Question

Question

सदिश विधि से निम्न रेखाओं के बीच की दूरी ज्ञात कीजिए ?

$\frac{x - 3}{3} = \frac{y - 8}{ - 1} = \frac{z - 3}{1} and \: \frac{x + 3}{3} = \frac{y + 7}{2} = \frac{z - 6}{4}$

note

➠class 12, vector algebra.
➠need quality answer.
➠need own good answer.
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Answer 1

Answer:

$\frac {x - 3} {3} = \frac {y - 8} { - 1} = \frac {z - 3} {1} and \: \frac {x + 3} {3} = \frac {y + 7} {2} = \frac {z - 6} {4}$

Answer 2

Answer:

3

3x−3

3x−3

3x−3 =

3x−3 = −1

3x−3 = −1y−8

3x−3 = −1y−8

3x−3 = −1y−8 =

3x−3 = −1y−8 = 1

3x−3 = −1y−8 = 1z−3

3x−3 = −1y−8 = 1z−3

3x−3 = −1y−8 = 1z−3 and

3x−3 = −1y−8 = 1z−3 and 3

3x−3 = −1y−8 = 1z−3 and 3x+3

3x−3 = −1y−8 = 1z−3 and 3x+3

3x−3 = −1y−8 = 1z−3 and 3x+3 =

3x−3 = −1y−8 = 1z−3 and 3x+3 = 2

3x−3 = −1y−8 = 1z−3 and 3x+3 = 2y+7

3x−3 = −1y−8 = 1z−3 and 3x+3 = 2y+7

3x−3 = −1y−8 = 1z−3 and 3x+3 = 2y+7 =

3x−3 = −1y−8 = 1z−3 and 3x+3 = 2y+7 = 4

3x−3 = −1y−8 = 1z−3 and 3x+3 = 2y+7 = 4z−6

3x−3 = −1y−8 = 1z−3 and 3x+3 = 2y+7 = 4z−6

i am in 8 th class...and you