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NCERT Exemplar · Q15

Q.Find the coordinate of the points which trisect the line segment joining the points A(2,1,−3)A(2,1,-3) and B(5,−8,3)B(5,-8,3).

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To trisect a line segment means to divide it into three equal parts using two points. Using the section formula in 3D with ratios 1:21:2 and 2:12:1, the trisecting points are P1(3,−2,−1)P_1(3, -2, -1) and P2(4,−5,1)P_2(4, -5, 1).

When we trisect a line segment, we're looking for the two points that divide it into three equal pieces. Think of cutting a rope into three equal lengths—you need two cuts, and those cut-points are what we're after.

The section formula is the natural tool here. If a point divides a line segment joining (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) in the ratio m:nm:n internally, its coordinates are:

(mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)

For trisection, the first point P1P_1 (closer to AA) divides ABAB in the ratio 1:21:2, and the second point P2P_2 divides ABAB in the ratio 2:12:1.

Finding the trisecting points

1. First trisecting point P1P_1 (ratio 1:21:2 from AA to BB)

Here m=1m = 1, n=2n = 2, with A(2,1,−3)A(2, 1, -3) and B(5,−8,3)B(5, -8, 3).

For the xx-coordinate:

x1=1⋅5+2⋅21+2=5+43=93=3x_1 = \frac{1 \cdot 5 + 2 \cdot 2}{1 + 2} = \frac{5 + 4}{3} = \frac{9}{3} = 3

For the yy-coordinate:

y1=1⋅(−8)+2⋅13=−8+23=−63=−2y_1 = \frac{1 \cdot (-8) + 2 \cdot 1}{3} = \frac{-8 + 2}{3} = \frac{-6}{3} = -2

For the zz-coordinate:

z1=1⋅3+2⋅(−3)3=3−63=−33=−1z_1 = \frac{1 \cdot 3 + 2 \cdot (-3)}{3} = \frac{3 - 6}{3} = \frac{-3}{3} = -1

So P1=(3,−2,−1)P_1 = (3, -2, -1).

2. Second trisecting point P2P_2 (ratio 2:12:1 from AA to BB)

Now m=2m = 2, n=1n = 1. …

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