The shortest distance between the linesx-17-y-1-6-z-11 andx-31-y-5-2-z-71 is

The shortest distance between the lines nbsp; =1+31+51+71 217 617× 2 1× 62+7×1 1×12+ 6×1 1× 22 nbsp; =4681 217 61 42+62+ 82=16+36+6416+36+64=116116=116=229 ...

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Standard XII

Mathematics

Shortest Distance between Two Skew Lines

The shortest ...

Question

The shortest distance between the lines
$\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$ and
$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$ is

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2 \sqrt{29}

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2 \sqrt{30}

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Solution

The correct option is C $2 \sqrt{29}$
The shortest distance between the lines
$= \frac{⎛ ⎜ ⎝ \begin{matrix} 1 + 3 & 1 + 5 & 1 + 7 1 & - 2 & 1 7 & - 6 & 1 \end{matrix} ∣ ∣ ∣ ∣}{\sqrt{{(7 \times (- 2) - 1 \times (- 6))}^{2} + {(7 \times 1 - 1 \times 1)}^{2} + {((- 6) \times 1 - 1 \times (- 2))}^{2}}}$

$= \frac{⎛ ⎜ ⎝ \begin{matrix} 4 & 6 & 8 1 & - 2 & 1 7 & - 6 & 1 \end{matrix} ∣ ∣ ∣ ∣}{\sqrt{{(- 4)}^{2} + {(6)}^{2} + {(- 8)}^{2}}} = \frac{16 + 36 + 64}{\sqrt{16 + 36 + 64}} = \frac{116}{\sqrt{116}} = \sqrt{116} = 2 \sqrt{29}$

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Similar questions

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. Find 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. 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 the following pairs of lines whose cartesian equations are:
(i)

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

(ii)

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

(iii)

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

(iv)

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

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}

is
(a)

\sqrt{30}

(b)

2 \sqrt{30}

(c)

5 \sqrt{30}

(d)

3 \sqrt{30}

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The shortest distance between the lines x+17=y+1−6=z+11 and x−31=y−5−2=z−71 is

The correct option is C 2√29The shortest distance between the lines =⎛⎜⎝1+31+51+71−217−61∣∣ ∣∣√(7×(−2)−1×(−6))2+(7×1−1×1)2+((−6)×1−1×(−2))2 =⎛⎜⎝4681−217−61∣∣ ∣∣√(−4)2+(6)2+(−8)2=16+36+64√16+36+64=116√116=√116=2√29

The shortest distance between the lines
$\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$ and
$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$ is