Find the vector equation of the line passing through the point (2, –1, –1) which is parallel to the line

The Cartesian equation of a line passing through a point (x₁, y₁, z₁) and having directional ratios proportional to a,b,c is given by,

The required line passes through the point (2,–1,–1), now we need to find the direction ratios of the line which are a,b,c . this equation of the required line is,

It is given that a line is parallel to the required line and has the Cartesian equation 6x–2 = 3y+1 = 2z–2, which can be further solved to it’s generalized form, which is ,

So we get the direction ratios as, = λ

a = 1λ, b = 2λ, c = 3λ

as we know that two parallel lines have their direction ratios, suppose a line has direction ratios a,b,c and the line parallel to this line will have direction ratios ka,kb,kc .

putting these values in the required line equation, we get,

To convert this Cartesian form to the vector equation form, first equate the Cartesian form to a scalar,

= λ

Now equate all parts to this scalar individually,

x–2 = λ, y+1 = 2λ, z+1 = 3λ

x = 2+λ, y = 2λ – 1, z = 3λ – 1

we know that = x+y+z = +λ

x+y+z = (2+λ) + (2λ – 1) + (3λ – 1)

x+y+z = (2–1–1) + λ(1+2+3)