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Question: The shortest distance between the lines \[\overrightarrow r = \left( {4\widehat i - \widehat j} \rig...

The shortest distance between the lines r→=(4i^−j^)+λ(i^+2j^−3k^)\overrightarrow r = \left( {4\widehat i - \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - 3\widehat k} \right) , λ∈R\lambda \in R and r→=(−i^−j^+2k^)+μ(2i^+4j^−5k^), μ∈R\overrightarrow r = \left( { - \widehat i - \widehat j + 2\widehat k} \right) + \mu \left( {2\widehat i + 4\widehat j - 5\widehat k} \right),{\text{ }}\mu \in R is
A. 65\dfrac{6}{5}
B. 15\dfrac{1}{{\sqrt 5 }}
C. 105\dfrac{{10}}{{\sqrt 5 }}
D. none of these

Explanation

Solution

This is a vector-equation in a 3-D plane. In this type of questions the shortest distance between any two vector lines in 3-D plane can be found out by using the formula given by Shortest distance = ∣(a2→−a1→)⋅(b1→×b2→)∣b1→×b2→∣∣{\text{Shortest distance = }}\left| {\dfrac{{\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right| in which the vectors a1→, a2→, b1→ and b2→\overrightarrow {{a_1}} ,\overrightarrow {{\text{ }}{a_2}} ,{\text{ }}\overrightarrow {{b_1}} {\text{ }}and{\text{ }}\overrightarrow {{b_2}} can be found out from the general form of the lines given as r1→=a1→+λb1→ and r2→=a2→+μb2→\overrightarrow {{r_1}} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} {\text{ and }}\overrightarrow {{r_2}} = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} respectively.

Complete step-by-step solution:
The given equation of the line is r→=(4i^−j^)+λ(i^+2j^−3k^)\overrightarrow r = \left( {4\widehat i - \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - 3\widehat k} \right) .
On comparing it with the general form of line i.e. r1→=a1→+λb1→\overrightarrow {{r_1}} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} ,
a1→=(4i^−j^)\overrightarrow {{a_1}} = \left( {4\widehat i - \widehat j} \right)
b1→=(i^+2j^−3k^)\overrightarrow {{b_1}} = \left( {\widehat i + 2\widehat j - 3\widehat k} \right)
Again,
The given second equation of line is r→=(−i^−j^+2k^)+μ(2i^+4j^−5k^)\overrightarrow r = \left( { - \widehat i - \widehat j + 2\widehat k} \right) + \mu \left( {2\widehat i + 4\widehat j - 5\widehat k} \right) .
On comparing with the general form of the line i.e. r2→=a2→+μb2→\overrightarrow {{r_2}} = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} ,
a2→=(−i^−j^+2k^)\overrightarrow {{a_2}} = \left( { - \widehat i - \widehat j + 2\widehat k} \right)
b2→=2i^+4j^−5k^\overrightarrow {{b_2}} = 2\widehat i + 4\widehat j - 5\widehat k
∵\because The shortest distance between any two vector lines is given by Shortest distance = ∣(a2→−a1→)⋅(b1→×b2→)∣b1→×b2→∣∣{\text{Shortest distance = }}\left| {\dfrac{{\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|
∴\therefore On putting the values of a1→, a2→, b1→ and b2→\overrightarrow {{a_1}} ,\overrightarrow {{\text{ }}{a_2}} ,{\text{ }}\overrightarrow {{b_1}} {\text{ }}and{\text{ }}\overrightarrow {{b_2}} ,
⇒\Rightarrow Shortest distance = ∣((−i^−j^+2k^)−(4i^−j^))⋅((i^+2j^−3k^)×(2i^+4j^−5k^))∣((i^+2j^−3k^)×(2i^+4j^−5k^))∣∣{\text{Shortest distance = }}\left| {\dfrac{{\left( {\left( { - \widehat i - \widehat j + 2\widehat k} \right) - \left( {4\widehat i - \widehat j} \right)} \right) \cdot \left( {\left( {\widehat i + 2\widehat j - 3\widehat k} \right) \times \left( {2\widehat i + 4\widehat j - 5\widehat k} \right)} \right)}}{{\left| {\left( {\left( {\widehat i + 2\widehat j - 3\widehat k} \right) \times \left( {2\widehat i + 4\widehat j - 5\widehat k} \right)} \right)} \right|}}} \right|

{\widehat i}&{\widehat j}&{\widehat k} \\\ 1&2&{ - 3} \\\ 2&4&{ - 5} \end{array}} \right|$$ On expanding along Row-1, $$ \Rightarrow \overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left( { - 10 + 12} \right)\widehat i - \left( { - 5 + 6} \right)\widehat j + \left( {4 - 4} \right)\widehat k$$ $$ \Rightarrow \overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = 2\widehat i - \widehat j$$ On taking modulus both sides, $$ \Rightarrow \left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \left| {2\widehat i - \widehat j} \right|$$ $\because \left| {x\widehat i + y\widehat j + z\widehat k} \right| = \sqrt {{x^2} + {y^2} + {z^2}} $ $$ \Rightarrow \left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \sqrt {{2^2} + {1^2}} $$ $$ \Rightarrow \left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \sqrt 5 $$$\therefore {\text{Shortest distance = }}\left| {\dfrac{{\left( {\left( { - \widehat i - \widehat j + 2\widehat k} \right) - \left( {4\widehat i - \widehat j} \right)} \right) \cdot \left( {\left( {\widehat i + 2\widehat j - 3\widehat k} \right) \times \left( {2\widehat i + 4\widehat j - 5\widehat k} \right)} \right)}}{{\left| {\left( {\left( {\widehat i + 2\widehat j - 3\widehat k} \right) \times \left( {2\widehat i + 4\widehat j - 5\widehat k} \right)} \right)} \right|}}} \right|{\text{ becomes}}$, $${\text{Shortest distance = }}\left| {\dfrac{{\left( {\left( { - \widehat i - \widehat j + 2\widehat k} \right) - \left( {4\widehat i - \widehat j} \right)} \right) \cdot \left( {2\widehat i - \widehat j} \right)}}{{\left| {\sqrt 5 } \right|}}} \right|$$ On simplifying further, $${\text{Shortest distance = }}\left| {\dfrac{{\left( { - 5\widehat i + 2\widehat k} \right) \cdot \left( {2\widehat i - \widehat j} \right)}}{{\left| {\sqrt 5 } \right|}}} \right|$$ $\because {\text{ }}\left( {{{\text{a}}_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k} \right) \cdot \left( {{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k} \right) = {{\text{a}}_1}{b_1} + {{\text{a}}_2}{b_2} + {{\text{a}}_3}{b_3}$ $ \Rightarrow {\text{Shortest distance = }}\left| {\dfrac{{\left( { - 5 \cdot 2} \right)}}{{\left| {\sqrt 5 } \right|}}} \right|$ $ \Rightarrow {\text{Shortest distance = }}\left| {\dfrac{{ - 10}}{{\left| {\sqrt 5 } \right|}}} \right|$ Finally, $ \Rightarrow {\text{Shortest distance = }}\dfrac{{10}}{{\sqrt 5 }}$ **$\therefore $ Option (C) is correct.** **Note:** Calculations should be done carefully to avoid any mistake. During multiplication the coefficients of the unit vectors are handled carefully. The cross-product of the vectors should be done carefully. The determinant of the cross-product of vectors can be simplified using elementary operations and then can be expanded to reduce the large multiplication jobs. The vectors $$\overrightarrow {{a_1}} ,\overrightarrow {{\text{ }}{a_2}} ,{\text{ }}\overrightarrow {{b_1}} {\text{ }}and{\text{ }}\overrightarrow {{b_2}} $$ is to be found out in accordance with the general form of the equation. If the final answer comes in fraction then it may be changed into decimal in rounded form. Always try to solve step by step so that the wrong step can be determined and could be changed. Always take the absolute value of the distance as the distance cannot be negative.