Question 1203668: Let e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) ∈ R 3. Find all real numbers c ∈ R such that the angle between the vectors −e1 + 2e2 + ke3 and −e1 + ke2 + 2e3 is π/2 (they are orthogonal).
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
It appears there might be a typo.
I think the portion Find all real numbers c ∈ R should be Find all real numbers k ∈ R
e1 = (1,0,0)
e2 = (0,1,0)
e3 = (0,0,1)
v1 = -1*e1 + 2*e2 + k*e3
v1 = -1*(1,0,0) + 2*(0,1,0) + k*(0,0,1)
v1 = (-1,0,0) + (0,2,0) + (0,0,k)
v1 = (-1+0+0, 0+2+0, 0+0+k)
v1 = (-1,2,k)
Follow similar steps to find that
v2 = -1*e1 + k*e2 + 2*e3
v2 = (-1,k,2)
We want to have vectors v1 and v2 to be orthogonal.
In other words, we want the vectors to be perpendicular to each other.
This occurs if and only if the dot product of said vectors is 0.
v1 dot v2 = (-1,2,k) dot (-1,k,2)
v1 dot v2 = (-1)*(-1) + 2*k + k*2
v1 dot v2 = 1 + 2k + 2k
v1 dot v2 = 1 + 4k
1 + 4k = 0
4k = -1
k = -1/4 is the final answer
Answer by ikleyn(52780)  (Show Source):
You can put this solution on YOUR website! .
Let e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) ∈   . Find all real numbers  k ∈ R such that the angle between
the vectors −e1 + 2e2 + ke3 and −e1 + ke2 + 2e3 is π/2 (they are orthogonal).
~~~~~~~~~~~~~~~~~~~~~
The given vectors are (-1,2,k) and (-1,k,2), in coordinate form.
They are orthogonal (perpendicular) if and only if their scalar product is zero.
Find the scalar product of these vectors, using their coordinate forms.
The scalsr product is (-1)*(-1) + 2k + 2k = 1 + 4k.
The vectors are orthogonal in if and only if
1 + 4k = 0, or k =  . ANSWER
Solved.
|
|
|
