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Shortest Distance between Two Skew Lines

Prove that th...

Question

Prove that the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ are intersecting to each other. Find their point of intersection.

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Solution

$\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ ......(i)
and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ .......(ii)
Condition for coplanar, then
$∣ ∣ ∣ ∣ \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} \end{matrix} ∣ ∣ ∣ ∣ = 0$
$∣ ∣ ∣ ∣ \begin{matrix} 2 + 1 & 4 + 3 & 6 + 5 3 & 5 & 7 1 & 3 & 5 \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} 3 & 7 & 11 3 & 5 & 7 1 & 3 & 5 \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow 3 (25 - 21) - 7 (15 - 7) + 11 (9 - 5)$
$\Rightarrow 3 (4) - 7 \times 8 + 11 (4)$
$\Rightarrow 12 - 56 + 44$
$\Rightarrow 56 - 56 = 0$
Let $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = r_{1}$
$x = 3 r_{1} - 1$
$y = 5 r_{1} - 3$
$z = 7 r_{1} - 5$
and from (ii) $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} = r_{2}$
$x = r_{2} + 2$
$y = 3 r_{2} + 4$
$z = 5 r_{2} + 6$
$3 r_{1} - 1 = r_{2} + 2$
$3 r_{1} - r_{2} = 2 + 1 = 3$ ....(i)
$5 r_{1} - 3 = 3 r_{2} + 4$
$5 r_{1} - 3 r_{2} = 7$ ......(ii)
$3 r_{1} - r_{2} = 3 \times 5$
$5 r_{1} - 3 r_{2} = 7 \times 4$
____________________
$15 r_{1} - 5 r_{2} = 15$
$15 r_{1} - 9 r_{2} = 21$
- + -
____________________
$4 r_{2} = - 6$
$r_{2} = \frac{- 6}{4} = \frac{- 3}{2}, r_{2} = \frac{- 3}{2}$
$3 r_{1} - r_{2} = 3$
$3 r_{1} + \frac{3}{2} = 3$
$3 r_{1} = 3 - \frac{3}{2} = \frac{3}{2}$
$r_{1} = \frac{3}{2} \times \frac{1}{3} = \frac{1}{2}$
$r_{1} = \frac{1}{2}, r_{2} = - \frac{3}{2}$
$x = 3 r_{1} - 1 = 3 \times \frac{1}{2} - 1$
$x = \frac{1}{2}$
$y = 5 r_{1} - 3$
$y = 5 \frac{1}{2} - 3 = \frac{5}{2} - 3$
$y = - \frac{1}{2}$
$z = 7 r_{1} - 5$
$z = 7 \times \frac{1}{2} - 5 = \frac{7}{2} - 5$
$z = - \frac{3}{2}$
$x = \frac{1}{2}$
$y = - \frac{1}{2}$
$z = - \frac{3}{2}$ .

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Similar questions

Q. Show that the lines

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

\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}

intersect. Also find their point of intersection.

Q. Show that the lines:

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

\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}

.
intersect each other. Find the co-ordinates of intersecting point also.

Q. Prove that the lines

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

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

intersects and find the point of intersection.

Q. Prove that the lines

[\frac{x + 3}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}] & [\frac{x + 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}]

not intersect at the point

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

?

Q. The point of intersection of the lines

\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} a n d \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}

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