Consider the lines L₁ and L₂ given by the following vector equations: L₁: r = ( i + j - k ) + (3 i + t j ) L₂: r = (4 i + a j - k ) + (2 i + 3 k ) If a = -2 and the lines intersect, then the value of 't' is:

Step-by-step solution

The general point on line L₁ is given by (1 + 3 ) i + (1 + t) j - k . Given a = -2 , the general point on line L₂ is given by (4 + 2 ) i - 2 j + (-1 + 3 ) k . Since the lines intersect, their corresponding components must be equal for some values of and . Equating the i , j , and k components, we get: 1 + 3 = 4 + 2 1 + t = -2 -1 = -1 + 3 From the third equation: 3 = 0 = 0 Substituting = 0 into the first equation: 1 + 3 = 4 + 0 3 = 3 = 1 Substituting = 1 into the second equation: 1 + (1)t = -2 t = -3 Answer: -3