If the lines x-42-y-54-z-1-3 and x-21-y-13-z2 intersects- find their point of intersection-

Points of intersection x 42 = x 21 x=0 y+54 = y+13 y = 11 z 1 3 = z2 z = 0.4 Hence, point of intersection is (0,11,0.4) #1 ...

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Byju's Answer

Standard XII

Mathematics

Shortest Distance between Two Skew Lines

If the lines ...

Question

If the lines $\frac{x - 4}{2} = \frac{y + 5}{4} = \frac{z - 1}{- 3}$ and $\frac{x - 2}{1} = \frac{y + 1}{3} = \frac{z}{2}$ intersects, find their point of intersection.

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Solution

Points of intersection-
$\frac{x - 4}{2} = \frac{x - 2}{1}$
$x = 0$
$\frac{y + 5}{4} = \frac{y + 1}{3}$
$y = 11$
$\frac{z - 1}{- 3} = \frac{z}{2}$
$z = 0.4$
Hence, point of intersection is- $(0, 11, 0.4)$

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If the lines x−42=y+54=z−1−3 and x−21=y+13=z2 intersects, find their point of intersection.

Points of intersection-x−42=x−21x=0y+54=y+13y=11z−1−3=z2z=0.4Hence, point of intersection is-(0,11,0.4)

If the lines $\frac{x - 4}{2} = \frac{y + 5}{4} = \frac{z - 1}{- 3}$ and $\frac{x - 2}{1} = \frac{y + 1}{3} = \frac{z}{2}$ intersects, find their point of intersection.

Points of intersection-
$\frac{x - 4}{2} = \frac{x - 2}{1}$
$x = 0$
$\frac{y + 5}{4} = \frac{y + 1}{3}$
$y = 11$
$\frac{z - 1}{- 3} = \frac{z}{2}$
$z = 0.4$
Hence, point of intersection is- $(0, 11, 0.4)$