Show that the lines x - 3 1 - y - 5 2 - z - 1 - 1 and x - 4 2 - y - 2 - 1 - z - 4 2 intersect each other- Also find the Coordinates of the point of intersection-

x 31=y 52=z 1 1=k x 42=y 2 1=z 42=m x=k+3;y=2k+5;z= k+1 x=2m+4;y= m+2; z=2m+4 k=x 3;y=2(x 3)+5 m= 2+x2;y=+2 x2+2 .y=2x 1 ...

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Byju's Answer

Standard XII

Mathematics

Equation of Perpendicular from a Point on a Line

Show that the...

Question

Show that the lines $\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{- 1}$ and $\frac{x - 4}{2} = \frac{y - 2}{- 1} = \frac{z - 4}{2}$ intersect each other. Also find the Coordinates of the point of intersection.

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Solution

$\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{- 1} = k$
$\frac{x - 4}{2} = \frac{y - 2}{- 1} = \frac{z - 4}{2} = m$
$x = k + 3; y = 2 k + 5; z = - k + 1$
$x = 2 m + 4; y = - m + 2; z = 2 m + 4$
$k = x - 3; y = 2 (x - 3) + 5$
$m = - 2 + \frac{x}{2}; y = + 2 - \frac{x}{2} + 2$
$\begin{matrix} \begin{matrix} y = 2 x - 1 y = \frac{- x}{2} + 4 \end{matrix} \end{matrix} ⎫ ⎬ ⎭ \begin{matrix} 2 x - 1 + \frac{x}{2} - 4 = 0 \frac{5 x}{2} = 5 \end{matrix}$
$x = 2; y = 3$
$2 = k + 3 \Rightarrow k = - 1$
$z = - (- 1) + 1 = 2$
$x = 2; y = 3; z = 2$ is point of intersection

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Show that the lines x−31=y−52=z−1−1 and x−42=y−2−1=z−42 intersect each other. Also find the Coordinates of the point of intersection.

x−31=y−52=z−1−1=kx−42=y−2−1=z−42=mx=k+3;y=2k+5;z=−k+1x=2m+4;y=−m+2;z=2m+4k=x−3;y=2(x−3)+5m=−2+x2;y=+2−x2+2y=2x−1y=−x2+4⎫⎬⎭2x−1+x2−4=05x2=5 x=2;y=32=k+3⇒k=−1z=−(−1)+1=2x=2;y=3;z=2 is point of intersection

Show that the lines $\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{- 1}$ and $\frac{x - 4}{2} = \frac{y - 2}{- 1} = \frac{z - 4}{2}$ intersect each other. Also find the Coordinates of the point of intersection.