How to calculate the angle between vectors, and how to find position vectors

Question

Firstly how can I

a)     Find the value of k such 2ki – (k² – 4)j is perpendicular to 3i + 2j.

b)     Calculate the angle between the vectors 2i – 3j and i + 2j

And secondly

a)     Given two non-zero vectors a and b, how can I show that if a + b and a – b are perpendicular then |a|= |b|.

b)     If a = i – 3j, b = -2i + 5j, c = 3i – j.

i)      How can I find the position vector of the point P dividing the line AB in the ratio 3:4 where a is the position vector of A, c is the position vector of C.

ii)      How can I Find the unit vector perpendicular to (AB)^→

Answer 1

a) Utilize the vector dot product knowing the magnitude is abcos(phi) and cos(90)=0. Solve the resulting quadratic equation to find k = -1 and +4 yield two vectors normal to 3i+2j.

b) Again utilize the vector dot product and the magnitude of a vector as sqrt of sum of squares finding the angle to be 119.7 degrees

c) The dot product (a+b)dot(a-b)=adota-adotb+bdota -b.b=0. Note that adota=a^2 and bdotb=b^2 and since a is normal to b adotb and bdota =0

d) Find vector AB = <-2-1, 8-(-3)>=<-3,8>. Note magnitude of AB= sqrt(73). Note that P is to be .75 of the length along AB, thus ABx=.75*-3 and ABy=.75*8. P will be at <-9/4,6>.

e) Utilize <-9/4,6>dot<ux,uy>=0 and ux^2+uy^2=1 and solve for ux=1.079 and uy=.197

Answer 2

I. a) 6k - 2(k² - 4) = 0; k² - 3k - 4 = 0; k = - 1 or k = 4.

b) cosθ = (2 - 6)/(√13·√5) = - 4/√65; θ = cos^-1(- 4/√65) = 119.745°.

II. a) (a + b) • (a - b) = a² - b² = IaI² - IbI² = 0; IaI = IbI

b) A = (1, - 3), B = (- 2, 5) and C = (3, - 1);

i)Vector AB = < - 3, 8>. Let P = (x, y). Then <x - 1, y + 3> =

3/7<- 3, 8>. Then x - 1 = - 9/7, y + 3 = 24/7; x = - 2/7, y = 3/7 and P = (- 2/7, 3/7)

ii) Vector perpendicular to AB = < - 3, 8> is <8, 3>. So unit vector perpendicular to AB is 1/ √73<8, 3>