How do I begin to solve this vector/cross product problem?

[rock.freak667]

Homework Statement

It is given that the soltuion of the vector equation y x a=b is

$$\underline{y}= \lambda \underline{a} + \frac{\underline{a} \times \underline{b}}{| \underline{a}|^2}$$

with a . b=0 and $\lambda$ is a scalar. Use this information to find the solution of the equation (x x a) + (x . b)c=d.

Where x is the unknown vector and [itex]\underline{a} \cdot \underline{c} \neq 0[/tex]<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> $$\underline{A} \cdot \underline{B}= |\underline{A}| |\underline{B}|cos\theta$$<br /> <br /> $$\underline{A} \times \underline{B} = |\underline{A}| |\underline{B}|sin\theta \hat{n}$$<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> This is my method of thinking.<br /> <br /> If $\underline{a} \cdot \underline{b} =0$ then this means that <u>a</u> and <u>b</u> are perpendicular<br /> <br /> $$y \times a =b$$<br /> <br /> $$a \time (y \times a)= a \times b$$<br /> <br /> $$= y(a \cdot c)-a(a \cdot y)= a \times b$$<br /> <br /> $$\Rightarrow = y(a \cdot c)=a(a \cdot y)+a \times b$$<br /> <br /> $$\div a \cdot c$$<br /> <br /> $$y= a \frac{a \cdot y}{a \cdot c} + \frac{a \times b}{a \cdot c}$$<br /> <br /> Comparing this with the given solution:<br /> <br /> $$\lambda = \frac{a \cdot y}{a \cdot c}$$<br /> <br /> AND<br /> <br /> $$|a|^2={a \cdot c}$$<br /> <br /> On the right track so far?<br /> <br /> <br /> $$(x \times a) + (x \cdot b)c=d$$<br /> <br /> $$a \times (x \times a)+ a \times (x \cdot b)c= a \times d$$<br /> <br /> $$x(a \cdot a)+ a \times c(x \cdot b)=a \times d$$<br /> <br /> $$x |a|^2 + a \times c(x \cdot b)= a \times d$$<br /> <br /> $$x(a \cdot c) + a \times c( x \cdot b)= a \times d$$<br /> <br /> and I am stuck here.[/itex]

 

[D H]

If $\underline{a} \cdot \underline{b} =0$ then this means that a and b are perpendicular

$$y \times a =b$$

Correct so far.

$$a \time (y \times a)= a \times b$$

OK, but you might want to expand $y\times a$ first. What is $\lambda a \times a$?

$$= y(a \cdot c)-a(a \cdot y)= a \times b$$

Whoa! Where did that c come from?

Hint: You can't solve for lambda. Its value is completely indeterminate here.

 

[rock.freak667]

OK, but you might want to expand $y\times a$ first. What is $\lambda a \times a$?

What do you mean by expanding $y\times a$ ? wouldn't $\lambda a \times a$ just be 0 since a x a =0?

Whoa! Where did that c come from?

ah that came from whenever I expanded a x (y x a), I wrote above it a x (b x c)= b(a.c)-c(a.b) and then forgot to replace the letters...dumb me...

but this is the correct thing, I hope:

$$y=\frac{a \times b}{|a|^2} + a \frac{a \cdot y}{|a|^2}$$

 

[D H]

What do you mean by expanding $y\times a$ ? wouldn't $\lambda a \times a$ just be 0 since a x a =0?

Exactly. The given equation yields absolutely no information about lambda. It can take on any value; you cannot solve for it.

Given that $a \cdot b = 0$, the equation for $y$ is an identity. You are supposed to use this identity to help solve the problem.

 

[rock.freak667]

Given that $a \cdot b = 0$, the equation for $y$ is an identity. You are supposed to use this identity to help solve the problem.

Well I do not see how I could use that. Unless I take $(x \times a) + (x \cdot b)c=d$ and then cross all of that with b

 

[D H]

Well I do not see how I could use that. Unless I take $(x \times a) + (x \cdot b)c=d$ and then cross all of that with b

Try taking the inner product with a instead.

 

[rock.freak667]

Try taking the inner product with a instead.

Then I should get:

$$a \cdot (x \times a) + a \cdot c (x \cdot b)=a \cdot d$$

$$x \cdot (a \times a)+ a \cdot c (x \cdot b)=a \cdot d$$

Since [itex]a \times a=0[/tex]<br /> <br /> $$x \cdot b=\frac{a \cdot d}{a \cdot c}$$<br /> <br /> <br /> I do not think I can isolate <u>x</u> from this.[/itex]

 

[D H]

So put this result back into your original equation for x. Can you use the given identity yet? If not, what can you do to make that identity usable?

BTW, you should now see why the problem statement asserts that $a\cdot c \ne 0$.

 

[rock.freak667]

So put this result back into your original equation for x. Can you use the given identity yet? If not, what can you do to make that identity usable?

BTW, you should now see why the problem statement asserts that $a\cdot c \ne 0$.

If I put it back into the original problem, I will get:

$$(x \times a) + \frac{a \cdot d}{a \cdot c} c= d$$

$$\Rightarrow x \times a = t \ Where \ t=d-\frac{a \cdot d}{a \cdot c} c$$

$$\Rightarrow x=\lambda a + \frac{a \times t}{|a|^2}$$

 

[D H]

Excellent.

 

[rock.freak667]

Excellent.

Thanks!


(I just wasted a bunch of time solving y x a =b though )