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Introduction to Three Dimensional Geometry — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSIntroduction to Three Dimensional Geometry2 Marks
Question
Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear.
✓

Answer

Let A (– 4, 6, 10), B (2, 4, 6) and C(14, 0, – 2) be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of the point P are
$\frac{2 k-4}{k+1}, \frac{4 k+6}{k+1}, \frac{6 k+10}{k+1}$    [using section formula]
Let us examine whether, for some value of k, the point P coincides with point C.
On putting $\frac{2 k-4}{k+1}=14$,  we get $k=-\frac{3}{2}$
When $k=-\frac{3}{2}, \text { then } \frac{4 k+6}{k+1}=\frac{4\left(-\frac{3}{2}\right)+6}{-\frac{3}{2}+1}=0$
and $\frac{6 k+10}{k+1}=\frac{6\left(-\frac{3}{2}\right)+10}{-\frac{3}{2}+1}=-2$
Therefore, C (14, 0, –2) is a point that divides AB externally in the ratio 3 : 2 and is the same as P.

Hence A, B, C are collinear.

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