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Step-by-Step Solution
Step 1: Identify the normals of the given planes
The two planes are given by:
1) $2x + 3y + z = 1$
2) $x + 3y + 2z = 2$
• The normal vector to the first plane is $ \langle 2,\,3,\,1 \rangle.$
• The normal vector to the second plane is $ \langle 1,\,3,\,2 \rangle.$
Step 2: Find the direction of the line of intersection
The direction vector of the line of intersection, $ \vec{d},$ is perpendicular to both normals. Hence we can find $ \vec{d}$ using the cross product:
$ \vec{d} = \langle 2,\,3,\,1 \rangle \times \langle 1,\,3,\,2 \rangle $
Compute the determinant:
$
\vec{d}
= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 1 \\
1 & 3 & 2
\end{vmatrix}
= \mathbf{i}\,(3 \cdot 2 - 1 \cdot 3) \;-\; \mathbf{j}\,(2 \cdot 2 - 1 \cdot 1)\;+\; \mathbf{k}\,(2 \cdot 3 - 3 \cdot 1).
$
$ = \mathbf{i}\,(6 - 3)\;-\;\mathbf{j}\,(4 - 1)\;+\;\mathbf{k}\,(6 - 3) $
$ = 3\,\mathbf{i} \;-\; 3\,\mathbf{j} \;+\; 3\,\mathbf{k} $
Thus, $ \vec{d} = \langle 3,\, -3,\, 3 \rangle. $
Step 3: Determine the direction cosines
The direction ratios are $ (3,\, -3,\, 3). $ The magnitude of $ \vec{d} $ is:
$
\|\vec{d}\| = \sqrt{3^2 + (-3)^2 + 3^2} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}.
$
Hence the direction cosines $ (l,\,m,\,n) $ are found by dividing each component by the magnitude:
$
l = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}, \quad
m = \frac{-3}{3\sqrt{3}} = -\frac{1}{\sqrt{3}}, \quad
n = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}.
$
Step 4: Find $ \cos\alpha $ with respect to the positive $x$-axis
If $ \alpha $ is the angle the line makes with the positive $x$-axis, then
$ \cos\alpha = l. $
From Step 3, we have
$ l = \frac{1}{\sqrt{3}}. $
Therefore,
$
\cos \alpha = \frac{1}{\sqrt{3}}.
$
Final Answer
$ \cos \alpha = \frac{1}{\sqrt{3}}.
$
