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Let A(2, -1, 4) and B(0, 2, -3) be two points and C be a point on AB produced such that 2 AC = 3 AB, then the coordinates of C are |
$\left(\frac{1}{2}, \frac{5}{4}, -\frac{5}{4}\right)$ $\left(-\frac{1}{2}, \frac{7}{4}, -\frac{13}{4}\right)$ $(6, -7, 18)$ none of these |
none of these |
We have, 2 AC = 3 AB
$⇒ 2AC = 3 (AC - BC) ⇒ AC = 3 BC ⇒ \frac{AC}{BC}= \frac{3}{1}$ ⇒ C divides AB externally in the ratio 3 : 1. ∴ Coordinates of C are $\left(\frac{3×0-2×1}{3-1},\frac{3×2-1×-1}{3-1}, \frac{3×(-3)-1×4}{3-1}\right) or, \left(-1, \frac{7}{2}, \frac{-13}{2}\right)$ |
