Show that the point $(1, - 1, 2)$ is common to the line which join $(6, - 7, 0)$ to $(16, - 19, - 4)$ and $(0, 3, - 6)$ to $(2, - 5, 10)$

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Solution

Line joining $(6, - 7, 0)$ to $(16, - 19, - 4)$ is:
$\frac{x - 6}{10} = \frac{y + 7}{- 12} = \frac{z - 0}{- 4} - (L_{1})$
Line joining $(0, 3, - 6)$ to $(2, - 5, 10)$ is:
$\frac{x - 0}{2} = \frac{y - 3}{- 8} = \frac{z + 6}{16} - (L_{2})$
Let $2 r, - 8 r + 3, 16 - 6$ be point on $L_{2}$ which is intersection point
$∴ \frac{2 r - 6}{10} = \frac{- 8 r + 3 + 7}{12} = \frac{16 r - 6}{- 4} \Rightarrow r = \frac{1}{2}$
$∴$ point of intersection $= (1, - 1, 2)$

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Show that the point (1,−1,2) is common to the line which join (6,−7,0) to (16,−19,−4) and (0,3,−6) to (2,−5,10)

Line joining (6,−7,0) to (16,−19,−4) is:x−610=y+7−12=z−0−4−(L1)Line joining (0,3,−6) to (2,−5,10) is:x−02=y−3−8=z+616−(L2)Let 2r,−8r+3,16−6 be point on L2 which is intersection point ∴2r−610=−8r+3+712=16r−6−4⇒r=12∴ point of intersection =(1,−1,2)

Show that the point $(1, - 1, 2)$ is common to the line which join $(6, - 7, 0)$ to $(16, - 19, - 4)$ and $(0, 3, - 6)$ to $(2, - 5, 10)$