Question

find the equations of the line though the point (1,2,4) parallel to the line 3x+2y-z=4 , x-2y-2z=5​

Answer 1

Answer The equation of the plane parallel to the lines

Step-by-step explanation:

pls see this

c994344508830ad494c3d598feea94fd.docx

Answer 2

❂Solution:-

The given line is the intersection of planes

3x+2y-z =4. ——( 1 )

and x-2y-2z=5. —–( 2 )

Let <a,b,c> be the direction number of this line , then

3a+2b-c=0

a-2b-2c=0. (Lines lies in both planes (1) and (2) )

$\sf \longrightarrow \: \dfrac{a}{ - 4 - 2} = \dfrac{b}{ - 1 + 6} = \dfrac{c}{ - 6 - 2} \\ \\ \sf \longrightarrow \: \dfrac{a}{ - 6} = \dfrac{b}{ 5} = \dfrac{c}{ -8} \\ \\ \sf \longrightarrow \: \dfrac{a}{ 6} = \dfrac{b}{ - 5} = \dfrac{c}{ 8} \mapsto \: a : b :c : : 6 : - 5 :8$

Since ,the line to be found is parallel to the above line passes through the point (1,2,4),therefore, its equations are

$\large \boxed { \bf \: \dfrac{x - 1}{6} = \dfrac{y - 2}{ - 5} = \dfrac{z - 4}{8} }$