Question

Find the vector and Cartesian equation of the line passing through (1,-2,1) and perpendicular to the lines x+3=2y=-12z and x/2=y+6/2=3z-9/1.

Answer 1

Answer:

(x-1)/2 =(y+2)/-2 =(z-2)/12

Step-by-step explanation:

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Answer 2

Answer:-

$\frac{x-1}{2}\ =\ \frac{y+2}{-3}\ =\ \frac{z-1}{6}$

Step-by-step explanation:

In this question,

We have been given that,

Line passing through (1,-2,1) is perpendicular to x+3 = 2y = -12z and $\frac{x}{2}\ =\ \frac{y+6}{2}\ =\ \frac{3z-9}{1}$

Let L,M,N be the direction of the line passing through (1,-2,1)

Direction cosines of the line $\frac{x+3}{1}\ =\ \frac{y}{\frac{1}{2}}\ =\ \frac{z}{-\frac{1}{12}}$ is ($1,\frac{1}{2},\frac{-1}{12}$)

Direction cosines of the line $\frac{x}{2}\ =\ \frac{y+6}{2}\ =\ \frac{z-3}{\frac{1}{3}}\ is\ (2,2,\frac{1}{3}$)

Since lines are perpendicular product of direction cosines will be 0

Therefore, $L\ +\ \frac{M}{2}\ -\ \frac{N}{12}\ =\ 0$ and,

                  $2L\ +\ \2M\ +\ \frac{N}{3}\ =\ 0$

On solving we get,

$\frac{L}{12}\ +\ \frac{M}{-18}\ +\ \frac{N}{36}$

Simplifying we get,

$\frac{L}{2}\ +\ \frac{M}{-3}\ +\ \frac{N}{6}$

Therefore equation of the line passing through will be (1,-2,1) will be

$\frac{x-1}{2}\ =\ \frac{y+2}{-3}\ =\ \frac{z-1}{6}$