The point on the line at a distance of 6 from the point (2,–3,

Engineering

Mathematics

Equation of Straight Line in 3D

Section Formulae and Centres of a Triangle

theorem in space

Question

The point on the line $\frac{x - 2}{1} = \frac{y + 3}{- 2} = \frac{z + 5}{- 2}$ at a distance of 6 from the point (2,–3,–5) is

(3,–5,–3)

(4,–7,–9)

(0,2,–1)

(–3,5,3)

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b. Direction cosines of the given line are

\frac{1}{3}, - \frac{2}{3}, - \frac{2}{3}

Hence, the equation of line can be point in the form

\frac{x - 2}{1 / 3} = \frac{y + 3}{- 2 / 3} = - \frac{z + 5}{- 2 / 3} = r

Therefore, any point on the line is

(2 + \frac{r}{3}, - 3 - \frac{2 r}{3}, - 5 - \frac{2 r}{3})

, where r=±6.

Points are(4,−7,−9) and (0,1,−1).