Find the co-ordinates of the points on the join of $(- 3, 7, - 13)$ and $(- 6, 1, - 10)$ which is nearest to the intersection of the planes: $3 x - y - 3 z + 32 = 0$ and $3 x + 2 y - 15 z - 8 = 0$

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Solution

$\frac{x + \frac{56}{9}}{7} = \frac{y - \frac{46}{3}}{12} = \frac{2}{3} = π$
$7 π - \frac{56}{9} = m - 3$
$14 π - \frac{- 112}{9} = m - 6$
$2 π - \frac{- 112}{9} - - \frac{40}{3} = - 13$
$12 π + - \frac{40}{9} = 24 + 7$

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Find the co-ordinates of the points on the join of (−3,7,−13) and (−6,1,−10) which is nearest to the intersection of the planes: 3x−y−3z+32=0 and 3x+2y−15z−8=0

x+5697=y−46312=23=π7π−569=m−314π−−1129=m−62π−−1129−−403=−1312π+−409=24+7

Find the co-ordinates of the points on the join of $(- 3, 7, - 13)$ and $(- 6, 1, - 10)$ which is nearest to the intersection of the planes: $3 x - y - 3 z + 32 = 0$ and $3 x + 2 y - 15 z - 8 = 0$

$\frac{x + \frac{56}{9}}{7} = \frac{y - \frac{46}{3}}{12} = \frac{2}{3} = π$
$7 π - \frac{56}{9} = m - 3$
$14 π - \frac{- 112}{9} = m - 6$
$2 π - \frac{- 112}{9} - - \frac{40}{3} = - 13$
$12 π + - \frac{40}{9} = 24 + 7$