Question

The Cartesian equation of a line passing through the point with position vector `veca = hati-hatj  "and parallel to the line"  vecr = hati + hatk + µ(2hati - hatj)`, is ______.

Options

• `(x-2)/1=(y+1)/0 = z/1`

• `(x-1)/2=(y+1)/-1 = z/0`

• `(x-1)/2=(y+1)/-1 = z/0`

• `(x-1)/2=(y+1)/-1 = z/0`

Answer 1

The Cartesian equation of a line passing through the point with position vector `veca = hati-hatj  "and parallel to the line"  vecr = hati + hatk + µ(2hati - hatj)`, is `underlinebb((x-1)/2=(y+1)/-1 = z/0)`.

Explanation:

Here, `veca=hati-hatj` and

`vecr=hati+hatk+µ(2hati-hatj)`

Let `vecm` be a line passing through `veca` and parallel to `vecr`

⇒ `vecm` be a line passing through `veca` and parallel to `(2hati-hatj)=vecb`

So we know that a line through a point with position vector `veca and "parallel to"  vecb` is given by the equation.

`vecm = veca+λvecb`

⇒ `(xhati+yhatj+zhatk)=(hati-hatj)+λ(2hati-hatj)`

= `hati(1+2λ)+hatj(-1-λ)`

So its Cartesian equation is

`(x-1)/2=(y+1)/-1=(z-0)/0`