Find the coordinates of those points on the line \frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{6} which is at a distance of 3 units from the point \left(\right. 1 ,

Engineering

Mathematics

Line of Greatest Slope

Question

Find the coordinates of those points on the line

\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{6}

which is at a distance of

3

units from the point

(1, - 2, 3)

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\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{6}

Let the point be

= (2 k - 1, 3 k - 2, 6 k + 3)

Distance from

(1, - 2, 3)

= \sqrt{(2 k - 1 - 1)^{2} + (3 k - 2 + 2)^{2} + (6 k + 3 - 3)^{2}}

= \sqrt{(2 k - 2^{2} + 9 k^{2} + 36 k^{2}} = 3

∴ 4 k^{2} + 4 - 8 k + 45 k^{2} = 9

∴ 49 k^{2} - 8 k - 5 = 0

∴ k = \frac{8 \pm \sqrt{64 + 20 \times 49}}{98}

∴ k = \frac{8 \pm \sqrt{65 + 980}}{98}

∴ k = \frac{8 \pm 32}{98} = \frac{40}{98} = \frac{20}{49}

∴

Point are

(\frac{- 9}{49}, \frac{- 38}{49}, \frac{267}{49})

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