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10 Nov 2019

Consider the vector between <-2,-2,2> and . <1,4,1>The vector is . The length of is . If the tail of is at <-4,-1,3> , then the tip is at . If the tip of is at <3,-4,2> then its tail is at . What vector has the same length as , but points in the opposite direction? . What vector has the same direction as , but is twice as long? .

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Lelia Lubowitz
Lelia LubowitzLv2
13 Sep 2019

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here is the problem it is already solved and this is the yahooanswer solution and my teachers solution as well. All i need is aGraph and for someone to explain how my teachergot 0,0 and 0,-2 as intresection points.

teachers solution:find all 4 points of intersections: (0,0),(0,-2), (-4sqrt(17sqrt(2)-24), 10sqrt(2)-14),(-4sqrt(17sqrt(2)-24), 10sqrt(2)-14).

yahoo answers:sin(theta) = r^2/4 ==> sin(theta)>= 0 . Hence 0 =< theta =< pi

r = 1 - sin(theta) ; therefore 0 =< r =< 1, because now weknow 0 =< sin(theta) =< 1

Therefore y = r sin(theta) >= 0 so the intersection points are in theupper half plane.

Also, sin(theta) = r^2/4 =< 1/4 < 1/sqrt(2) = sin pi/4 =sin(3pi/4),

so 0 =< theta < pi/4 or 3pi/4 < theta =< pi.

---
At an intersection point, sin(theta) = r^2/4 = 1 - r

r^2 + 4r - 4 = 0

Roots: r = -- 2 +/- 2 sqrt(2); but r >= 0, so r = 2 sqrt(2) - 2= 2(sqrt(2) - 1)

sin(theta) = 1 - r = 3 - 2 sqrt(2)

There is just one value of y, namely

y = r sin(theta) = 2 (sqrt(2) - 1)(3 - 2 sqrt(2) ) = 2 (5 sqrt(2) -7) = 10 sqrt(2) - 14

( or y = 10 (sqrt(2) - 1.4 ) = 0.142135...

However, there are two intersection points, with x^2 = r^2 -y^2

x = +/- sqrt( r^2 - y^2 )

x^2 = r^2 - y^2 = 4 (sin theta) -- y^2 = 4 (3 - 2 sqrt(2)) -- [2(5sqrt(2) - 7)]^2

x^2 = 12 - 8 sqrt(2) -4 [ 99 - 70 sqrt(2)] = 272 sqrt(2) - 384 (=17 x 2^4 sqrt(2) - 3 x 2^7)

x = +/- 4 sqrt(17 sqrt(2) - 24) [ = +/- 0.8161427...]

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