# Shortest Distance between Two Lines

• The shortest distance between the two skew lines \vec{r}={\vec{a}_1}+\lambda {\vec{b}_1} and \vec{r}={\vec{a}_2}+\mu {\vec{b}_2} is \frac{|(\vec{a}_{2}-\vec{a}_{1}) \cdot (\vec{b}_{1}\times \vec{b}_{2})|}{|\vec{b}_{1}\times \vec{b}_{2}|}
• The shortest distance between the two skew lines \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}} is = \begin{vmatrix}\frac{\begin{vmatrix}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1}\\ a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\end{vmatrix}}{\sqrt{(a_{1}b_{2}-a_{2}b_{1})^{2}+(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}}}\end{vmatrix}
• The shortest distance between the two parallel lines \vec{r}=\vec{a}_{1}+\lambda \vec{b} and \vec{r}=\vec{a}_{2}+\mu \vec{b} is \frac{|(\vec{a}_{2}-\vec{a}_{1})\times \vec{b}|}{|\vec{b}|}
• The shortest distance between the lines \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} and \frac{x-x_{2}}{a}=\frac{y-y_{2}}{b}=\frac{z-z_{2}}{c} is \frac{\begin{vmatrix}\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1}\\ a & b & c\end{vmatrix} \end{vmatrix}}{\sqrt{a^{2}+b^{2}+c^{2}}}
• The distance from the point to the line is nothing but the distance between the given point and foot of the perpendicular.
• The equation of the line of shortest distance between the line \vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} , \vec{r}=\vec{a}_{2}+\mu \vec{b}_{2} is \vec{r}=\vec{a}_{1}+\lambda (\vec{b}_{1}\times \vec{b}_{2}) (or) \vec{r}=\vec{a}_{2}+\lambda (\vec{b}_{1}\times \vec{b}_{2})
• If the projections of a line of length d on the coordinate axes are d1, d2 and d3 respectively, then d^{2}=d_1^2+d_2^2+d_3^2
• If the projection of a line of length d on the coordinate planes are d1, d2 and d3 respectively, then d_1^2+d_2^2+d_3^2=2d^{2}
• Area of the triangle formed by origin 'O' and the points A(x1, y1, z1) B(x2, y2, z2) is =\frac{1}{2}\sqrt{(x_{1}y_{2}-x_{2}y_{1})^{2} + (y_{1}z_{2}-y_{2}z_{1})^{2} + (z_{2}x_{1}-z_{1}x_{2})^{2}}

### View the Topic in this video From 01:40 To 23:00

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1. Shortest distance between \vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} and \vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}, is \begin{vmatrix}\frac{(\vec{b}_{1} \times \vec{b}_{2})\cdot (\vec{a}_{2}- \vec{a}_{1})}{|\vec{b}_{1}\times \vec{b}_{2}|} \end{vmatrix}

2. Shortest distance between the lines: \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}} is

\frac{\begin{vmatrix}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\a_{1} & b_{1} & c_{1}\\a_{2} & b_{2} & c_{2}\end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}

3. Distance between parallel lines \vec{r}=\vec{a}_{1}+\lambda \vec{b} and \vec{r}=\vec{a}_{2}+\mu \vec{b}, is \begin{vmatrix}\frac{\vec{b}\times (\vec{a}_{2}-\vec{a}_{1})}{|\vec{b}|} \end{vmatrix}