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

Short Answer

Expert verified
Answer: The points of intersection are `(-6, 0, 0)` for the x-axis, `(0, -4, 0)` for the y-axis, and `(0, 0, 3)` for the z-axis.

Step by step solution

01

Find the intersection with the x-axis (y=0, z=0)

To find the intersection with the x-axis, we need to set y and z equal to zero in the equation of the plane: $$-2x - 3(0) + 4(0) = 12.$$ Now, we can solve for x: $$-2x = 12 \Rightarrow x = -6.$$ So, the point of intersection with the x-axis is `(-6, 0, 0)`.
02

Find the intersection with the y-axis (x=0, z=0)

To find the intersection with the y-axis, we need to set x and z equal to zero in the equation of the plane: $$-2(0) - 3y + 4(0) = 12.$$ Now, we can solve for y: $$-3y = 12 \Rightarrow y = -4.$$ So, the point of intersection with the y-axis is `(0, -4, 0)`.
03

Find the intersection with the z-axis (x=0, y=0)

To find the intersection with the z-axis, we need to set x and y equal to zero in the equation of the plane: $$-2(0) - 3(0) + 4z = 12.$$ Now, we can solve for z: $$4z = 12 \Rightarrow z = 3.$$ So, the point of intersection with the z-axis is `(0, 0, 3)`.
04

Final Answer

The points where the plane intersects the coordinate axes are `(-6, 0, 0)` for the x-axis, `(0, -4, 0)` for the y-axis, and `(0, 0, 3)` for the z-axis.

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