UPSC 2016 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

Shortest distance between two skew lines · Analytic Geometry · asked 5× in 14 yrs · Read the full method →

Question

Find the shortest distance between the lines $\dfrac{x-1}{2}=\dfrac{y-2}{4}=z-3$ and $y-mx=z=0$ . For what value of $m$ will the two lines intersect?

Technique

Skew-line shortest distance $=\dfrac{|(\mathbf{P_2}-\mathbf{P_1})\cdot(\mathbf d_1\times\mathbf d_2)|}{\|\mathbf d_1\times\mathbf d_2\|}$ ; intersection $\iff$ numerator $=0$ .

Solution

Step 1 — Identify points and direction vectors

Line $L_1$ : through $P_1=(1,2,3)$ with direction $\mathbf d_1=(2,4,1)$ . Line $L_2$ : $z=0$ and $y=mx$ , i.e. through $P_2=(0,0,0)$ with direction $\mathbf d_2=(1,m,0)$ .

Step 2 — Common perpendicular direction

\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&4&1\\1&m&0\end{vmatrix}=(4\cdot0-1\cdot m,\ 1\cdot1-2\cdot0,\ 2m-4)=(-m,\ 1,\ 2m-4).

Step 3 — Shortest distance formula

With $\mathbf w=P_2-P_1=(-1,-2,-3)$ ,

\text{S.D.}=\frac{|\,\mathbf w\cdot(\mathbf d_1\times\mathbf d_2)\,|}{\|\mathbf d_1\times\mathbf d_2\|}.

Numerator: $\mathbf w\cdot(-m,1,2m-4)=(-1)(-m)+(-2)(1)+(-3)(2m-4)=m-2-6m+12=10-5m.$ Denominator: $\sqrt{m^2+1+(2m-4)^2}=\sqrt{5m^2-16m+17}.$

Answer

\boxed{\;\text{S.D.}=\frac{|10-5m|}{\sqrt{5m^2-16m+17}}=\frac{5\,|2-m|}{\sqrt{5m^2-16m+17}}.\;}