Question

The vector equation of the line whose cartesian equations are \(x = 2,2y-3z+7 = 0\)

• A. \(\overset{⃗}{r} = (2\hat{i}+2\hat{j}-3\hat{k})+λ(2\hat{j}-3\hat{k})\)
• B. \(\overset{⃗}{r} = (2\hat{i}+2\hat{j}-3\hat{k})+λ(2\hat{i}-3\hat{j})\)
• C. \(\overset{⃗}{r} = (2\hat{i}+\frac{7}{3}\hat{k})+λ(3\hat{j}+2\hat{k})\)
• D. \(\overset{⃗}{r} = (-2\hat{i}+\frac{7}{3}\hat{k})+λ(3\hat{j}+2\hat{k})\)

Hint:

Find a point on the line and the direction as the cross product of the two plane normals.

Answer 1

The Correct Option is C

Step 1: Parametrise:
Let $y = 3\lambda$. Then $3z = 2y + 7 = 6\lambda + 7$, so $z = 2\lambda + \tfrac73$, and $x = 2$.

Step 2: Vector form:
$(x, y, z) = (2, 0, \tfrac73) + \lambda(0, 3, 2)$.

Step 3: Match:
Option (C) is exactly this. Option (D) starts at $x = -2$, which violates $x = 2$.

Final Answer:
The line passes through (2, 0, 7/3) along (0, 3, 2). \[ \boxed{\text{(C) }\vec r=\left(2\hat i+\tfrac73\hat k\right)+\lambda\left(3\hat j+2\hat k\right)} \]