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Find the shortest distance between lines \vec{r}=(6 \hat{i}+2 \hat{j}+2 \hat{k})+\lambda(1 \hat{i}-2 \hat{j}+\hat{2}) \text { and } \overrightarrow{1}=(-4 \hat{i}-\hat{k})+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})
CBSE | Class 12 | Maths | Miscellaneous Exercise | NCERT | Three Dimensional Geometry
Find the shortest distance between lines \vec{r}=(6 \hat{i}+2 \hat{j}+2 \hat{k})+\lambda(1 \hat{i}-2 \hat{j}+\hat{2}) \text { and } \overrightarrow{1}=(-4 \hat{i}-\hat{k})+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})

Solution:

It is known to us that the shortest distance between lines with vector equations \vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} and \vec{r}=\overrightarrow{a_{2}}+\lambda \overrightarrow{b_{2}} is given as
\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|

Given:
\overrightarrow{\mathrm{r}}=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(1 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{2})
Let’s compare it with \vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}, we get \overrightarrow{\mathrm{a}_{1}}=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}_{1}}=(1 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{2})
In the similar way,
\overrightarrow{\mathrm{r}}=(-4 \hat{\mathrm{i}}-\hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})

Let’s compare it with \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{2}}+\lambda \overrightarrow{\mathrm{b}_{2}}, we obtain \overrightarrow{\mathrm{a}_{2}}=(-4 \hat{\mathrm{i}}-\hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}_{2}}=(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{2})

Now,
\begin{aligned} \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) =(-4 \hat{i}-\hat{k})-(6 \hat{i}+2 \hat{j}+2 \hat{k}) \\ =((-4-6) \hat{i}+(0-2) \hat{j}+(-1-2) \hat{k}) \\ =(-10 \hat{i}-2 \hat{j}-3 \hat{k}) \end{aligned}

And,
\begin{array}{l} \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{array}\right| \\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{array}\right| \\ =\hat{i}[(-2 \times-2)-(-2 \times 2)]-\hat{j}[(1 \times-2)-(3 \times 2)]+\hat{k}[(1 \times-2)-(3 \times-2)] \\ =\hat{i}[4+4]-\hat{j}[-2-6]+\hat{k}[-2+6] \end{array}

=8 \hat{i}+8 \hat{j}+4 \hat{k} So, Magnitude of \overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right| \begin{array}{l}=\sqrt{8^{2}+8^{2}+4^{2}} \\ =\sqrt{144}\end{array}=\sqrt{64+64+16} =
=12

Also,
\begin{aligned} \left(\vec{b}_{1} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) =(8 \hat{i}+8 \hat{j}+4 \hat{k}) \cdot(-10 \hat{i}-2 \hat{j}-3 \hat{k}) \\ =-80+(-16)+(-12) \\ =-108 \end{aligned}

Therefore the shortest distance is given as
\begin{aligned} =\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|=\left|\frac{-108}{12}\right| =|-9| \\ =9 \end{aligned}
As a result, the shortest distance between the given two lines is 9 .

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