Solution of Vector equation

Question

If $\mathbf{a} \wedge \mathbf{x}$= $\mathbf{a} \wedge \mathbf{b}$ and $\mathbf{a}\cdot \mathbf{(x-b)}=2a^2$, show that $\mathbf{x=2a+b}$.

I take cross product throughout

$$\mathbf{a} \wedge \mathbf{a} \wedge \mathbf{x} = \mathbf{a} \wedge \mathbf{a} \wedge \mathbf{b}$$

$$\mathbf{(a.x).a-(a.a).x}=\mathbf{(a.b).a-(a.a).b}$$

$$\mathbf{(a.x).a-|a|^{2}.x}=\mathbf{(a.b).a-|a|^2.b}$$

$$\mathbf{|a|^2(x-b)-(a.x).a-(a.b).a}=0$$

Then,I do not know how to proceed.

A second part in the question is

Given $\mathbf{x}$^$\mathbf{a}$=$\mathbf{b-x}$

Prove

$(a^2+1)\mathbf{x}=\mathbf{a}$^$\mathbf{b+b(a.b)a}$

I have been using this same dot product and crossproduct method, but don't get it.

Answer 1

Answer for part(i)

$\mathbf{a} \wedge \mathbf{x}$= $\mathbf{a} \wedge \mathbf{b}$ [EQN(1)]

$\mathbf{a}\cdot \mathbf{(x-b)}=2a^2$ [EQN(2)]

R.T.S

$\mathbf{x=2a+b}$.

Consider EQN(2)

$a^2=\mathbf{a.a}$

$\mathbf{a}\cdot \mathbf{(x-b)}=2\mathbf{a.a}$

This implies $\mathbf{x-b=2a}$ Hence, we make $\mathbf{x}$ subject of formula.

Answer for part(ii)

Given $\mathbf{xΛa=b-x}$

RTS

$(a^2+1)\mathbf{x}=\mathbf{a}$^$\mathbf{b+b(a.b)a}$

Cross with $\mathbf{a}$

$\mathbf{aΛ(xΛa)=aΛb-aΛx}$

$\mathbf{(a.a)x-(a.x)a=aΛb-aΛx}$

$\mathbf{aΛx=-(xΛa)=-(b-x)}$ [Cross-product is anti-commutative]

$\mathbf{(a.a)x-(a.x)a=aΛb+b-x}$

$\mathbf{(a^2)x+x=aΛb+b+(a.x)a}$

$\mathbf{(a^2+1)x=aΛb+b+(a.x)a}$

Dot product with $\mathbf{a}$ to eliminate $\mathbf{(a.x)}$ /EQN 3

$\mathbf{(a^2+1)(x.a)=aΛb.a+b.a+(a.x)a.a}$

$\mathbf{(a^2+1)(x.a)=aΛb.a+b.a+(x.a)a^2}$ [Dot product is commutative]

$\mathbf{(a^2+1)(x.a)=b.a+(x.a)a^2}$

$\mathbf{(a^2+1-a^2)(x.a)=b.a}$

$\mathbf{a.x=b.a=a.b}$

Hence replace

$\mathbf{a.x=b.a=a.b}$ in EQN 3 to get the answer.

Answer 2

$$ \mathbf{a}\times (\mathbf{x}-\mathbf{b}) = \mathbf{0}\\ \mathbf{a}\cdot (\mathbf{x}-\mathbf{b}) = 2a^2 $$ From the first equation you know that $$ (\mathbf{x}-\mathbf{b}) = \lambda \mathbf{a} $$ from the second we have (since the brackets terms must be parallel to $\mathbf{a}$) $$ \mathbf{a}\cdot (\mathbf{x}-\mathbf{b}) =\mathbf{a}\cdot \lambda\mathbf{a} = \lambda \mathbf{a}\cdot \mathbf{a} = 2a^2 \implies \lambda = 2 $$ thus we have $$ \mathbf{x} = \lambda \mathbf{a} + \mathbf{b} = 2\mathbf{a} +\mathbf{b} $$

Answer 3

By the expulsion formula,

$$\mathbf a\wedge (\mathbf a\wedge (\mathbf x-\mathbf b))=(\mathbf a(\mathbf x-\mathbf b))\mathbf a-\mathbf a^2(\mathbf x-\mathbf b)=2\mathbf a^2\mathbf a-\mathbf a^2(\mathbf x-\mathbf b)=0.$$

Obviously, unless $\mathbf a=\mathbf 0$,

$$\mathbf x-\mathbf b=2\mathbf a.$$