Recent content by CaliforniaRoll88
-
Find the coordinates of a point in 3-space
##d^2=(x_Q-x_P)^2+(y_Q-y_P)^2## ##100=(3k-1)^2+(4k-2)^2## ##100=(3k-1)^2+(4k-2)^2## ##(a-b)^2=a^2-2ab+b^2## ##(3k-1)^2=(3k)^2-2(3k)(-1)+(-1)^2=9k^2+6k+1## ##(4k-2)^2=(4k)^2-2(4k)(-2)+(-2)^2=16k^2+16k+4## ##(3k-1)^2+(4k-2)^2=25k^2+22k+5## @Steve4Physics, I can see that I am using the quadratic...- CaliforniaRoll88
- Post #51
- Forum: Introductory Physics Homework Help
-
Find the coordinates of a point in 3-space
##10^2=5+25k-22k## ##95=3k## ##k=\frac {95}{3}## ##\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>## Is this right @Steve4Physics, @PeroK? Edit: I am sure it's wrong.- CaliforniaRoll88
- Post #47
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
That's awesome. I have my own surface XD. Thank you.- CaliforniaRoll88
- Post #33
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Is that compatible with windows?- CaliforniaRoll88
- Post #30
- Forum: Introductory Physics Homework Help
-
Find the coordinates of a point in 3-space
##\vec P=P\left<1,2\right>## ##\vec Q=Q\frac{1}{5}\left<3,4\right>## Let ##k=\frac{1}{5}Q## ##\vec Q=k\left<3,4\right>=\left<3k,4k\right>## ##\vec P\cdot\vec Q=1\cdot3k+2\cdot4k=11k## ##\vec R## is the resultant of ##\vec P## and ##\vec Q## Law of Cosines: ##R^2=P^2+Q^2-2\vec P\cdot\vec Q##...- CaliforniaRoll88
- Post #44
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Part d) ##\vec G=24xy\hat a_x+12(x^2+2)\hat a_y+18z^2\hat a_z## ##60^2=(24xy)^2+[12(x^2+2)]^2+(18z^2)^2## ##60^2=24^2x^2y^2+[12x^2+24]^2+18^2z^4## ##60^2=24^2x^2y^2+12^2x^4+2(12x^2)(24)+24^2+18^2z^4## ##60^2=24^2x^2y^2+12^2x^4+24^2x^2+24^2+18^2z^4##...- CaliforniaRoll88
- Post #29
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Sum the squares of the components of the vector and takes it's square root.- CaliforniaRoll88
- Post #26
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Could you please advise me on how to start part d)?- CaliforniaRoll88
- Post #24
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
##\vec {PQ}=\left< 1-(-2),2-1,-1-3 \right>## ##\vec {PQ}=\left< 3,1,-4 \right>## ##a_{\vec {PQ}}=\frac {\vec {PQ}}{PQ}=\left< \frac {3}{\sqrt 26},\frac {1}{\sqrt 26},\frac {-4}{\sqrt 26} \right>=\left<.588,.196,-.784\right>##- CaliforniaRoll88
- Post #22
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
I am stumped. Can you please give me a hint?- CaliforniaRoll88
- Post #19
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Aren't I creating a displacement vector between the two points and finding it's unit vector?- CaliforniaRoll88
- Post #17
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Part c) ##\vec G(P)=48\hat a_x+36\hat a_y+18\hat a_z## ##\vec G(Q)=-48\hat a_x+72\hat a_y+162\hat a_z## ##\vec G(P)-\vec G(Q)=[48-(-48)]\hat a_x+(36-72)\hat a_y+(18-162)\hat a_z=96\hat a_x-36\hat a_y+144\hat a_z## ##\hat a_{\vec G(P)-\vec G(Q)}=\frac{96\hat a_x-36\hat a_y+144\hat a_z}{\sqrt...- CaliforniaRoll88
- Post #15
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Wow that's fast. How did you input the formulas so fast?- CaliforniaRoll88
- Post #13
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Correction: ##\vec G(Q)=-48\hat a_x+72\hat a_y+162\hat a_z## ##\hat a_{\vec G(Q)}=\frac{-48\hat a_x+72\hat a_y+162\hat a_z}{\sqrt{48^2+72^2+162^2}}## ##\hat a_{\vec G(Q)}=\frac{-48\hat a_x+72\hat a_y+162\hat a_z}{\sqrt{33732}}## ##\hat a_{\vec G(Q)}=-0.261\hat a_x+0.392\hat a_y+0.882\hat a_z##- CaliforniaRoll88
- Post #12
- Forum: Introductory Physics Homework Help
-
Multi-part question involving a vector field
Correction: ##\vec G(P)=-48\hat a_x+48\hat a_y+162\hat a_z## Part b) Continued: ##\hat a_{\vec G(P)}=\frac{-48\hat a_x+48\hat a_y+162\hat a_z}{\sqrt{48^2+48^2+162^2}}## ##\hat a_{\vec G(P)}=\frac{-48\hat a_x+48\hat a_y+162\hat a_z}{\sqrt{30852}}## ##\hat a_{\vec G(P)}=-0.2745\hat a_x+0.2745\hat...- CaliforniaRoll88
- Post #9
- Forum: Introductory Physics Homework Help