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Shortest distance between lines $\vec{r}=(2+2\lambda)\hat{i}+(4-2\lambda)\hat{j}+4\lambda\hat{k}$ and $\vec{r}=(4-3\lambda)\hat{i}+(4+4\lambda)\hat{j}+(1-2\lambda)\hat{k}$ is

Option 1)
$\frac{9}{\sqrt{53}}$

Option 2)
$\frac{11}{\sqrt{53}}$

Option 3)
$\frac{13}{\sqrt{53}}$

Option 4)
$\frac{15}{\sqrt{53}}$

Answers (1)

best_answer

As we have learned

Shortest distance between two skew lines (vector form) -

Shortest distance between $\vec{r}=\vec{a}+\lambda \vec{b}\, \: and \,\: \vec{r}=\vec{a_{1}}+\mu\vec{b}$ is given by

$\left | \frac{\left ( \vec{b} \times \vec{b_{1}}\right )\cdot \left ( \vec{a} -\vec{a_{1}}\right )}{\left | \vec{b} \times \vec{b_{1}} \right |} \right |$

- wherein

shortest distance is among the line which is perpendicular to both

$LM= \vec{b}\times \vec{b_{1}}$

shortest distance will be projection of PQ = $\vec{a}- \vec{a_{1}}$ on LM

$\vec{r}=(2\hat{i}+4\hat{j})+\lambda(2\hat{i}-2\hat{j}+4\hat{k})$ and

$\vec{r}=(4\hat{i}+4\hat{j}+\hat{k})+\lambda(-3\hat{i}+4\hat{j}-2\hat{k})$

Here $\vec{a}=2\hat{i}+4\hat{j}, \vec{b}=2\hat{i}-2\hat{j}+4\hat{k}$

$\vec{a_{1}}=4\hat{i}+4\hat{j}+\hat{k}, \vec{b_{1}}=-3\hat{i}+4\hat{j}-2\hat{k}$

$\therefore \vec{b}\times\vec{b}_{1}=\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 2 & -2& 4\\ -3& 4 & -2 \end{vmatrix}= \hat{i}(-12)-\hat{j}(8)+\hat{k}2= 12\hat{i}-8\hat{j}+2\hat{k}$

$(\vec{a}-\vec{a_1})= -2\hat{i}-\hat{k}$

$(\vec{a}-\vec{a_1})\cdot ( \vec{b}\times \vec{b_1})= 24-2=22$

shortest distance = $= \frac{22}{|\vec{b}\times \vec{b_1}|}= \frac{22}{\sqrt{144+64+4}}=\frac{11}{\sqrt{53}}$

Option 1)

$\frac{9}{\sqrt{53}}$

Option 2)

$\frac{11}{\sqrt{53}}$

Option 3)

$\frac{13}{\sqrt{53}}$

Option 4)

$\frac{15}{\sqrt{53}}$

Posted by

Himanshu

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