Find the shortest distance between lines- x-31-y-52-z-71 and x-17-y-1-6-z-11

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Shortest Distance between Two Skew Lines

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Question

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

\frac{46}{5 \sqrt{3}}

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\frac{46}{5 \sqrt{6}}

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\frac{6}{5 \sqrt{7}}

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\frac{4}{8 \sqrt{8}}

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Solution

The correct option is B $\frac{46}{5 \sqrt{3}}$
required option is not in option It may be typing mistake

Given lines
$l_{1} : \frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 7}{1}$
$l_{2} : \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$
position vector of line $l_{1}$
$\to a = 3^i + 5^j + 7^k$
position vector of line $l_{2}$
$\to c = -^i -^j -^k$
normal vector of line $l_{1}$
$_{1} =^i + 2^j +^k$
normal vector of line $l_{2}$
$_{2} = 7^i - 6^j +^k$
so line are skews line
$S D = ∣ ∣ ∣ \frac{(\to c - \to a) \cdot (_{1} \times_{2})}{| (_{1} \times_{2}) |} ∣ ∣ ∣$

$\to c - \to a = -^i -^j -^k - 3^i - 5^j - 7^k$
$\to c - \to a = - 4^i - 6^j - 8^k$
$_{1} \times_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 2 & 1 7 & - 6 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$_{1} \times_{2} =^i (2 + 6) -^j (1 - 7) +^k (- 6 - 14)$
$_{1} \times_{2} = 8^i + 6^j - 20^k$

$|_{1} \times_{2} | = \sqrt{8^{2} + 6^{2} + (- 20)^{2}}$

$|_{1} \times_{2} | = \sqrt{300}$

putting $\to c - \to a,_{1} \times_{2}, |_{1} \times_{2} |$ in formula

$S D = ∣ ∣ ∣ ∣ \frac{(- 4^i - 6^j - 8^k) \cdot (8^i + 6^j - 20^k}{\sqrt{300}} ∣ ∣ ∣ ∣$

$S D = ∣ ∣ ∣ \frac{- 32 - 36 + 160}{\sqrt{300}} ∣ ∣ ∣$

$S D = ∣ ∣ ∣ \frac{92}{\sqrt{25 \times 4 \times 3}} ∣ ∣ ∣$

$S D = \frac{2 \times 46}{5 \times 2 \sqrt{3}}$

$S D = \frac{46}{5 \sqrt{3}}$
Hence it is correct answer
option A is not correct

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Similar questions

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}

Q. Find the shortest distance between 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 lines

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

and

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

approximately is

Find the shortest distance between the lines
$\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} a n d \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$

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}

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Find the shortest distance between lines: x−31=y−52=z−71 and x+17=y+1−6=z+11

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