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A symmetrical form of the line of intersection of the planes $x=a y+b$ and $z=c y+d$ is
Options:
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Verified Answer
The correct answer is:
$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$
$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$
Given two planes:
$$
x-a y-b=0 \text { and } c y-z+d=0
$$
Let, $l, m, n$ be the direction ratio of the required line. Since the required line is perpendicular to normal of both the plane, therefore $l-a m=0$ and $c m-n=0$ $\Rightarrow l-a m+0 . n=0$ and $0 . l+c m-n=0$
$$
\therefore \quad \frac{l}{a-0}=\frac{m}{0+1}=\frac{n}{c-0}
$$
Hence, d.R of the required line are $a, 1$, $c$.
Hence, options (c) and (d) are rejected. Now, the point $(a+b, 1, c+d)$ satisfy the equation of the two given planes.
$\therefore$ Option (b) is correct.
$$
x-a y-b=0 \text { and } c y-z+d=0
$$
Let, $l, m, n$ be the direction ratio of the required line. Since the required line is perpendicular to normal of both the plane, therefore $l-a m=0$ and $c m-n=0$ $\Rightarrow l-a m+0 . n=0$ and $0 . l+c m-n=0$
$$
\therefore \quad \frac{l}{a-0}=\frac{m}{0+1}=\frac{n}{c-0}
$$
Hence, d.R of the required line are $a, 1$, $c$.
Hence, options (c) and (d) are rejected. Now, the point $(a+b, 1, c+d)$ satisfy the equation of the two given planes.
$\therefore$ Option (b) is correct.
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