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The co-ordinates of the point at which the line x33=y+12=z42\frac{x-3}{3} = \frac{y+1}{2} = \frac{z-4}{-2} crosses x-y plane, are

Solution

Correct Option: 3

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


The line is given as:

x33=y+12=z42\frac{x-3}{3} = \frac{y+1}{2} = \frac{z-4}{-2}

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


x33=y+12=z42=k\frac{x-3}{3} = \frac{y+1}{2} = \frac{z-4}{-2} = k

This gives:

x3=3kx - 3 = 3k

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

y+1=2ky + 1 = 2k

y=2k1y = 2k - 1

z4=2kz - 4 = -2k

z=2k+4z = -2k + 4


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

2k+4=0-2k + 4 = 0

2k=4-2k = -4

k=2k = 2


Substituting k=2k = 2:

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

x=6+3x = 6 + 3

x=9x = 9

y=2(2)1y = 2(2) - 1

y=41y = 4 - 1

y=3y = 3

z=0z = 0

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

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