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Find the shortest distance between the lines whose vector equations are \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k}) and \vec{r}=4 \hat{i}+5 \hat{j}-6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})
CBSE | Class 12 | Exercise 11.2 | Maths | NCERT | Three Dimensional Geometry
Find the shortest distance between the lines whose vector equations are \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k}) and \vec{r}=4 \hat{i}+5 \hat{j}-6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})

Solution:

It is known to us that shortest distance between two lines
\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} and \vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} is given as:
\mathrm{d}=\left|\frac{\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)}{\left|\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right|}\right|
On comparing the equations we obtain,
\begin{array}{l} \overrightarrow{a_{1}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{b_{1}}=\hat{i}-3 \hat{j}+2 \hat{k} \\ \overrightarrow{a_{2}}=4 \hat{i}+5 \hat{j}+6 \hat{k}, \overrightarrow{b_{2}}=2 \hat{i}+3 \hat{j}+\hat{k} \end{array}
Subtracting the above equations we obtain,
\left(\mathrm{x}_{1} \hat{\mathrm{i}}+\mathrm{y}_{1} \hat{\mathrm{j}}+\mathrm{z}_{1} \hat{\mathrm{k}}\right)-\left(\mathrm{x}_{2} \hat{\mathrm{i}}+\mathrm{y}_{2} \hat{\mathrm{j}}+\mathrm{z}_{2} \hat{\mathrm{k}}\right)=\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right) \hat{\mathrm{i}}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right) \hat{\mathrm{j}}+\left(\mathrm{z}_{1}-\mathrm{z}_{2}\right) \hat{\mathrm{k}}
\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}=(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} And,

\begin{array}{l} \overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ =\mid \begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ =\left|\begin{array}{ccc} 1 & -3 & 2 \\ 2 & 3 & 1 \end{array}\right| \\ =-9 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+9 \hat{\mathrm{k}} \\ \Rightarrow & \overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=-9 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+9 \hat{\mathrm{k}} \\ \Rightarrow\left|\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right|=\sqrt{(-9)^{2}+3^{2}+9^{2}}=\sqrt{81+9+81}=\sqrt{171}=3 \sqrt{19} \end{array} \end{array}

Multiplying eq. (2) and eq. (3) we get,
\begin{array}{l} \left(\mathrm{a}_{1} \hat{\mathrm{i}}+\mathrm{b}_{1} \hat{\mathrm{j}}+\mathrm{c}_{1} \hat{\mathrm{k}}\right) \cdot\left(\mathrm{a}_{2} \hat{\mathrm{i}}+\mathrm{b}_{2} \hat{\mathrm{j}}+\mathrm{c}_{2} \hat{\mathrm{k}}\right)=\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2} \\ \left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)=(-9 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=-27+9+27=9 \end{array}
On substituting all the values in eq. (1), we get
The shortest distance between the two lines,
\mathrm{d}=\left|\frac{9}{3 \sqrt{19}}\right|=\frac{9}{3 \sqrt{19}}=\frac{3}{\sqrt{19}}
As a result, the shortest distance is 3 \sqrt{19}

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