Question

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THREE DIMENSIONAL GEOMETRY. ​

Answer 1

$\bold {\huge {Your ~answer :-}}$

$\frac{x - 7}{164} = \frac{y - 4}{31} = \frac{z + 1}{134}$

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EXPLAINATION ➣

Given -

A(0, 6, -9) & B(-3, -6, 3)

We can write $\green{\texttt{equation of line}}$

$\frac{x}{3} = \frac{y - 6}{12} = \frac{z + 9}{ - 6}$

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Let (3r, 12r + 2, - 6r - 8) be the $\green{\texttt{foot of perpendicular}}$

from C(7, 4, - 1)

→Therefore, (3r - 7, 12r + 2, - 6r + 8) is perpendicular to AB

⇒(3r-7)3 + 12(12r +2) - 6(-6r-8) =0

On solving

⇒r = - 17/63

→Therefore the cordinates of point D

=(-17/21, 53/21, - 155/21)

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→$\green{\texttt{Equation of line CD }}$

$\frac{x - 7}{7 + \frac{17}{21} } = \frac{y - 4}{4 - \frac{53}{21} } = \frac{z + 1}{ - 1 + \frac{155}{21} } \\ further \: solving\\ = > \frac{x - 7}{164} = \frac{y - 4}{31} = \frac{z + 1}{134}$

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