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13. If a line $\mathrm{L}$ is the line of intersection of the planes $2 x+3 y+z=1$ and $x+3 y+2 z=2$. If line $\mathrm{L}$ makes an angle $\alpha$ with the positive $\mathrm{X}$-axis, then the value of $\sec \alpha$ is
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The correct answer is:
$\sqrt{3}$
Given equations:
$\begin{aligned}
& 2 x+3 y+z=1 \\
& 2 x+3 y=1-z ... (i)\\
& x+3 y+2 z=2 \\
& x+3 y=2-2 z ... (ii)
\end{aligned}$
Subtracting (ii) from (i), we get
$\begin{aligned}
& 2 x+3 y-x-3 y=1-z-2+2 z \\
& x=-1+z \\
& z=\frac{x+1}{1} ... (iii)
\end{aligned}$
Putting value of $x$ in equation (ii), we get
$\begin{aligned}
& -1+z+3 y=2-2 z \\
& 3 z=3-3 y \\
& z=\frac{y-1}{-1} ... (iv)
\end{aligned}$
From (iii), (iv)
$\frac{x+1}{1}=\frac{y-1}{-1}=\frac{z}{1}$
Thus, angle between above line and $\mathrm{X}$-axis having Direction Ratio's $(1,0,0)$ is given as
$\begin{aligned}
\cos \alpha & =\left|\frac{1 \cdot(1)+0+0}{\sqrt{1+1+1} \cdot \sqrt{1}}\right|=\frac{1}{\sqrt{3}} \\
\therefore \quad \sec \alpha & =\sqrt{3}
\end{aligned}$
$\begin{aligned}
& 2 x+3 y+z=1 \\
& 2 x+3 y=1-z ... (i)\\
& x+3 y+2 z=2 \\
& x+3 y=2-2 z ... (ii)
\end{aligned}$
Subtracting (ii) from (i), we get
$\begin{aligned}
& 2 x+3 y-x-3 y=1-z-2+2 z \\
& x=-1+z \\
& z=\frac{x+1}{1} ... (iii)
\end{aligned}$
Putting value of $x$ in equation (ii), we get
$\begin{aligned}
& -1+z+3 y=2-2 z \\
& 3 z=3-3 y \\
& z=\frac{y-1}{-1} ... (iv)
\end{aligned}$
From (iii), (iv)
$\frac{x+1}{1}=\frac{y-1}{-1}=\frac{z}{1}$
Thus, angle between above line and $\mathrm{X}$-axis having Direction Ratio's $(1,0,0)$ is given as
$\begin{aligned}
\cos \alpha & =\left|\frac{1 \cdot(1)+0+0}{\sqrt{1+1+1} \cdot \sqrt{1}}\right|=\frac{1}{\sqrt{3}} \\
\therefore \quad \sec \alpha & =\sqrt{3}
\end{aligned}$
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