Solving a Vector Problem: v1 and v2 Sum to <-4,1,1> with Given Conditions

  • Thread starter Thread starter Loppyfoot
  • Start date Start date
  • Tags Tags
    Vector
Loppyfoot
Messages
192
Reaction score
0

Homework Statement



Find two vectors, v1, and v2, whose sum is <-4,1,1>, where v1 is parallel to <2,5,-4>, and where v2 is perpendicular to <2,5,-4>



Homework Equations



I am guessing I use the cross product for this equation, but I'm confused about how to start this problem.

The Attempt at a Solution



any hlep at all would be greatly appreciated!
 
Physics news on Phys.org
Loppyfoot said:

Homework Statement



Find two vectors, v1, and v2, whose sum is <-4,1,1>, where v1 is parallel to <2,5,-4>, and where v2 is perpendicular to <2,5,-4>



Homework Equations



I am guessing I use the cross product for this equation, but I'm confused about how to start this problem.
Don't use the cross product.

Write v1 and v2 using coordinates. You're given that they add to <-4, 1, 1>. You're also given that v1 is parallel to <2, 5, -4>, which means that v1 is some constant multiple of <2, 5, -4>. You're also given that v2 is perpendicular to <2, 5, -4>. Two vectors being perpendicular should suggest a particular operation.
 
So, first I realized that v1= K<2,5,-4>

Then I figured that since v1+v2=sum, ;
then:

<4,1,1> - K<2,5,-4> = v2

Now, what process would I use to find K?
 
Loppyfoot said:
So, first I realized that v1= K<2,5,-4>
Then what are the coordinates of v1?

Loppyfoot said:
Then I figured that since v1+v2=sum, ;
then:

<4,1,1> - K<2,5,-4> = v2

Now, what process would I use to find K?
How about finding coordinates for v2 first?

Also what can you do with the given information that v2 is perpendicular to <2, 5, -4>?
 
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...
Back
Top