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If the three planes $x=5,2 x-5 a y+3 z-2=0$ and $3 b x+y-3 z=0$ contain a common line, then $(a, b)$ is equal to
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Verified Answer
The correct answer is:
$\left(\frac{1}{5},-\frac{8}{15}\right)$
$\left(\frac{1}{5},-\frac{8}{15}\right)$
Let the direction ratios of the common line be $\ell, m$ and $n$.
$$
\begin{aligned}
& \therefore \ell \times 1+\mathrm{m} \times 0+\mathrm{n} \times 0=0 \Rightarrow \ell=0 \\
& 2 \ell-5 m a+3 n=0 \Rightarrow 5 m a-3 n=0 \\
& 3 \ell b+m-3 n=0 \Rightarrow m-3 n=0
\end{aligned}
$$
Subtracting (3) from (1), we get $m(5 a-1)=0$
Now, value of $m$ can not be zero because if $m=0$ then $n=0$
$\Rightarrow \ell=m=n=0$ which is not possible.
Hence, $5 a-1=0 \Rightarrow a=\frac{1}{5}$
$$
\begin{aligned}
& \therefore \ell \times 1+\mathrm{m} \times 0+\mathrm{n} \times 0=0 \Rightarrow \ell=0 \\
& 2 \ell-5 m a+3 n=0 \Rightarrow 5 m a-3 n=0 \\
& 3 \ell b+m-3 n=0 \Rightarrow m-3 n=0
\end{aligned}
$$
Subtracting (3) from (1), we get $m(5 a-1)=0$
Now, value of $m$ can not be zero because if $m=0$ then $n=0$
$\Rightarrow \ell=m=n=0$ which is not possible.
Hence, $5 a-1=0 \Rightarrow a=\frac{1}{5}$
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