Browse Questions

# Find the shortest distance between the lines : $\overrightarrow r = 6\hat i + 2 \hat j +2\hat k + \lambda(\hat i-2\hat j+2\hat k)\: and \: \overrightarrow r = -4\hat i -\hat k + \mu(3\hat i-2\hat j-2\hat k)$

Toolbox:
• The shortest distance between two lines is
• $d=\begin{vmatrix}\large\frac{(\overrightarrow b_1\times \overrightarrow b_2).(\overrightarrow a_2-\overrightarrow a_1)}{\overrightarrow b_1\times \overrightarrow b_2}\end{vmatrix}$
Step 1:
Let the given lines be
$\overrightarrow r=6\hat i+2\hat j+2\hat k+\lambda(\hat i-2\hat j+2\hat k)$----(1)
$\overrightarrow r=-4\hat i+\hat k+\mu(3\hat i-2\hat j-2\hat k)$----(2)
We know that the shortest distance between two lines,$\overrightarrow r=\overrightarrow a_1+\lambda \overrightarrow b_1$ and $\overrightarrow r=\overrightarrow a_2+\lambda \overrightarrow b_2$ is given by
$d=\begin{vmatrix}\large\frac{(\overrightarrow b_1\times \overrightarrow b_2).(\overrightarrow a_2-\overrightarrow a_1)}{\overrightarrow b_1\times \overrightarrow b_2}\end{vmatrix}$
Step 2:
From equ(1) and equ(2) we can determine,
$\overrightarrow a_1=6\hat i+2\hat j+2\hat k$
$\overrightarrow b_1=\hat i-2\hat j+2\hat k$
$\overrightarrow a_2=-4\hat i-\hat k$
$\overrightarrow b_2=3\hat i-2\hat j-2\hat k$
Let us determine $(\overrightarrow a_2-\overrightarrow a_1)$
$(\overrightarrow a_2-\overrightarrow a_1)=(-4\hat i-\hat k)-(6\hat i+2\hat j+2\hat k)$
$\qquad\qquad\;\;=-10\hat i-2\hat j-3\hat k$
Step 3:
Next let us determine $\overrightarrow b_1\times \overrightarrow b_2=\begin{vmatrix}\hat i &\hat j&\hat k\\1 & -2 & 2\\3 &-2 & -2\end{vmatrix}$
On expanding we get,
$\hat i(4+4)-\hat j(-2-6)+\hat k(-2+6)$
$\Rightarrow 8\hat i+8\hat j+4\hat k$
$\overrightarrow b_1\times \overrightarrow b_2=8\hat i+8\hat j+4\hat k$
Step 4:
$\mid \overrightarrow b_1\times \overrightarrow b_2\mid=\sqrt{8^2+8^2+4^2}$
$\qquad\qquad\;\;=\sqrt{64+64+16}$
$\qquad\qquad\;\;=\sqrt{144}$
$\qquad\qquad\;\;=12$
Step 5:
Now substituting the respective value we get
$d=\begin{vmatrix}\large\frac{(8\hat i+8\hat j+4\hat k).(-10\hat i-2\hat j-3\hat k)}{12}\end{vmatrix}$
$d=\begin{vmatrix}\large\frac{-80-16-12}{12}\end{vmatrix}$
$\;\;\;=\large\frac{108}{12}$
$\;\;\;=9$ units.
Hence the shortest distance between the given two lines is $9$ units.