The line of shortest distance will be along \(\vec {b_1}\)×\(\vec{b_2}\)
Where,
\(\vec {b_1}\)=\(\hat j\)+\(\hat k\)
and
\(\vec {b_2}\)=2\(\hat i\)+2\(\hat j\)+\(\hat k\)
\(\vec {b_1}\)×\(\vec {b_2}\)=\(\begin{vmatrix} \hat i &\hat j &\hat k \\ 0&0 &1 \\ 2&2 &1 \end{vmatrix}\)=−\(\hat i\)+2\(\hat j\)−2\(\hat k\)
Angle between \(\vec {b_1}\)×\(\vec {b_2}\) and plane P,
sinθ=|\(\frac{-a-2+2}{3.\sqrt{a^2+2}}\)|=\(\frac{5}{\sqrt{27}}\)
⇒\(\frac{|a|}{\sqrt{a^2+2}}\)=\(\frac{5}{\sqrt3}\)
⇒ a2=-\(\frac{25}{11}\)(not possible)
Formula to find distance between two parallel line:
Consider two parallel lines are shown in the following form :
\(y = mx + c_1\) …(i)
\(y = mx + c_2\) ….(ii)
Here, m = slope of line
Then, the formula for shortest distance can be written as given below:
\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)
If the equations of two parallel lines are demonstrated in the following way :
\(ax + by + d_1 = 0\)
\(ax + by + d_2 = 0\)
then there is a little change in the formula.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)