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Question: Calculate the shortest distance between the lines \( \vec r = \left( {4\hat i - \hat j} \right) + \l...

Calculate the shortest distance between the lines r⃗=(4i^−j^)+λ(i^+2j^−3k^)\vec r = \left( {4\hat i - \hat j} \right) + \lambda \left( {\hat i + 2\hat j - 3\hat k} \right) and r⃗=(i^−j^+2k^)+μ(2i^+4j^−5k^)\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( {2\hat i + 4\hat j - 5\hat k} \right) .

Explanation

Solution

Hint : The given question is related to the concept of shortest distance between the lines. Here, in this question we are given two lines in vector form and we have to find the shortest distance between these two lines. We will proceed solving this question by first comparing the given lines with r⃗=a^1+λb^1\vec r = {\hat a_1} + \lambda {\hat b_1} and r⃗=a^2+λb^2\vec r = {\hat a_2} + \lambda {\hat b_2} respectively. Then, find out the various values which are included in the formula and then substitute the obtained values in the shortest distance formula.
Formula used:
r⃗=a^1+λb^1\vec r = {\hat a_1} + \lambda {\hat b_1}
r⃗=a^2+λb^2\vec r = {\hat a_2} + \lambda {\hat b_2}
d=∣(a⃗2−a⃗1)×(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣d = \left| {\dfrac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right) \times \left( {{{\vec b}_1} \times {{\vec b}_2}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|

Complete step by step solution:
Given equation of lines in vector form are r⃗=(4i^−j^)+λ(i^+2j^−3k^)\vec r = \left( {4\hat i - \hat j} \right) + \lambda \left( {\hat i + 2\hat j - 3\hat k} \right) and r⃗=(i^−j^+2k^)+μ(2i^+4j^−5k^)\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( {2\hat i + 4\hat j - 5\hat k} \right) .
As we know that the shortest distance between the lines r⃗=a^1+λb^1\vec r = {\hat a_1} + \lambda {\hat b_1} and r⃗=a^2+λb^2\vec r = {\hat a_2} + \lambda {\hat b_2} is d=∣(a⃗2−a⃗1)×(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣d = \left| {\dfrac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right) \times \left( {{{\vec b}_1} \times {{\vec b}_2}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|
Comparing the given lines with r⃗=a^1+λb^1\vec r = {\hat a_1} + \lambda {\hat b_1} and r⃗=a^2+λb^2\vec r = {\hat a_2} + \lambda {\hat b_2} respectively, we get
a^1=4i^−j^ a^2=i^−j^+2k^ b^1=i^+2j^−3k^ b^2=2i^+4j^−5k^  {{\hat a}_1} = 4\hat i - \hat j \\\ {{\hat a}_2} = \hat i - \hat j + 2\hat k \\\ {{\hat b}_1} = \hat i + 2\hat j - 3\hat k \\\ {{\hat b}_2} = 2\hat i + 4\hat j - 5\hat k \\\
Now, we will calculate the value of a⃗2−a⃗1{\vec a_2} - {\vec a_1}
⇒a^2−a^1=(i^−j^+2k^)−(4i^−j^) ⇒a^2−a^1=(i^−j^+2k^)−4i^+j^ ⇒a^2−a^1=−3i^+0j^+2k^  \Rightarrow {{\hat a}_2} - {{\hat a}_1} = \left( {\hat i - \hat j + 2\hat k} \right) - \left( {4\hat i - \hat j} \right) \\\ \Rightarrow {{\hat a}_2} - {{\hat a}_1} = \left( {\hat i - \hat j + 2\hat k} \right) - 4\hat i + \hat j \\\ \Rightarrow {{\hat a}_2} - {{\hat a}_1} = - 3\hat i + 0\hat j + 2\hat k \\\
Next, we calculate the value of b⃗1×b⃗2{\vec b_1} \times {\vec b_2}
\Rightarrow {{\vec b}_1} \times {{\vec b}_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 1&2&{ - 3} \\\ 2&4&{ - 5} \end{array}} \right| \\\ \Rightarrow {{\vec b}_1} \times {{\vec b}_2} = \left( {\left( {2 \times - 5} \right) - \left( { - 3 \times 4} \right)} \right)\hat i - \left( {\left( {1 \times - 5} \right) - \left( { - 3 \times 2} \right)} \right)\hat j + \left( {\left( {1 \times 4} \right) - \left( {2 \times 2} \right)} \right)\hat k \\\ \Rightarrow {{\vec b}_1} \times \vec b = \left( { - 10 + 12} \right)\hat i - \left( { - 5 + 6} \right)\hat j + \left( {4 - 4} \right)\hat k \\\ \Rightarrow {{\vec b}_1} \times \vec b = 2\hat i - \hat j + 0\hat k \;
The value of ∣b⃗1×b⃗2∣\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|
⇒∣b⃗1×b⃗2∣=(2)2+(−1)2+(0)2 ⇒∣b⃗1×b⃗2∣=4+1+0 ⇒∣b⃗1×b⃗2∣=5   \Rightarrow \left| {{{\vec b}_1} \times {{\vec b}_2}} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 0 \right)}^2}} \\\ \Rightarrow \left| {{{\vec b}_1} \times {{\vec b}_2}} \right| = \sqrt {4 + 1 + 0} \\\ \Rightarrow \left| {{{\vec b}_1} \times {{\vec b}_2}} \right| = \sqrt 5 \;
The value obtained after (a⃗2−a⃗1)×(b⃗1×b⃗2)\left( {{{\vec a}_2} - {{\vec a}_1}} \right) \times \left( {{{\vec b}_1} \times {{\vec b}_2}} \right) is
⇒(a⃗2−a⃗1)×(b⃗1×b⃗2)=(−3i^+0j^+2k^)×(2i^−j^+0k^) ⇒(a⃗2−a⃗1)×(b⃗1×b⃗2)=−6   \Rightarrow \left( {{{\vec a}_2} - {{\vec a}_1}} \right) \times \left( {{{\vec b}_1} \times {{\vec b}_2}} \right) = \left( { - 3\hat i + 0\hat j + 2\hat k} \right) \times \left( {2\hat i - \hat j + 0\hat k} \right) \\\ \Rightarrow \left( {{{\vec a}_2} - {{\vec a}_1}} \right) \times \left( {{{\vec b}_1} \times {{\vec b}_2}} \right) = - 6 \;
Now, we will substitute all the values obtained in the shortest distance formula and get,
⇒d=∣(a⃗2−a⃗1)×(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣ ⇒d=∣−65∣ ⇒d=65   \Rightarrow d = \left| {\dfrac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right) \times \left( {{{\vec b}_1} \times {{\vec b}_2}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right| \\\ \Rightarrow d = \left| {\dfrac{{ - 6}}{{\sqrt 5 }}} \right| \\\ \Rightarrow d = \dfrac{6}{{\sqrt 5 }} \;
Therefore, the shortest distance between the given equation of lines is 65\dfrac{6}{{\sqrt 5 }} .

Note : The given question was an easy one. While solving such questions, students should keep in mind the direction ratios. Here, in the above question we couldn’t take the negative answer as the distance can never be negative that’s why there is a modulus in the formula of shortest distance between two lines. Students should avoid making calculation mistakes while solving determinants. In these types of questions, the vector sign is very important as it signifies the direction of the vector.