Question

The point of intersection of the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-11}{4}$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ is ________.

• A. $(-2,\frac{11}{2},5)$
• B. $(-2,\frac{11}{2},-5)$
• C. $(-2,\frac{-11}{2},-5)$
• D. $(-2,\frac{11}{4},-5)$
• E. $(-2,\frac{-11}{5},\frac{-5}{2})$

Hint:

To find the intersection of two lines, convert them to parametric form and solve for the parameters $r$ and $k$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
To find the point of intersection of two lines in 3D space, we need to find a point \((x, y, z)\) that lies on both lines simultaneously. We can do this by writing the parametric equations for each line and then equating the corresponding coordinates.
Step 2: Key Formula or Approach
Let the first line be \(L_1\) and the second line be \(L_2\).
For \(L_1\): \(\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-11}{4} = \lambda\).
A general point on \(L_1\) is \(P_1 = (2\lambda+1, 3\lambda-1, 4\lambda+11)\).
For \(L_2\): \(\frac{x-3}{1} = \frac{y-2}{2} = \frac{z}{1} = \mu\).
A general point on \(L_2\) is \(P_2 = (\mu+3, 2\mu+2, \mu)\).
If the lines intersect, there must exist values of \(\lambda\) and \(\mu\) such that \(P_1 = P_2\).
Step 3: Detailed Explanation
1. Set up the system of equations.
Equating the coordinates of \(P_1\) and \(P_2\):
(i) \(2\lambda + 1 = \mu + 3\)
(ii) \(3\lambda - 1 = 2\mu + 2\)
(iii) \(4\lambda + 11 = \mu\)
2. Solve the system of equations.
We have a system of three equations with two variables. We will solve two of them and then check if the solution satisfies the third equation.
Let's use equation (iii) to substitute for \(\mu\) in equation (i).
Substitute \(\mu = 4\lambda + 11\) into \(2\lambda + 1 = \mu + 3\):
\[ 2\lambda + 1 = (4\lambda + 11) + 3 \] \[ 2\lambda + 1 = 4\lambda + 14 \] \[ -13 = 2\lambda \] \[ \lambda = -\frac{13}{2} \] Now find \(\mu\) using equation (iii):
\[ \mu = 4\left(-\frac{13}{2}\right) + 11 = 2(-13) + 11 = -26 + 11 = -15 \] 3. Check the solution in the third equation.
Now we check if these values of \(\lambda = -13/2\) and \(\mu = -15\) satisfy equation (ii): \(3\lambda - 1 = 2\mu + 2\).
- Left-Hand Side (LHS): \(3(-\frac{13}{2}) - 1 = -\frac{39}{2} - \frac{2}{2} = -\frac{41}{2}\)
- Right-Hand Side (RHS): \(2(-15) + 2 = -30 + 2 = -28\)
Since LHS \((-\frac{41}{2} = -20.5)\) is not equal to RHS (-28), the system is inconsistent.
4. Conclusion.
Because there are no values of \(\lambda\) and \(\mu\) that satisfy all three equations simultaneously, the lines do not intersect. They are skew lines. Therefore, the question is flawed and has no solution. This is why it was cancelled.
Step 4: Final Answer
The lines do not intersect, so there is no point of intersection. The question is cancelled.