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If the shortest distance between the lines \vec{r}= \mathrm{\left ( -\hat{i}+3\hat{k} \right )+\lambda \left ( \hat{i}-a\hat{j} \right )\: \: and\; \; }\vec{r}= \mathrm{\left ( -\hat{j}+2\hat{k} \right )+\mu \left ( \hat{i}-\hat{j}+\hat{k} \right )\: is\sqrt{\frac{2}{3}}}, then the integral value of a is equal to _______.

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\vec{r}= \mathrm{\left ( -\hat{i}+3\hat{k} \right )+\lambda \left ( \hat{i}-a\hat{j} \right )}
\vec{r}= \mathrm{\left ( -\hat{j}+2\hat{k} \right )+\mu \left ( \hat{i}-\hat{j}+\hat{k} \right )}

\mathrm{shortest\: distance= \frac{\left [(\bar{a}-\bar{b}) \: \: \bar{c}\,\,\,\bar{d}\right ]}{\left | \bar{c}\times\bar{d} \right |}}

\mathrm{\begin{vmatrix} -1& 1 &1 \\ 1& -a& 0\\ 1& -1 & 1 \end{vmatrix}= a-1-1+a= 2a-2}

\mathrm{\left ( \bar{c} \times \bar{d} \right )= -a \hat{i}-\hat{j}+\left ( a-1 \right )\hat{k}}

\mathrm{\therefore \: \sqrt{\frac{2}{3}}= \pm \frac{2a-2}{\sqrt{a^{2}+1+\left ( a-1 \right )^{2}}}}
        \mathrm{\frac{2}{3}= \frac{4a^{2}+4-8a}{a^{2}+1+a^{2}+1-2a}}

\mathrm{4a^{2}+4-4a= 12a^{2}+12-24a}
\mathrm{8a^{2}-20a+8= 0}
\mathrm{2a^{2}-5a+2= 0}
\mathrm{a= \frac{1}{2},2}
\mathrm{\therefore }   integral value of \mathrm{a = 2 }

Posted by

vishal kumar

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