Vector Path

This article is about vector path. Vector path is a outline which represents the vector lines instead of dots.  Vector path are independent of resolution. Path created by vector shapes are also obtained. Vector is mentioned with magnitude and dimensions only.  Vector path can be helped by the tutor vista website in online at any time. Vector path is simply a path that are created by the vectors. Online tutors of tutor vista are highly qualified tutors to help the students any time. Below we can see more about the vector path.

vector path

Problem 1: Find a unit vector normal to the plane `vecr.(2hati - 3 hatj + 6 hatk)` + 14 = 0
Solution
The equation of the plane is `vecr.(2hati - 3 hatj + 6 hatk)` + 14 = 0
`vecr.(2hati - 3 hatj + 6 hatk)` = - 14
`vecr.(-2hati + 3 hatj - 6 hatk)` = 14
`vecr.vecn = 14` where `vecn = (-2hati + 3 hatj - 6 hatk)`
`vecr. vecn/|vecn| = 14/|vecn|` ,
where `|vecn| = sqrt ((-2)^2 + 3^2 +(-6)^2`    = 7
`vecr. (-2hati + 3 hatj - 6 hatk)/7` = `14/7`
`vecr. (-2hati + 3 hatj - 6 hatk)/7`=2
Hence the unit vector` hatn = (-2/7hati + 3/7 hatj - 6/7 hatk)`

Problem2: The vector equation of a plane is `vecr.(2 hati-hatj+2hatk)` = 9. Reduace it to the normal vector .
Solution
The equation of the plane is `vecr.vecn` = 9 where `vecn` = `(2 hati-hatj+2hatk)`
`|vecn| = sqrt ((2^2) +(-1)+(2^2))`   =3
`vecr. (2hati - hatj +2hatk)/3 = 3 `
`vecr. (2/3 hati - 1/3 hatj + 2/3 hatk ) = 3. `
Now `vecr . vecn` = 9
`vecr . vecn/|vecn| = 9 /|vecn|`
`vecr. (2hati - hatj + 2 hatk)/3 = 3`
`vecr . (2/3 hati - 1/3 hatj +2/3 hatk) = 3`

vector path

Direction vector
       If the modulus is unity then it is a unit vector. The direction vector `hata` =`|hata|`= 1. The unit vectors parallel to `veca` are `+-hata`
Result:  ` veca = |veca|hata`         [i.e. any vector = (its modulus) x ( unit vector in that direction)]
 ` hata = veca/|veca| ; (veca vecO)`
Direction vector = `"(vector in any direction)"/"(modulus of the vector)"`
3.Find the unit vector in the direction of `3vecp+4vecq -12 vecr`
Solution
Let `vecx = 3vecp+4vecq -12 vecr`
    `|vecx| = |vecp+4vecy -12 vecz|`
               `= sqrt ((3)^2 + (4)^2 + (-12)^2`
              ` = sqrt (9+16+144)`
               `= sqrt (169)`
              = 13
Unit vector of `vecx` is  = `vecx / |vecx|`
                                                                  = `(3vecp+4vecq -12 vecr)/13`

4.Find the unit vectors parallel to the vector` -3vecc + 4 vecd`
Solution
Let `veca = -3vecc + 4 vecd`
       `|veca| = |-3vecc + 4 vecd|`
                 ` =sqrt ((-3)^2 + 4^2)`
                  `= sqrt (9+16)`
                  `= sqrt (25)`
                  = 5
Unit vector =` veca/|veca|`
                    = `1/5 (-3vecc + 4 vecd)`
Units vectors parallel to `veca ` are `+-((-3)/5 vecc + 4/5vecd)`