#vectors
55 messages · Page 1 of 1 (latest)
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Is “firkant” quadrilateral, or parallelogram? I don’t think the translator is correct.
You took the dot product, right?
Hint: Properties of quadrilaterals come into play.
hii, yess firkant means quadrilateral and i have to find out what type of quadrilateral it is
for b?
For a). But yeah, you can get information about the quadrilateral from part a).
Ok.
Well, we know BC || AD.
So what does that tell us about ABCD?
that it is a parallelogram
or wait
two of the sides are parallel
but how do i know if the length of these two are equal?
i know that they are parallel but if the length of then aren’t the same then it might not be parallelogram
right?
or am i wrong
The info given, and part a) (I am not saying that they are equal.)
aaah
for part a i got 4
so they are equal
so now i know that three sides have the length that equals to four which means that the last one would also equal to 4, so DC
so it could have been a parallelogram but since all the sides are equal then i would say its a rhombus maybe
a parallelogram would have two equal sides and the two other would also equal each other - so for instance 2y sides and 2x sides where the angle is not equal to 90
however i am not sure if my conclusion would be correct..
Yes.
but i am not sure how to solve for d..
What does it represent here?
what do you mean?
The dot product.
And also, consider the angles that a rhombus/parallelogram has.
i would say that BA is for instance -vector a
Yes.
-a
and BC would be b
a*b would equal to -8
and so i dont understand why -a*b would equal to 8
^ You tried drawing it out, correct?
yess i did
You should then get that ||the angle between BA and BC is 60 degrees||.
i would have just done it like this
ohh i drew it incorrectly - i wrote C where i wanted to write D
so the calculation doesn’t make any sense
now its correct i think
g) $\vec{BP}=\frac12\vec{BD}=\frac12(0\vec{i}-4\sqrt{3}\vec{j})=-2\sqrt{3}\vec{j}$\
since $AC=4\vec{i}$, and we know that $BD$ and $AC$ are perpendicular, we must prove that $k=\frac12$ for $\vec{AP}=k\vec{AC}$\
we can do this with the dot product and thus since $\vec{AC}\cdot\vec{BD}=0$, we have $2\vec{BP}\cdot \frac1k\vec{AP}=0$ and thus $0\times 4-2\sqrt{3}\times 0=0$, so they are indeed perpendicular as well and thus $k=\frac12$
vengeance
@civic canopy
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