The shortest distance between the lines frac{ | vedclass

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Question

(A) The given lines are in the form $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}

Vedclass · Accepted Answer

$\frac{187}{\sqrt{563}}$

Vedclass · Answer

(A) The given lines are in the form $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$.From the equations,the points on the lines are $A(3, -7, 1)$ and $B(5, 9, -2)$.The direction vectors are $\vec{p} = 4\hat{i} - 11\hat{j} + 5\hat{k}$ and $\vec{q} = 3\hat{i} - 6\hat{j} + 1\hat{k}$.First,find the vector $\vec{AB} = (5-3)\hat{i} + (9-(-7))\hat{j} + (-2-1)\hat{k} = 2\hat{i} + 16\hat{j} - 3\hat{k}$.Next,find the cross product $\vec{n} = \vec{p} 	imes \vec{q}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -11 & 5 \ 3 & -6 & 1 \end{vmatrix} = \hat{i}(-11 + 30) - \hat{j}(4 - 15) + \hat{k}(-24 + 33) = 19\hat{i} + 11\hat{j} + 9\hat{k}$.The shortest distance ($S$.d.) is the projection of $\vec{AB}$ on $\vec{n}$:$S.d. = \left| \frac{\vec{AB} \cdot \vec{n}}{|\vec{n}|} ight| = \left| \frac{(2\hat{i} + 16\hat{j} - 3\hat{k}) \cdot (19\hat{i} + 11\hat{j} + 9\hat{k})}{\sqrt{19^2 + 11^2 + 9^2}} ight|$$S.d. = \left| \frac{38 + 176 - 27}{\sqrt{361 + 121 + 81}} ight| = \frac{187}{\sqrt{563}}$.

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is:

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