Vector question?

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2 Answers
May 8, 2018

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 modvec(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

May 8, 2018

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}