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If the distance between planes, $4 x-2 y-4 z+1=$ 0 and $4 x-2 y-4 z+d=0$ is 7 , then $d$ is:
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Verified Answer
The correct answer is:
$-41$ or 43
$-41$ or 43
Given planes are
$$
4 x-2 y-4 z+1=0
$$
and $4 x-2 y-4 z+d=0$
They are parallel.
Distance between them is $\pm 7$
$=\frac{d-1}{\sqrt{16+4+16}}$
$$
\begin{aligned}
&\Rightarrow \frac{d-1}{6}=\pm 7 \\
&\Rightarrow d=42+1 \\
&\text { or }-42+1 \\
&\text { i.e. } d=-41 \text { or } 43 .
\end{aligned}
$$
$$
4 x-2 y-4 z+1=0
$$
and $4 x-2 y-4 z+d=0$
They are parallel.
Distance between them is $\pm 7$
$=\frac{d-1}{\sqrt{16+4+16}}$
$$
\begin{aligned}
&\Rightarrow \frac{d-1}{6}=\pm 7 \\
&\Rightarrow d=42+1 \\
&\text { or }-42+1 \\
&\text { i.e. } d=-41 \text { or } 43 .
\end{aligned}
$$
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