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: Understanding the Concept:
The line is the intersection of the planes \(x = 2\) and \(2y - 3z + 7 = 0\). Its direction is perpendicular to both normals.

Step 2: Direction:
Normals: \(\vec n_1 = (1, 0, 0)\) and \(\vec n_2 = (0, 2, -3)\).
\[ \vec n_1\times\vec n_2 = (0\cdot(-3) - 0\cdot2,\ 0\cdot0 - 1\cdot(-3),\ 1\cdot2 - 0) = (0, 3, 2) \]

Step 3: A point on the line:
Take \(y = 0\): \(-3z + 7 = 0\), so \(z = \tfrac73\). The point is \((2, 0, \tfrac73)\).

Step 4: Write the equation:
\[ \vec r = 2\hat i + \tfrac73\hat k + \lambda(3\hat j + 2\hat k) \]
This is option (C). Option (A) and (B) use a point like \((2, 2, -3)\), which gives \(4 + 9 + 7 \neq 0\) in \(2y - 3z + 7\), so it is not on the line.

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