Question

Vector question?

Answer 1

See process below

Explanation:

i) ABC are in straight line, then are linearly dependents `vec(AB)= kvec(BC)`

Calculate `vec(AB)=(2-0,5-2,-2+3)=(2,3,1)` and
`vec(BC)=(1,p-5,q+2)`. Then

`(2,3,1)=k(1,p-5,q+2)=(k,kp-5k,kq+2k)` from here

`2=k`
`3=kp-5k=2p-10=` then `p=13/2`
`1=q+2` then `q=-1`

ii) If BAC is 90º, then `vec(AB)·vec(AC)=0`

`vec(AB)=(2,3,1)`
`vec(AC)=(3,p-2,q+3)`

`vec(AB)·vec(AC)=6+3p-6+q+3=3p+q+3=0` then

`q=-3-3p`

iii)If `p=3` the mod`vec(AB)=sqrt(4+9+1)=sqrt14`

`vec(AC)=(3,1,q+3)` then mod `vec(AC)=sqrt(9+1+(q+3)^2)`

Then squaring both sides `14=10+(q+3)^2=10+q^2+6q+4` or

`q^2+6q=q(q+6)=0` which give `q=0` and `q=-6`

Answer 2

Please see the explanation below.

Explanation:

`Question (i)`

If `ABC` is a straight line, this means that `A, B " and " C` are colinear then

`vec(AB)=kvec(BC)` where `k in RR`

`((2-0),(5-2),(-2-(-3)))=k*((3-2),(p-5),(q+2))`

`((2),(3),(1))=k*((1),(p-5),(q+2))`

`{(k=2),(3=2p-10),(1=2q+4):}`

`<=>`, `{(k=2),(p=13/2),(q=-3/2):}`

`Question (ii)`

If the angle `hat(BAC)=90^@`, then

The dot product is

`vec(AB).vec(AC)=0`

`((2-0),(5-2),(-2-(-3))).((3-0),(p-2),(q-(-3)))=0`

`((2),(3),(1)).((3),(p-2),(q+3))=0`

`2*3+3(p-2)+(q+3)=0`

`6+3p-6+q+3=0`

`3p+q=-3`

*`Question (iii)`*

`vec(AB)=((2),(3),(1))`

`vec(AC)=((3),(p-2),(q+3))`

If `AB=AC`, then

`sqrt(2^2+3^2+1^2)=sqrt(3^2+(p-2)^2+(q+3)^2)`

`14=9+(p-2)^2+(q+3)^2`

`(p-2)^2+(q+3)^2=5`

If `p=3`, then

`1+(q+3)^2=5`

`q^2+6q+10-5=0`

`q^2+6q+5=0`

Solving this quadratic equatio in `q`

`q=(-6+-sqrt(36-20))/(2)=(-6+-sqrt16)/2`

`=-3+-2`

The possible values of `q` are

`S={-1, -5}`