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Question

Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).

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Solution

(i) We know that the z-coordinate of every point on the xy-plane is zero.
So, let P (x, y, 0) be a point on the xy-plane such that PA = PB = PC
Now, PA = PB
PA2=PB2x-12+y+12+0-02 =x-22+y-12+0-22x2-2x+1+y2+2y+1=x2-4x+4+y2-2y+1+4-2x+4x+2y+2y+2-9=02x+4y-7=02x+4y=7 ...(1) PB=PC PB2=PC2x-22+y-12+0-22=x-32+y-22+0+12x2-4x+4+y2-2y+1+4=x2-6x+9+y2-4y+4+1-4x+6x-2y+4y+9-14=02x+2y-5=0x+y=52x=52-y ...2Putting the value of x in equation 1: 252-y+4y=7 5-2y+4y=7 5+2y=7 2y=2 y=22y=1Putting the value of y in equation 2:x=52-1x=5-22x=32Hence, the required point is 32, 1, 0.

(ii) We know that the x-coordinate of every point on the yz-plane is zero.
So, let P (0, y, z) be a point on the yz-plane such that PA = PB = PC
Now, PA = PB
0-12+y+12+z-02=0-22+y-12+z-22
1+y2+2y+1+z2=4+y2-2y+1+z2-4z+42y+2=-2y-4z+92y+2y-4z=9-24y-4z=7y-z=74 ... 1PB=PCPB2=PC20-22+y-12+z-22=0-32+y-22+z+124+y2-2y+1+z2-4z+4=9+y2-4y+4+z2+2z+1-2y-4z+9=-4y+2z+14-2y+4y-4z-2z=14-92y-6z=5y-3z=52 y=52+3z 2Putting the value of y in equation 1: y-z=7452+3z-z=742z=74-522z=7-1042z=-34 z=-38Putting the value of z in equation 2: y=52+3zy=52+3-38y=52-98y=20-98 y=118
Hence, the required point is 0,118,-38.

(iii) We know that the y-coordinate of every point on the zx-plane is zero.
So, let P (x, 0, z) be a point on the zx-plane such that PA = PB = PC
Now, PA = PB
x-12+0+12+z-02=x-22+0-12+x-22
x2+1-2x+1+z2=x2-4x+4+1+z2-4z+4-2x+2=-4x-4z+9-2x+4x-4z=72x-4z=7x-2z=72 ...1 PB=PCPB2=PC2x-22+0-12+z-22=x-32+0-22+z+12x2-4x+4+1+z2-4z+4=x2-6x+9+4+z2+2z+1-4x-4z+9=-6x+2z+14-4x+6x-4z-2z=14-92x-6z=5x-3z=52 x=52+3z ...2Putting the value of x in equation 1: x-2z=7252+3z-2z=7252+z=72z=72-52z=7-52z=22 z=1Putting the value of z in equation 2: x=52+3zx=52+31x=52+3x=5+62 x=112
Hence, the required point is 112, 0, 1.

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