Prove that the lines x - 2/1 = y - 4/4 = z - 6/7 and x + 1/3 = y + 3/5 = z + 5/7 are coplanar. Also find the equation of the plane

Question

Prove that the lines \(\frac{x-2}1\) = \(\frac{y-4}4\) = \(\frac{z-6}7\) and \(\frac{x+1}3\) = \(\frac{y+3}5\) = \(\frac{z+5}7\) are coplanar. Also find the equation of the plane containing these lines.

Answer 1

Given : Equations of lines –

Line 1 :\(\frac{x-2}1\) = \(\frac{y-4}4\) = \(\frac{z-6}7\) 

Line 2 :  = \(\frac{y+3}5\) = \(\frac{z+5}7\)

 To Prove : Line 1 & line 2 are coplanar. 

To Find : Equation of plane. 

Formulae : 1) Coplanarity of two lines : 

If two lines are given by,

2) Equation of plane : 

The equation of plane containing two coplanar lines \(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\) = \(\frac{z-z_1}{c_1}\)

Answer : 

Given lines –

Line 1 :\(\frac{x-2}1\) = \(\frac{y-4}4\) = \(\frac{z-6}7\) 

Line 2 :  = \(\frac{y+3}5\) = \(\frac{z+5}7\)

Here, x₁ = 2 , y₁ = 4 , z₁ = 6 , a₁ = 1 , b₁ = 4 , c₁ = 7 

x₂ = -1 , y₂ = -3 , z₂ = -5 , a₂ = 3 , b₂ = 5 , c₂ = 7 

Now,

Hence, given two lines are coplanar. 

Equation of plane passing through line 1 and line 2 is given by

7x + 14 + 14y - 56 – 7z + 42 = 0 

- 7x + 14y – 7z = 0 

x – 2y + z = 0 

Therefore, equation of plane is 

x – 2y + z = 0