Find the coordinates of those points on the line which is at a distance of 3 units from the point (1,

Engineering

Mathematics

Section Formulae and Centres of a Triangle

Distance from a Plane

Equation of Straight Line in 3D and Distance Of a Point From Line

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

= (2k –1,3k –2, 6k +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

∴ 4k² + 4 – 8k + 45k² = 9

∴ 49k² – 8k – 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})