- #1

Ursa

- 11

- 2

- Homework Statement
- In a meeting of mimes, mime 1 goes through a displacement d1 = (7.07 m)i + (3.94 m)j and and mime 2 goes through a displacement d2 = (-7.03 m)i + (4.1 m)j. What are (a)|d1 × d2|, (b)d1 · d2, (c)(d1 + d2) · d2, and (d) the component of d1 along the direction of d2?

- Relevant Equations
- a · b = ab *cos ϕ

a · b = ax*bx+ ay*by+az*bz

a = sqrt(ax^2+ay^2)

rx=ax+bx

rx=ax+bx

(a) I did (7.07*4.1)-(-7.03*3.94)=56.7 with this method I got this answer correct in my first attempt.

(b) This where I seem to have gone wrong. I used a · b = (a

(c) I first solved for (d1 + d2) with a

(d) This one I didn't understand at all, I gave it a shot by trying to solve d

cos

d

d

cos

Then 8.14 cos 37.4= 6.43 mI just fundamentally don't understand vectors and really don't get how to do math with them. Any analyses of what I'm doing wrong, explanation and/or drawings would be appreciated.

(b) This where I seem to have gone wrong. I used a · b = (a

_{x}b_{x}+a_{y}b_{y}) then I used a = sqrt(ax^{2}+ay^{2}) to get a single number for the answer. Filling in the numbers 7.07*-7.03 + 4.94*4.1 = -49.7i+16.2j. Then sqrt(-49.^{2}+16.2^{2})=52.3(c) I first solved for (d1 + d2) with a

_{x}+b_{x}7.07+-7.03 + 4.94+4.1 = -0.04i+8.04j. Then used those unit vectors in the same way as in part (b) to get 32.97(d) This one I didn't understand at all, I gave it a shot by trying to solve d

_{1}*cos ϕ. first trying to find ϕ by a · b = ab *cos ϕcos

^{-1}((a · b)/ab). Finding d_{1}and d_{2}by a = sqrt(ax^{2}+ay^{2}d

_{1}=8.09d

_{2}= 8.14cos

^{-1}(52.3/8.09 * 8.14)=37.4Then 8.14 cos 37.4= 6.43 mI just fundamentally don't understand vectors and really don't get how to do math with them. Any analyses of what I'm doing wrong, explanation and/or drawings would be appreciated.

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