If the shortest distance between the lines\frac{x-1}{\alpha }=\frac{y+1}{-1}= \frac{z}{1},\left ( \alpha \neq -1 \right )and

x+y+z+1=0=2x-y+z+3 \: is\: \frac{1}{\sqrt{3}},then
a value of  \alpha  is :

 

  • Option 1)

    -\frac{16}{19}

  • Option 2)

    -\frac{19}{16}

  • Option 3)

    \frac{32}{19}

  • Option 4)

    \frac{19}{32}

 

Answers (1)
V Vakul

As we learnt in 

Shortest distance between two skew lines (Cartesian form) -

Shortest distance between

\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 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 | Where 

\vec{a}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}

\vec{a_{1}}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}

\vec{b}= a_{1}\hat{i}+b_{1}\hat{j}+c_{1}\hat{k}

\vec{b_{1}}= a_{2}\hat{i}+b_{2}\hat{j}+c_{2}\hat{k}

 

-

 

 Normal vector of line is L_{2} is

\begin{vmatrix} \hat{i}& \hat{j} &\hat{k} \\ 1 & 1&1 \\ 2 & -1& 1 \end{vmatrix}=2\hat{i}+\hat{j}-3\hat{k}

b_{1}^{-1}\times b_{2}^{-1}=(\alpha\hat{i}-\hat{j}+\hat{k})\times (2\hat{i}+\hat{j}-\hat{3k})

=\begin{bmatrix} \hat{i} & \hat{j} &\hat{k} \\ \alpha &-1 & 1\\ 2 & 1 & -3 \end{bmatrix}=2\hat{i}+\hat{j}(3\alpha+2)+\hat{k}(\alpha+2)

One point on L_{2} is (0,1,-2) and on L_{1} is (1,-1,0)

Distance =\left | \frac{(\hat{i}-2\hat{j}+2\hat{k}). (2\hat{1}+(3\alpha+2)\hat{j})}{\sqrt{4+(3\alpha+2)^{2}+(\alpha+2)^{2}}} \right |=\frac{1}{\sqrt{3}}

3(2-6\alpha-4+2\alpha+4)^{2}=10\alpha^{2}+12+16\alpha

3\times 4(2\alpha-1)^{2}=2(5\alpha^{2}+8\alpha+6)

24\alpha^{2}+6-24\alpha=5\alpha^{2}+8\alpha+6

19\alpha^{2}-32\alpha=0\Rightarrow \alpha=\frac{32}{19}


Option 1)

-\frac{16}{19}

This option is incorrect

Option 2)

-\frac{19}{16}

This option is incorrect

Option 3)

\frac{32}{19}

This option is correct

Option 4)

\frac{19}{32}

This option is incorrect

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