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The foot of the perpendicular drawn from the point (1,8,4) on the line joining the point (0,-11,4) and (2,-3,1) is
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Verified Answer
The correct answer is:
(4,5,-2)
Equation of line joining points (0,-11,4) and (2,-3,1) is
$$
\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda
$$
Let $P$ is any point of the above line then coordinate of $P$ is $(2 \lambda+2,8 \lambda-3,-3 \lambda+1)$
$\therefore \mathrm{DR}^{\prime} \mathrm{s}$ of $P Q$ is $(2 \lambda+1,8 \lambda-11,-3 \lambda-3$
Now, $(2 \lambda+1)(4+(8 \lambda-11)(8)+(-3 \lambda-3) (-3)=0$
$\Rightarrow \quad 4 \lambda+2+64 \lambda-88+9 \lambda+9=0$
$77 \lambda-77=0$
$\therefore$ Required foot of perpendicular.
$$
P(4,5,-2)
$$
$$
\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda
$$
Let $P$ is any point of the above line then coordinate of $P$ is $(2 \lambda+2,8 \lambda-3,-3 \lambda+1)$
$\therefore \mathrm{DR}^{\prime} \mathrm{s}$ of $P Q$ is $(2 \lambda+1,8 \lambda-11,-3 \lambda-3$
Now, $(2 \lambda+1)(4+(8 \lambda-11)(8)+(-3 \lambda-3) (-3)=0$
$\Rightarrow \quad 4 \lambda+2+64 \lambda-88+9 \lambda+9=0$
$77 \lambda-77=0$
$\therefore$ Required foot of perpendicular.
$$
P(4,5,-2)
$$
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