Finding a vector using scalar and vector projections

user8899
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Homework Statement



Determine the vector(s) whose vector projection on u =< 1,2,2 > is v =< 3,6,6 > and its
scalar projection on w =< 1,1,1 > is √3.

Homework Equations


Vector Projection of b onto a: (|b.a| \ |a|) * (1/ |a|) * a
Scalar Projection: (|b.a| \ |a|)


The Attempt at a Solution


First started by finding the vector <a,b,3-b-c> (using the scalar projection equation), but don't know what else to do from there. Help please?
 
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welcome to pf!

hi user8899! welcome to pf! :smile:
user8899 said:
First started by finding the vector <a,b,3-b-c> (using the scalar projection equation) …

isn't the scalar projection just the magnitude of the vector projection? :wink:
 
Hi, Thank you!

well I substituted <a,b,3-b-c> into the vector projection equation, but I think my problem is the algebra... I'm not sure
 
what is your vector projection equation? :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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