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Standard XII

Mathematics

Equation of Line: Symmetrical Form

A line whose ...

Question

A line whose direction cosines are proportional to $2, 1, 2$ , meets with the lines $x = y + a = z$ and $x + a = 2 y = 2 z$ at $A$ and $B$ respectively. If the distance between $A$ and $B$ is $d$ , then $\frac{12 d}{| a |}$ is

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Solution

For $A$ ,
let $x = y + a = z = t$
$\Rightarrow A \equiv (t, t - a, t)$
For $B$ ,
let $\frac{x + a}{2} = y = z = k$
$\Rightarrow B \equiv (2 k - a, k, k)$

Now, the direction ratios of $A B$ are $(t + a - 2 k, t - a - k, t - k)$
Since, direction ratios are parallel to $(2, 1, 2),$
$\frac{t + a - 2 k}{2} = \frac{t - a - k}{1} = \frac{t - k}{2}$
$\Rightarrow t = 3 a, k = a$
$∴ A \equiv (3 a, 2 a, 3 a), B \equiv (a, a, a)$
$\Rightarrow d = 3 | a |$

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Equation of Line: Symmetrical Form

Standard XII Mathematics

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A line whose direction cosines are proportional to 2,1,2, meets with the lines x=y+a=z and x+a=2y=2z at A and B respectively. If the distance between A and B is d, then 12d|a| is

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A line whose direction cosines are proportional to $2, 1, 2$ , meets with the lines $x = y + a = z$ and $x + a = 2 y = 2 z$ at $A$ and $B$ respectively. If the distance between $A$ and $B$ is $d$ , then $\frac{12 d}{| a |}$ is