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Let \lambda be an integer. If the shortest distance between the lines x-\lambda =2y-1=-2z and x= y+2\lambda =z-\lambda is \frac{\sqrt{7}}{2\sqrt{2}}, then the value of \left | \lambda \right | is ________.
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 4

Answers (1)

best_answer

 

\\\frac{x-\lambda}{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z-0}{-\frac{1}{2}} \\ \\\frac{x-0}{1}=\frac{y+2 \lambda}{1}=\frac{z-\lambda}{1}

\text {Shortest distance }=\frac{\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)}{\left|b_{1} \times b_{2}\right|}

\mathrm{b}_{1} \times \mathrm{b}_{2}=\left|\begin{array}{ccc} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 1 & 1 \end{array}\right|

\\=\hat{\mathrm{i}}\left(\frac{1}{2}+\frac{1}{2}\right)-\hat{\mathrm{j}}\left(1+\frac{1}{2}\right)+\hat{\mathrm{k}}\left(1-\frac{1}{2}\right) \\\\ =\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+\frac{\hat{\mathrm{k}}}{2}=\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{2}

\frac{\mathrm{b}_{1} \times \mathrm{b}_{2}}{\left|\mathrm{~b}_{1} \times \mathrm{b}_{2}\right|}=\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}} \\\\ \frac{\left(\mathrm{a}_{2}-\mathrm{a}_{1}\right) \cdot\left(\mathrm{b}_{1} \times \mathrm{b}_{2}\right)}{\left|\mathrm{b}_{1} \times \mathrm{b}_{2}\right|}=\left(-\lambda \hat{\mathrm{i}}+\left(-2 \lambda+\frac{1}{2}\right)+\lambda \hat{\mathrm{k}}\right) \cdot \left(\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\right)

\\=\left|\frac{-2 \lambda+6 \lambda-\frac{3}{2}+\lambda}{\sqrt{14}}\right|=\frac{\sqrt{7}}{2 \sqrt{2}} \\ \\\left|5 \lambda-\frac{3}{2}\right|=\frac{7}{2} \\ \\5 \lambda=\frac{3}{2} \pm \frac{7}{2} \\ \\5 \lambda=5,-2 \\ \\\lambda=1,-\frac{2}{5}

Posted by

Suraj Bhandari

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