The shortest distance between the lines {r_1} | vedclass

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(D) The given lines are in the form ${r_1} = {a_1} + \lambda {b_1}$ and ${r_2} = {a_2} + \mu {b_2}$.Here,${a_1} = 4i - 3j - k$,${b_1} = i - 4j + 7k$,$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) The given lines are in the form ${r_1} = {a_1} + \lambda {b_1}$ and ${r_2} = {a_2} + \mu {b_2}$.Here,${a_1} = 4i - 3j - k$,${b_1} = i - 4j + 7k$,${a_2} = i - j - 10k$,and ${b_2} = 2i - 3j + 8k$.First,we calculate the cross product ${b_1} 	imes {b_2}$:${b_1} 	imes {b_2} = \begin{vmatrix} i & j & k \ 1 & -4 & 7 \ 2 & -3 & 8 \end{vmatrix} = i(-32 + 21) - j(8 - 14) + k(-3 + 8) = -11i + 6j + 5k$.Next,we find the vector ${a_2} - {a_1} = (1-4)i + (-1 - (-3))j + (-10 - (-1))k = -3i + 2j - 9k$.The shortest distance is given by $d = \frac{|({a_2} - {a_1}) \cdot ({b_1} 	imes {b_2})|}{|{b_1} 	imes {b_2}|}$.Calculating the dot product: $({a_2} - {a_1}) \cdot ({b_1} 	imes {b_2}) = (-3)(-11) + (2)(6) + (-9)(5) = 33 + 12 - 45 = 0$.Since the scalar triple product is $0$,the lines are coplanar and intersect,so the shortest distance is $0$.

The shortest distance between the lines ${r_1} = 4i - 3j - k + \lambda (i - 4j + 7k)$ and ${r_2} = i - j - 10k + \mu (2i - 3j + 8k)$ is

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