Question

The vector equation of plane in parametric form, passing through the points (-1, 2, 0), (2, 2, -1) and parallel to the line \(\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}\) is

• A. \(\overset{̄}{r} = (-\hat{i}+2\hat{j})+λ(3\hat{i}-\hat{k})+μ(\hat{i}+\hat{j}-\hat{k})\)
• B. \(\overset{̄}{r} = (-\hat{i}+2\hat{j})+λ(3\hat{i}-\hat{k})+μ(\hat{i}+2\hat{j}-\hat{k})\)
• C. \(\overset{̄}{r} = (\hat{i}-2\hat{j})+λ(3\hat{i}+\hat{k})+μ(\hat{i}+\hat{j}-\hat{k})\)
• D. \(\overset{̄}{r} = (-\hat{i}+2\hat{j})+λ(3\hat{i}-\hat{k})+μ(\hat{i}-\hat{j}+\hat{k})\)

Hint:

Use one point as position vector, the line joining the two points as one direction, and the given line as the other.

Answer 1

The Correct Option is A

Step 1: Plane needs two directions
A plane through a point can be written $\vec r = \vec a+\lambda\vec p+\mu\vec q$, where $\vec p,\vec q$ lie in the plane.

Step 2: Choose
$\vec a = -\hat i+2\hat j$, $\vec p = (2,2,-1)-(-1,2,0) = 3\hat i-\hat k$, and $\vec q$ is parallel to the given line. Rewriting $\frac{2y+1}{2} = y+\frac12$, we get $\vec q = \hat i+\hat j-\hat k$.

Step 3: Answer
Option (A).

Final Answer:
Option A. \[ \boxed{\text{(A)}\ \vec r=(-\hat i+2\hat j)+\lambda(3\hat i-\hat k)+\mu(\hat i+\hat j-\hat k)} \]