Question

The point of intersection of the line x + 1 = \(\frac{y+3}{3}=\frac{-z+2}{2}\) with the plane 3x + 4y + 5z = 10 is

• A. (2, 6, -4)
• B. (-2, 6, -4)
• C. (2, 6, 4)
• D. (2, -6, -4)

Hint:

To find the point of intersection of a line and a plane, parametrize the line and substitute the parametric equations for \( x \), \( y \), and \( z \) into the plane equation. Solve for the parameter and then substitute back to find the coordinates of the intersection point.

Answer 1

The Correct Option is A

The correct answer is: (A): (2, 6, -4)

We are tasked with finding the point of intersection of the line \( x + 1 = \frac{y + 3}{3} = \frac{-z + 2}{2} \) with the plane \( 3x + 4y + 5z = 10 \).

Step 1: Parametrize the line equation
The given line equation is symmetric, so we introduce a parameter \( t \) and express \( x \), \( y \), and \( z \) in terms of \( t \):

\( x + 1 = t, \quad \frac{y + 3}{3} = t, \quad \frac{-z + 2}{2} = t \)

Now solve for \( x \), \( y \), and \( z \) in terms of \( t \):

\( x = t - 1, \quad y = 3t - 3, \quad z = 2 - 2t \)

Step 2: Substitute into the plane equation
The equation of the plane is \( 3x + 4y + 5z = 10 \). Substitute \( x = t - 1 \), \( y = 3t - 3 \), and \( z = 2 - 2t \) into this equation:

\( 3(t - 1) + 4(3t - 3) + 5(2 - 2t) = 10 \)

Step 3: Simplify the equation
Expand each term:

\( 3t - 3 + 12t - 12 + 10 - 10t = 10 \)

Now, combine like terms:

\( 3t + 12t - 10t - 3 - 12 + 10 = 10 \)

\( 5t - 5 = 10 \)

Step 4: Solve for \( t \)
Now, solve for \( t \):

\( 5t = 15 \quad \Rightarrow \quad t = 3 \)

Step 5: Find the coordinates of the point of intersection
Substitute \( t = 3 \) back into the parametric equations for \( x \), \( y \), and \( z \):

\( x = 3 - 1 = 2, \quad y = 3(3) - 3 = 6, \quad z = 2 - 2(3) = -4 \)

Conclusion:
The point of intersection is \( (2, 6, -4) \), so the correct answer is (A): (2, 6, -4).