Problem 3 - Calculus Early Transcendentals

Question

Where does the plane \(-2 x-3 y+4 z=12\) intersect the coordinate axes?

Step-by-Step Solution

Answer

Question: Determine the points of intersection of the plane -2x - 3y + 4z = 12 with the x, y, and z coordinate axes. Answer: The points of intersection are (-6, 0, 0), (0, -4, 0), and (0, 0, 3).

Step 1: Intersection with the x-axis

To find the point where the plane intersects the x-axis, set y and z to zero in the equation of the plane: \(-2x - 3(0) + 4(0) = 12\) Now solve for x: \(-2x = 12\) \(x = -6\) So, the intersection of the plane with the x-axis is at point (-6, 0, 0).

Step 2: Intersection with the y-axis

To find the point where the plane intersects the y-axis, set x and z to zero in the equation of the plane: \(-2(0) - 3y + 4(0) = 12\) Now solve for y: \(-3y = 12\) \(y = -4\) So, the intersection of the plane with the y-axis is at point (0, -4, 0).

Step 3: Intersection with the z-axis

To find the point where the plane intersects the z-axis, set x and y to zero in the equation of the plane: \(-2(0) - 3(0) + 4z = 12\) Now solve for z: \(4z = 12\) \(z = 3\) So, the intersection of the plane with the z-axis is at point (0, 0, 3). The points of intersection of the plane \(-2x-3y+4z=12\) with the x, y, and z coordinate axes are (-6, 0, 0), (0, -4, 0), and (0, 0, 3), respectively.