# Take a point P in R3 and two non-zero vectors v and w. Consider the vector equation`Q = P + s(vec v) + t(vec w)` for any real s and t:Let S be the set of all such points Q. (c) Determine whether...

Take a point P in R3 and two non-zero vectors v and w. Consider the vector equation

`Q = P + s(vec v) + t(vec w)` for any real s and t:

Let S be the set of all such points Q.

(c) Determine whether the following vector equations give you a point, a line or a plane.

i. (x, y, z) = (1, 4, 2) + s< 4,-5, 1 > +t< -8, 10, 3 >

ii. (x, y, z) = (-1, 0, 1) + s< -5, 10, 5 > + t< 3,-6,-3 >

iii. (x, y, z) = (5, 1, 2) + s< 0, 0, 0 > + t< 0, 0, 0 >

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To determine wether set of ponts `(x,y,z)` is a point, a line or a plane all we need to do find wether the vectors `v` and `w` are linearly independent or not. If they are independent there cannot exist scalar `a ne 0` such that `av=w`

**i.** `a<<4,-5,1>>=<<-8,10,3>>` **(1)**

From this we get 3 equations:

`4a=-8`

`-5a=10`

`a=3`

From the last equation we see that `a=3` but that does not satisfy first or even second equation. So the equation (1) will be satisfied if `a=0.` Hence, vectors are independent and `sv+tw` is **a plane** and so is the set of points `(x,y,z).`

**ii.** `a<<-5,10,5>>=<<3,-6,-3>>`

From this we get 3 equations:

`-5a=3`

`10a=-6`

`5a=-3`

We can see that `a=-3/5` is solution to all 3 equations. So we have find non-zero scalar `a` such that `av=w` hence set of points `(x,y,z)` **is a line**.

**iii.** In this case `v` and `w` are both null vectors so they can't be independent. Also no matter what `s` and `t` do we choose we will always get null vector so the set of points `(x,y,z)` ia actually only **one point**. To be precise `(x,y,z)=(5,1,2)`.