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have to solve it by dot products in calc 3
Mertsj has already figured out the lengths of these vectors
yea but how do you go about to dove this? i have no clue? i am trying all kinds of different things
First, let me fix it. we want PQR \[(\overrightarrow P -\overrightarrow Q)\cdot (\overrightarrow R -\overrightarrow Q)= |\overrightarrow P -\overrightarrow Q||\overrightarrow R-\overrightarrow Q|\cos\theta\]
I assume you know how to find P-Q and R-Q : subtract corresponding elements I'll do P-Q (1,-4) - (2,7)= (1-2, -4-7)= (-1, -11)
the length of a vector (x,y) = | (x,y)| = sqrt(x*x + y*y)
a dot product of (x,y) and (a,b)= ax+by
Did you find R-Q ?
yea its (-4,-5)
Did you find the magnitude of R-Q (and P-Q) ?
you just square them right?
the length of a vector (x,y) = | (x,y)| = sqrt(x*x + y*y) that means you take the first number , square it, square the 2nd number, add together then take the square root
p-q = sqrt (121) r-q = sqrt (41)
double check p-q
oh its sqrt (122)
ok , now the dot product of (p-q) dot (r-q) (-1,-11) dot (-4, -5)
dot product of (x,y) and (a,b)= ax+by
so its (-1*-4 + -11 * -5)
yes, but that is just a number=?
it = 59?
yes. now put it all together (p-q) dot (r-q) = |p-q| | r-q| cos A we now everything except A.
but that will only give us one angle right?
This gives us the angle formed by p-q and r-q p-q is a vector with its base at q. same for r-q |dw:1328106371363:dw|
yea but we need three angles angle P=? , angle Q=? , and angle R =?.
find the approximate measurements of angle PQR this means the angle from P to Q to R in other words, Q is the vertex . Just 1 angle
this is problem 60 in the book. problem 61 is the same question but P: (0, -1, 3), Q = (2,2,1), R= (-2, 2, 4) and answer given in the back is p=78.8 degrees , q= 47.2 degrees and r = 54.0 degree
thats's why i am confused
First, what did you get for angle pqr?
i can't solve it. sorry i am really bad at it
so its 59 = sqrt (122) * sqrt (41) cos a
the contraction of "it is" is "it's" (sorry , pet peeve!)
to find cos a, divide both sides of the equation by (sqrt(122)*sqrt(41))
lol sorry i am nervous right now
i get 0.83
you can type acos(59/(sqrt(122)*sqrt(41))) in degrees in the google search window to find the angle or use a calculator.
i get 0.58
that's in radians. I would do 2 things. first, use a few more digits in your answer to 59/(sqrt(122)*sqrt(41)) (because your book is finding the answer rounded to tenths of a degree) then use degree mode with the calculator or type in degrees when using google
oh k i get 33.5 degrees
good, matches what I got.
but thats just one angle. how do i find the other ones
If you want the angle with vertex P, which we would name as angle QPR (or angle RPQ), form the vectors (q-p) and (r-p) and do the same thing: (q-p) dot (r-p) = |(q-p)| |(r-p)| cos P
would you help me with other question that i posted?
and to find the third angle, i could add two angles and make it equal to 180 right?
btw, notice that (q-p) = -(p-q), so just negate each element in your (p-q) vector , which you already know. And of course the length of (q-p) = length (p-q)
Yes, the third angle = 180 - sum(other two) but it doesn't hurt to do the dot product.
as a check, and to get comfortable with the procedure
yes i need practice.
when you post your answers I'll double check them.
i get 31.8 degrees
what would be the setup to find the last angle?
yes, looks good. for the last angle, the vertex is R look at how you did the previous two, and look for the pattern. post your setup
(p-r) * (q-r) = |p-r| |q-r| cos a ?
i get 114.8 for the last one
notice that 180 - (31.8+33.5)= 114.7 which matches your result within rounding error.
can you please help me with my other problem?
repost it. this one is long enough