Question

Let L be the line of intersection of the planes \[2x + 3y + z = 1\] and \[x + 3y + 2z = 2,\] if L makes an angle $ \alpha $ with the positive x-axis, then cos alpha equals to
A) $ 1 $
B) $\dfrac{1}{{\sqrt 2 }} $
C) $\dfrac{1}{{\sqrt 3 }} $
D) $\dfrac{1}{2} $

Answer 1

Hint: To solve this question, we will first evaluate the normal vector of the equation of both the planes, then we will find the direction vector of the line, by doing cross multiplication. Now, by using the condition that L makes an angle $ \alpha $ with the positive x-axis, we will find the value of cos alpha by applying the formula, hence, we will get our required answer.

Complete step by step solution:
We have been given that the L is the line of intersection of the planes \[2x + 3y + z = 1\] and \[x + 3y + 2z = 2,\] it is given that L makes an angle $\alpha $ with the positive x-axis, we need to find the value of \[cos\alpha .\]
So, the two given planes are \[2x + 3y + z = 1\] and \[x + 3y + 2z = 2.\]
The normal vector of first plane is, \[\overrightarrow {{n_1}} = 2\widehat i + 3\widehat j + \hat k\] and the normal vector of second plane is, $\overrightarrow {{n_2}} = \widehat i + 3\widehat j + 2\hat k$
We know that, the direction vector of line, $\overrightarrow a = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $
So, on doing cross multiplication we get,
$
  \overrightarrow a = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  2&3&1 \\
  1&3&2
\end{array}} \right| \\
   = \widehat i(6 - 3) - \widehat j(4 - 1) + \hat k(6 - 3) \\
   = 3\widehat i - 3\widehat j + 3\hat k......(1) \\
$
Now, it is given that, line L makes an angle alpha with the positive x-axis thus, \[cos\alpha = \dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }}\]
From equation (1) we get that, l is the component of i, i.e., \[l = 3,\] m is the component of j, i.e., \[m = 3,\] and n is the component of k, i.e., \[n = 3.\]
On putting the values of \[l = 3,{\text{ }}m = 3\] and \[n = 3,\] in the formula above, we get
$
  \cos \alpha = \dfrac{3}{{\sqrt {{{(3)}^2} + {{(3)}^2} + {{(3)}^2}} }} \\
   \Rightarrow \cos \alpha = \dfrac{3}{{3\sqrt 3 }} \\
   \Rightarrow \cos \alpha = \dfrac{1}{{\sqrt 3 }} \\
 $
So, the value of \[cos\alpha \] is $\dfrac{1}{{\sqrt 3 }}.$ Thus, option (C) $\dfrac{1}{{\sqrt 3 }}$, is correct.
So, the correct answer is “Option C”.

Note: Students should note that the cross product of vectors always gives a vector answer, and is also called the vector product, but the dot product of vectors gives a scalar answer, and is also called the scalar product.