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NCERT Exemplar · Q18

Q.Show that the three points A(2,3,4)A(2,3,4), B(−1,2,−3)B(-1,2,-3) and C(−4,1,−10)C(-4,1,-10) are collinear and find the ratio in which C divides AB.

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Three points are collinear if the vectors joining them are parallel (one is a scalar multiple of the other). Here AC⃗=2AB⃗\vec{AC} = 2\vec{AB}, so the points lie on a line and CC divides ABAB externally in the ratio 2:12:1.

Why the vector approach works

When three points lie on the same straight line, any vector connecting two of them must be parallel to the vector connecting any other pair. Parallel vectors differ only by a scalar multiple: if AC⃗=kAB⃗\vec{AC} = k\vec{AB}, then CC lies on the line through AA and BB. The value of kk tells us exactly where CC sits relative to AA and BB, and from that we can extract the division ratio.

The beauty of this method is that it simultaneously proves collinearity and gives us the ratio in one calculation.


Step-by-step solution

1. Find the position vectors and compute AB⃗\vec{AB}

The vector from AA to BB is

AB⃗=B−A=(−1−2, 2−3, −3−4)=(−3,−1,−7).\vec{AB} = B - A = (-1-2,\, 2-3,\, -3-4) = (-3, -1, -7).

2. Compute AC⃗\vec{AC}

Similarly, the vector from AA to CC is

AC⃗=C−A=(−4−2, 1−3, −10−4)=(−6,−2,−14).\vec{AC} = C - A = (-4-2,\, 1-3,\, -10-4) = (-6, -2, -14).

3. Check if AC⃗\vec{AC} is a scalar multiple of AB⃗\vec{AB}

Compare component by component:

−6−3=2,−2−1=2,−14−7=2.\frac{-6}{-3} = 2, \quad \frac{-2}{-1} = 2, \quad \frac{-14}{-7} = 2.

All three ratios are equal, so

AC⃗=2AB⃗.\vec{AC} = 2\vec{AB}.

This confirms that AC⃗\vec{AC} and AB⃗\vec{AB} are parallel, and since they share the common point AA, the three points are collinear.

Tip

If even one component ratio had differed, the points would not be collinear. Always check all coordinates.

4. Interpret the scalar multiple to find the division ratio …

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