The shortest distance between the lines \;x-3/2=\;y-2/3=\;z-1/-1 and \;x+3/2=\;y-6/1=\;z-5/3 is :

Correct Answer is : 185\;{18}{{5}}518

Correct Answer is : 185\;{18}{{5}}518

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BITSAT 2023 Shift 1 Paper

Section: Mathematics

Question : 129 of 130

Marks: +1, -0

\;\frac{x-3}{2}=\;\frac{y-2}{3}=\;\frac{z-1}{-1}

\;\frac{x+3}{2}=\;\frac{y-6}{1}=\;\frac{z-5}{3}

Solution: 👈: Video Solution

\;\frac{x+3}{2}=\;\frac{y-6}{1}=\;\frac{z-5}{3}

Lines passes through the points

\;\overset{\rightarrow_1}{a}=(3,2,1) \;\text{ and }\; \overset{\rightarrow_2}{a}=(-3,6,5),

\;\overset{\rightarrow_1}{b}=2\hat{i}+3\hat{j}-\hat{k}

\;\overset{\rightarrow_1}{b}=2\hat{i}+\hat{j}-3k, \overset{\rightarrow_2}{a}-\overset{\rightarrow_1}{a}=6\hat{i}-4j-4\hat{k}

Shortest distance

=\;\frac{\overset{\rightarrow_2}{\left| \left(a-\overset{\rightarrow_1}{a}\right)\left(\overset{\rightarrow_1}{b} \times \overset{\rightarrow_2}{b}\right) \right|}}{\overset{\rightarrow_1}{\left| \left(\overset{\rightarrow_1}{b} \times \overset{\rightarrow_2}{b}\right) \right|}}

\overset{\rightarrow_1}{b} \times \overset{\rightarrow_2}{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix}=10\hat{i}-8\hat{j}-4\hat{k}

\; \left(\overset{\rightarrow_2}{a}-\overset{\rightarrow_1}{a}\right) \cdot \left(\overset{\rightarrow_1}{b} \times \overset{\rightarrow_2}{b}\right) =60+32+16=108

\;\overset{\rightarrow_1}{\left| b \times \overset{\rightarrow_2}{b} \right|}= \sqrt{100+64+16}= \sqrt{180}

\;\text{S.D.}=\;\frac{108}{\sqrt{180}}=\;\frac{108}{6\sqrt{5}}=\;\frac{18}{\sqrt{5}}

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