CUET UG 2025 — Mathematics Geometry

The x-y plane is where $z = 0$z=0. The point where the line crosses this plane has $z = 0$z=0.

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The line is given as:

$$\frac{x-3}{3} = \frac{y+1}{2} = \frac{z-4}{-2}$$3x−3​=2y+1​=−2z−4​

This is the symmetric form of a line. All three fractions are equal to some parameter $k$k.

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$$\frac{x-3}{3} = \frac{y+1}{2} = \frac{z-4}{-2} = k$$3x−3​=2y+1​=−2z−4​=k

This gives:

$$x - 3 = 3k$$x−3=3k

$$x = 3k + 3$$x=3k+3

$$y + 1 = 2k$$y+1=2k

$$y = 2k - 1$$y=2k−1

$$z - 4 = -2k$$z−4=−2k

$$z = -2k + 4$$z=−2k+4

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On the x-y plane, $z = 0$z=0:

$$-2k + 4 = 0$$−2k+4=0

$$-2k = -4$$−2k=−4

$$k = 2$$k=2

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Substituting $k = 2$k=2:

$$x = 3(2) + 3$$x=3(2)+3

$$x = 6 + 3$$x=6+3

$$x = 9$$x=9

$$y = 2(2) - 1$$y=2(2)−1

$$y = 4 - 1$$y=4−1

$$y = 3$$y=3

$$z = 0$$z=0

Therefore, the point where the line crosses the x-y plane is $(9, 3, 0)$(9,3,0).