Question:

If the line of intersection of the planes \(2x+3y+z=1\) and \(x+3y+2z=2\) makes an angle \( \alpha \) with the positive x-axis, then \( \cos \alpha = \)

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The line of intersection of two planes \( \vec{r} \cdot \vec{n_1} = d_1 \) and \( \vec{r} \cdot \vec{n_2} = d_2 \) has a direction vector parallel to \( \vec{n_1} \times \vec{n_2} \). If the direction ratios of a line are (a,b,c), its direction cosines are \( (\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}) \). If \( l,m,n \) are direction cosines, \( l = \cos \alpha \), \( m = \cos \beta \), \( n = \cos \gamma \), where \( \alpha, \beta, \gamma \) are angles with positive x,y,z axes respectively.
Updated On: Jun 5, 2025
  • \( \frac{1}{\sqrt{3}} \)
  • \( \frac{1}{\sqrt{2}} \)
  • \( \frac{1}{2} \)
  • \( \frac{\sqrt{3}}{2} \)
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The Correct Option is A

Solution and Explanation

The direction vector of the line of intersection of two planes is perpendicular to the normal vectors of both planes.
Normal vector of plane 1 (\(P_1: 2x+3y+z=1\)) is \( \vec{n_1} = (2,3,1) \).
Normal vector of plane 2 (\(P_2: x+3y+2z=2\)) is \( \vec{n_2} = (1,3,2) \).
The direction vector \( \vec{d} \) of the line of intersection is \( \vec{d} = \vec{n_1} \times \vec{n_2} \).
\[ \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \vec{i}(3 \cdot 2 - 1 \cdot 3) - \vec{j}(2 \cdot 2 - 1 \cdot 1) + \vec{k}(2 \cdot 3 - 3 \cdot 1) \] \[ = \vec{i}(6-3) - \vec{j}(4-1) + \vec{k}(6-3) = 3\vec{i} - 3\vec{j} + 3\vec{k} \] So the direction ratios of the line are (3, -3, 3), or equivalently (1, -1, 1).
Let \( \vec{d'} = (1,-1,1) \).
The angle \( \alpha \) is the angle this line makes with the positive x-axis.
The direction cosines (l,m,n) of the line are: \( l = \frac{1}{\sqrt{1^2+(-1)^2+1^2}} = \frac{1}{\sqrt{3}} \) \( m = \frac{-1}{\sqrt{1^2+(-1)^2+1^2}} = \frac{-1}{\sqrt{3}} \) \( n = \frac{1}{\sqrt{1^2+(-1)^2+1^2}} = \frac{1}{\sqrt{3}} \) The cosine of the angle made with the positive x-axis is \( l \).
So, \( \cos \alpha = l = \frac{1}{\sqrt{3}} \).
This matches option (1).
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