As we know that the vector equation of a line is of form r = a + mb where a is the position vector through which line is passing, b is a vector parallel to line and m is a constant.
If two equations of line
r = a1 + mb1
r = a2 + nb2 are given,
Then to find the distance between these lines first we have to find,
b1 x b2
a2 – a1
|b1 x b2|
On comparing the given equation with general equation, we get
\(a_1 = \hat i + \hat j\)
\(a_2 = 2\hat i + \hat j - \hat k\)
\(b_1 = 2\hat i - \hat j + \hat k\)
\(b_2 = 3\hat i - 5\hat j + 2\hat k\)
And, the distance between 2 lines is given by
\(D = \left|\frac{(b_1 \times b_2).(a_2 - a_1)}{|b_1 \times b_2|}\right|\)
So,
\(b_1 \times b_2 = \begin{vmatrix} \hat i & \hat j & \hat k\\2&-1&1\\3&-5&2\end{vmatrix}\)
On solving the determinant along row
\(= \hat i (-2 + 5) - \hat j(4- 3) + \hat k (-10 + 3)\)
\(= 3\hat i - \hat j - 7\hat k\)
⇒ \(a_2 - a_1 = 2\hat i - \hat i + \hat j - \hat j - \hat k\)
\(= \hat i - \hat k\)
⇒ \(|b_1 \times b_2| = \sqrt{(3)^2 + (-1)^2 + (-7)^2}\)
\(= \sqrt{9 + 1 + 49}\)
\(= \sqrt{59}\)
Now,
\(D = \left| \frac{(3\hat i - \hat j - 7\hat k).(\hat i - \hat k)}{\sqrt{59}}\right|\)
\(D = \left|\frac{3 + 7}{\sqrt {59}}\right|\)
\(= \frac{10}{\sqrt{59}}\) units
