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  4. Line L meets lines L 1: ( x /1)=( y /2)=( z /3) and L 2: ( x -2/2)=( y -1/4)=( z -4/5) orthogonally at points P and Q. (P Q)2 is D. DRs of line L are (a, b, c) a, b, c ∈ I then least value of (5 D/3)+|a|+|b|+|c|.

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Q. Line $L$ meets lines $L _1: \frac{ x }{1}=\frac{ y }{2}=\frac{ z }{3}$ and $L _2: \frac{ x -2}{2}=\frac{ y -1}{4}=\frac{ z -4}{5}$ orthogonally at points $P$ and Q. $(P Q)^2$ is D. DRs of line $L$ are $(a, b, c)\{a, b, c \in I\}$, then least value of $\frac{5 D}{3}+|a|+|b|+|c|$.

Vector Algebra

Solution:

PQ distance is shortest distance between $L _1$ and $L _2$.
S.D. $=\frac{|[2 \hat{i}+\hat{j}+4 \hat{k} \hat{i}+2 \hat{j}+3 \hat{k} 2 \hat{i}+4 \hat{j}+5 \hat{k}]|}{|(\hat{i}+2 \hat{j}+3 \hat{k}) \times(2 \hat{i}+4 \hat{j}+5 \hat{k})|}$
image
$=\frac{3}{|-2 \hat{i}+\hat{j}|}=\frac{3}{\sqrt{5}}$
or $( PQ )^2=\frac{9}{5} \Rightarrow 5 D =9$
DRs of line $L$ are proportional to $(-2,1,0)$.