The shortest distance between the lines r=3i+ | vedclass

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(D) The given lines are $r=a_1+\lambda b_1$ and $r=a_2+\mu b_2$,where:$a_1 = 3i+5j+7k, b_1 = i+2j+k$$a_2 = -i-j-k, b_2 = 7i-6j+k$The shortest distance

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$\frac{46}{5\sqrt{5}}$

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(D) The given lines are $r=a_1+\lambda b_1$ and $r=a_2+\mu b_2$,where:$a_1 = 3i+5j+7k, b_1 = i+2j+k$$a_2 = -i-j-k, b_2 = 7i-6j+k$The shortest distance $d$ is given by the formula $d = \frac{|(a_2-a_1) \cdot (b_1 	imes b_2)|}{|b_1 	imes b_2|}$.First,calculate the cross product $b_1 	imes b_2$:$b_1 	imes b_2 = \begin{vmatrix} i & j & k \ 1 & 2 & 1 \ 7 & -6 & 1 \end{vmatrix} = i(2+6) - j(1-7) + k(-6-14) = 8i + 6j - 20k$.The magnitude $|b_1 	imes b_2| = \sqrt{8^2 + 6^2 + (-20)^2} = \sqrt{64 + 36 + 400} = \sqrt{500} = 10\sqrt{5}$.Next,calculate $a_2 - a_1 = (-i-j-k) - (3i+5j+7k) = -4i - 6j - 8k$.Now,calculate the dot product $(a_2-a_1) \cdot (b_1 	imes b_2) = (-4i - 6j - 8k) \cdot (8i + 6j - 20k) = (-4)(8) + (-6)(6) + (-8)(-20) = -32 - 36 + 160 = 92$.The shortest distance $d = \frac{|92|}{10\sqrt{5}} = \frac{92}{10\sqrt{5}} = \frac{46}{5\sqrt{5}}$.

The shortest distance between the lines $r=3i+5j+7k+\lambda(i+2j+k)$ and $r=-i-j-k+\mu(7i-6j+k)$ is

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