 + t(2 - 0) = 2t, \quad z = 0 + t(3 - 0) = 3t. \] Thus, the parametric form of \( L_1 \) is \( (x, y, z) = (t, 2t, 3t) \). For the line \( L_2 \) joining \( (2, 3, 4) \) and \( (2, 3, 4) \), the parametric equations are: \[ x = 2 + s(2 - 2) = 2, \quad y = 3 + s(3 - 3) = 3, \quad z = 4 + s(4 - 4) = 4. \] Since these lines are collinear and intersect at a fixed point, the correct point would be \( (1, 2, 3) \).

The correct option is (B) : \((1,2,3)\)

Answer 2

The line L₁ passes through (0, 0, 0) and (1, 2, 3). The direction vector for L₁ is (1-0, 2-0, 3-0) = (1, 2, 3). The parametric equation of L₁ is \((x, y, z) = (0, 0, 0) + t(1, 2, 3) = (t, 2t, 3t)\) for some parameter t.

The line L₂ passes through (2, 3, 4) and (3, 4, 5). The direction vector for L₂ is (3-2, 4-3, 5-4) = (1, 1, 1). The parametric equation of L₂ is \((x, y, z) = (2, 3, 4) + s(1, 1, 1) = (2+s, 3+s, 4+s)\) for some parameter s.

If the lines intersect, there must be values of t and s such that the coordinates are equal. So we must have:

\(t = 2 + s\)

\(2t = 3 + s\)

\(3t = 4 + s\)

From the first equation, \(s = t - 2\). Substituting this into the second equation, we get:

\(2t = 3 + (t - 2)\)

\(2t = 1 + t\)

\(t = 1\)

Then, \(s = 1 - 2 = -1\). We can check this with the third equation:

\(3t = 3(1) = 3\)

\(4 + s = 4 + (-1) = 3\)

Since all three equations are satisfied, the lines intersect. The point of intersection is on L₁ at \((t, 2t, 3t) = (1, 2(1), 3(1)) = (1, 2, 3)\).