The shortest distance between the linesx-33 - y-8-1 - z-31 andx-3-3 - y-72 - z-64

Given lines are x 3/3 = y 8/ 1 = z 3/1 = r_1 (say) amp; x+3/ 3 = y+7/2 = z 6/4 = r_2 Any point on first line is P(3r_1 + 3, 8 r_1, r_1+3) and on second lin ...

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Standard XIII

Mathematics

Shortest Distance between Two Skew Lines

The shortest ...

Question

The shortest distance between the lines
$\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and
$\frac{x - 3}{- 3} = \frac{y - 7}{2} = \frac{z - 6}{4}$

\sqrt{10}

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2 \sqrt{20}

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3 \sqrt{30}

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4 \sqrt{40}

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Solution

The correct option is C $3 \sqrt{30}$
Given lines are $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} = r_{1}$ (say) & $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} = r_{2}$
Any point on first line is $P (3 r_{1} + 3, 8 - r_{1}, r_{1} + 3)$ and on second line is $Q (- 3 - 3 r_{2}, 2 r_{2} - 7, 4 r_{2} + 6) .$ If PQ is the shortest distance, then direction ratios of $P Q = (3 r_{1} + 3 - (- 3 - 3 r_{2}), 8 - r_{1} - (2 r_{2} - 7), r_{1} + 3 - (4 r_{2} + 6)) .$
$= (3 r_{1} + 3 r_{2} + 6, - r_{2} - 2 r_{2} + 15, r_{1} - 4 r_{2} - 3) .$
As PQ is perpendicular to first line and second line -
$3 (3 r_{1} + 3 r_{2} + 6) + (- 1) (- r_{1} - 2 r_{2} + 15) + 1 (r_{1} - 4 r_{2} - 3) = 0$
Or $11 r_{1} + 7 r_{2} = 0..... (2)$
$- 3 (3 r_{1} + 3 r_{2} + 6) + (2) (- r_{1} - 2 r_{2} + 15) + 4 (r_{1} - 4 r_{2} - 3) = 0$
Or $7 r_{1} + 11 r_{2} = 0..... (2)$
On solving equation 1 & 2 we get,
$r_{1} = r_{2} = 0$
So, point $P (3, 8, 3)$ and $Q (- 3, - 7, 6)$
And length of shortest distance PQ =
$\sqrt{{(- 3 - 3)^{2} + (- 7 - 8)^{2} + (6 - 3)^{2}}}$
$= 3 \sqrt{30}$

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Similar questions

Q. The shortest distance between the lines

\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}

and

\frac{x - 3}{- 3} = \frac{y - 7}{2} = \frac{z - 6}{4}

Q. Find the shortest distance between lines:

\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}

Q. The shortest distance between the lines

\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

and

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}

Q. If two lines

\frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - 3}{5}

and

\frac{x - 9}{2} = \frac{y - 8}{4} = \frac{z - 7}{8}

lie in the same plane then the equation of that plane is -

Q. Find the shortest distance between the lines:

\frac{x + 1}{4} = \frac{y - 3}{- 6} = \frac{z + 1}{1}

and

\frac{x + 3}{3} = \frac{y - 5}{2} = \frac{z - 7}{6}

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The shortest distance between the lines x−33=y−8−1=z−31 and x−3−3=y−72=z−64

The shortest distance between the lines
$\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and
$\frac{x - 3}{- 3} = \frac{y - 7}{2} = \frac{z - 6}{4}$