The shortest distance between the lines [(x-3)/(2)=(y+15)/(-7)=(z-9)/(5)] and [(x+1)/(2)=(…

Question

The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is

Accepted Answer

Given two lines are represented as:

$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$

and

$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$

The formula for the shortest distance between two lines is:

$d=\frac{|({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})|}{|{\vec{b}}_{1}	imes {\vec{b}}_{2}|}$

From the given lines:

${\vec{a}}_{1}=(3,-15,9)$

${\vec{a}}_{2}=(-1,1,9)$

${\vec{b}}_{1}=(2,-7,5)$

${\vec{b}}_{2}=(2,1,-3)$

Calculate the difference:

${\vec{a}}_{2}-{\vec{a}}_{1}=(-1-3,1+15,9-9)=(-4,16,0)$

Next, compute the cross product:

${\vec{b}}_{1}	imes {\vec{b}}_{2}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 2 \quad  -7 \quad  5 \ 2 \quad  1 \quad  -3 \end{vmatrix}=(16\hat{i}+16\hat{j}+16\hat{k})$

The magnitude of the cross product is:

$|{\vec{b}}_{1}	imes {\vec{b}}_{2}|=|(16\hat{i}+16\hat{j}+16\hat{k})|=16\sqrt{3}$

Calculate the dot product:

$({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})=(-4\hat{i}+16\hat{j}+0\hat{k})\cdot (16\hat{i}+16\hat{j}+16\hat{k})=-64+256+0=192$

Finally, find the shortest distance:

$d=\frac{|192|}{16\sqrt{3}}=\frac{192}{16\sqrt{3}}=4\sqrt{3}$

Thus, the shortest distance is:

$4\sqrt{3}$

Solutions · 1