I have a question about vectors if anyone has an idea.?

Question

I have started learning about vectors for A level and cant really remember anything about GCSE math.

I have to find an angle of 2 vectors. I have done the equation and am left with -1 / 6. The example says the answer is 99.6 and I am trying to figure out why?

here is the example

a = 2i + j - k b = i -j + 2k

cos = (2) (1) + (1) (-1) + (-1) (2)
(4 + 1 +1) (1 + 1 +4)

cos = -1 / 6
cos = 99.6

Answer 1

how were you shown how to find the angle between two vectors? the most obvious way would be to use the scalar product. this is defined as being for two vectors a & b :-

a . b = |a| |b| cos(theta)

here |a|,|b| denote the magnitudes of vectors a and b, and theta is the angle between them.

you can find the scalar product of two vectors in cartsian form (i,j,k, unit vector form) from :-

a . b = a1b1 + a2b2 + a3b3

if a = a1i + a2j + a3k
b = b1i + b2j + b3k


now firstly finding a.b and then |a| and |b| will allow you to easily find cos(theta), from which you can find theta from the Arccosine.

hope this helps

Answer 2

Find the angle θ between the vectors:

a = 2i + j - k = <2, 1, -1>
b = i - j + 2k = <1, -1, 2>

First calculate the dot product of the vectors and the magnitude of each vector individually.

a • b = <2, 1, -1> • <1, -1, 2> = 2 - 1 - 2 = -1

|| a || = √[2² + 1² + (-1)²] = √(4 + 1 + 1) = √6
|| b || = √[1² + (-1)² + 2²] = √(1 + 1 + 4) = √6

Now calculate the dot product using another definition.

a • b = || a || || b || cosθ

cosθ = (a • b) / (|| a || || b ||) = -1 / [(√6) (√6)] = -1/6

θ = arccos(-1/6) ≈ 99.6°

Answer 3

yeah, before i start, i will advise u to get a textbook on vectors or visit net regularly and read up these issues, they are very simplified on the net. your question is so simple. it goes like this

to find an angle bet. 2 vectors, the formula is
A.B divide by the magnitude of A times the magnitude of B

cos theta=a.b divide sq rt of a x sq rt of b

in your case
a= 2i + j - k
b= i - j + 2k
a.b=2-1-2 (multiply each component)= -1

magnitude of a=sq rt 2 squared+1 squared+ 1 squared=square rt 6
mag of b=sq rt 1squared+1squared +2 squared =sqare rt 6

now, sq rt of 6 =2.44
i.e 2.44 x2.44=5.95 (denominator) remember sq rt of a x sq rt of b
cos theta= -1/5.95 = -0.16 (remember numerator is -1)

now inverse of cos -0.16=99.6 degrees

this clearly shows that the two vetors are parallel ( in case you are ask to prove if it is parallel or peperndicular) 2 vectors are perpendicular if their dot product is 0 and 2 vectors are parall if thier cross product is 0, and 1 of the vector MUST be a scalar multiple of the other and the angle between them MUST be either 1 or -1 making the cos to be between 0 and 180 degs. here since the cos is negative , it shows that they point in opposite direction and therefore obey LEFT- HAND RULE. vectors are interesting topics in maths and sciences, pls read them up for more understanding. cheers

Answer 4

the formula for finding the angle is
cos A = [scalar product of the 2 vectors]/[the product of the magnitudes of the 2 vectors]
if your vectors are [ x i + y j ] and [ x' i + y'j] then their scalar product is x . x' +y . y'
the magnitude of a vector is the square root of the sum of the squares of its components along the x axis and y axis
firstly check your calculation
secondly you have probably found the answer in radians- convert it into degrees.

Answer 5

what are the vectors?