Question

If \(P(x,y,z)\) is the intersection of the lines \[ \vec r=(2\hat i-2\hat j+3\hat k)+t(\hat i-3\hat j+\hat k) \] and \[ \vec r=(2\hat j-\hat k)+s(\hat i-\hat j+3\hat k), \] then \(x+y+z=\)

• A. \(2\)
• B. \(5\)
• C. \(3\)
• D. \(4\)

Hint:

To find the intersection of two vector equations of lines, equate the corresponding coordinates and solve the resulting system of equations for the parameters.

Answer 1

The Correct Option is D

Step 1: Write the parametric equations. For the first line, \[ (x,y,z) = (2+t,\,-2-3t,\;3+t). \] For the second line, \[ (x,y,z) = (s,\;2-s,\;-1+3s). \]

Step 2: Find the point of intersection. Equating corresponding coordinates, \[ 2+t=s, \] \[ -2-3t=2-s, \] \[ 3+t=-1+3s. \] Using \[ s=2+t, \] in the second equation, \[ -2-3t=2-(2+t), \] \[ -2-3t=-t, \] \[ t=-1. \] Hence, \[ s=2+(-1)=1. \] The third equation is also satisfied: \[ 3+(-1)=2=-1+3(1). \] Therefore, \[ P=(1,1,2). \]

Step 3: Find the required sum. Thus, \[ x+y+z = 1+1+2 = 4. \] Hence, \[ \boxed{4}. \] Thus, \[ \boxed{(D)} \] is the correct answer.