Show that the straight lines x+1=2y=-12z and x=y+2=6z-6 are skew and hence find the shortest distance between

Exercise 6.5 · Q5

x+1=2y=-12z

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Parametrise each line by a common variable to extract direction ratios and a base point, then apply the skew-lines shortest-distance formula.

Step 1. Line 1: $x+1=2y=-12z=t$ . $x=t-1,\ y=t/2,\ z=-t/12$ ; direction $\propto(1,1/2,-1/12)$ , clear denominators ( $\times12$ ): $(12,6,-1)$ . At $t=0$ : point $\vec a=(-1,0,0)$ .

Step 2. Line 2: $x=y+2=6z-6=s$ . $x=s,\ y=s-2,\ z=(s+6)/6$ ; direction $\propto(1,1,1/6)$ , clear denominators ( $\times6$ ): $(6,6,1)$ . At $s=0$ : point $\vec c=(0,-2,1)$ .

Step 3. Check parallel. $\dfrac{12}6=2,\ \dfrac66=1$ — not equal, so not parallel.

Step 4. Compute $\vec b\times\vec d$ with $\vec b=(12,6,-1),\vec d=(6,6,1)$ :

\vec b\times\vec d=\begin{vmatrix}\hat i&\hat j&\hat k\\12&6&-1\\6&6&1\end{vmatrix}=\hat i(6+6)-\hat j(12+6)+\hat k(72-36)=12\hat i-18\hat j+36\hat k.

$|\vec b\times\vec d|=\sqrt{144+324+1296}=\sqrt{1764}=42.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2016Set ANNUAL1 markMCQ
$\dfrac{x-3}{4}=\dfrac{y-1}{2}=\dfrac{z-5}{-3}$
(a) $3$
(b) $2$
(c) $1$
(d) $0$
›Reveal solutionSolution
Using the shortest-distance-between-parallel-lines formula, the perpendicular distance works out exactly to $3$ .

Line 1: $\dfrac{x-3}{4}=\dfrac{y-1}{2}=\dfrac{z-5}{-3}$ passes through $A_1(3,1,5)$ with direction $\vec b=4\vec i+2\vec j-3\vec k$ .

Line 2: $\dfrac{x-1}{4}=\dfrac{y-2}{2}=\dfrac{z-3}{-3}$ passes through $A_2(1,2,3)$ with the same direction $\vec b$ , confirming the lines are parallel.

Vector joining the two points: $\vec{A_1A_2}=(1-3)\vec i+(2-1)\vec j+(3-5)\vec k=-2\vec i+\vec j-2\vec k$ .

Shortest distance between parallel lines: $d=\dfrac{\left|\vec{A_1A_2}\times\vec b\right|}{|\vec b|}$ .

Compute the cross product: $\vec{A_1A_2}\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\-2&1&-2\\4&2&-3\end{vmatrix}=(1)(1\vec i)+\ldots$ Component-wise: $i$ -comp $=(1)(-3)-(-2)(2)=-3+4=1$ ; $j$ -comp $=-[(-2)(-3)-(-2)(4)]=-[6+8]=-14$ ; $k$ -comp $=(-2)(2)-(1)(4)=-4-4=-8$ . So the cross product is $(1,-14,-8)$ . …
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Q.Show that the straight lines $x+1=2y=-12z$ and $x=y+2=6z-6$ are skew and hence find the shortest distance between them.

Concept understanding — Skew Lines and Shortest Distance

Q.Show that the straight lines x+1=2y=−12zx+1=2y=-12zx+1=2y=−12z and x=y+2=6z−6x=y+2=6z-6x=y+2=6z−6 are skew and hence find the shortest distance between them.

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Q.Show that the straight lines $x+1=2y=-12z$ and $x=y+2=6z-6$ are skew and hence find the shortest distance between them.

Unlock everything free for 14 days