रेखाओं `(x +2)/(-4) = (y)/(1) = (z - 7)/( 1 )` और ` ( x + 3 )/(-4) = ( y - 6 )/(3) = (z)/(2) ` के बीच की न्यूनतम दुरी ज्ञात कीजिए |

Explanation

यहाँ ` x _ 1 = -2, y _ 1 = 0 , z _ 1 = 7, a _ 1 = - 4, b _1 = 1 , c _1 = 1 `
` x _2 = - 3, y _ 2 = 6, z _ 2 =0, a _ 2 = -4, b _ 2 = 3, c _ 2 = 2 `
अब ` D = ( a _ 1 b _2 - a _2 b _ 1 ) ^(2) + ( b _1 c _ 2 - b _ 2 c_1 )^(2) + ( a _ 2 c_ 1 - a _ 1 c_2 )^(2) `
` = (-12 + 4 ) ^(2) + ( 2 - 3 ) ^(2) + ( - 4 + 8 )^(2) `
= ` 64 + 1 + 16 = 81 `
और न्यूनतम दुरी = ` (1)/(sqrtD) |{:(x_2 - x_1,, y_2 - y _ 1,, z _ 2 - z _ 1 ),(a _ 1,, b _ 1,, c_ 1 ),(a_2 ,,b_2,, c_2 ):}| `
` = (1)/(sqrt(81)) |{:(-1,,6,, -7),(-4,,1,,1),(-4,,3,,2):}|`
` = (1)/(9) [ - 1 (2 -3) - 6( - 8 + 4 ) - 7(-12 + 4)] `
` = (1)/(9) [ 1 + 24 + 56] = (81)/(9) = 9 ` मात्रक