The point on the line x-2-1 - y-3-2 - z-5-2 at a distance of 6 from the point 2- -3- -5 is

The correct option is B (4, 7, 9)DC’s of the given line is 1/3, 2/3, 2/3 Hence, the equation of line can be point in the form x 2/1/3 = y+3/ 2/3 = z+5/ 2/3 ...

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Standard XII

Mathematics

Equation of Line: Symmetrical Form

The point on ...

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)

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(4, -7, -9)

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(0, 2, -1)

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(-3, 5, 3)

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Solution

The correct option is B (4, -7, -9)
DC’s of the given line is $\frac{1}{3}, - \frac{2}{3}, - \frac{2}{3}$
Hence, the equation of line can be point in the form $\frac{x - 2}{\frac{1}{3}} = \frac{y + 3}{\frac{- 2}{3}} = \frac{z + 5}{\frac{- 2}{3}} = r$
$∴$ Point is $(2 + \frac{r}{3}, - 3 - \frac{2 r}{3}, - 5 - \frac{2 r}{3})$
$∴ r = \pm 6$
Points are (4, -7, -9) and (0, 1, -1)

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Similar questions

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The point on the line x−21=y+3−2=z+5−2 at a distance of 6 from the point (2, -3, -5) is

The correct option is B (4, -7, -9)DC’s of the given line is 13,−23,−23 Hence, the equation of line can be point in the form x−213=y+3−23=z+5−23=r ∴ Point is (2+r3,−3−2r3,−5−2r3) ∴r=±6 Points are (4, -7, -9) and (0, 1, -1)

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