Equation with vectors.

Question

Let $\vec{a}$ and $\vec{b}$ be unit vectors separated by an angle of $\pi/4$. Solve the equation for $\vec{u}$: $$ \left(\vec{a} \cdot \vec{u} \right) \vec{b} + 4\vec{a} = 2\vec{u}. $$

I' am trying for about three hours but I can't find the solution. Thanks for the help

Answer 1

HINT: Write $\vec u = s\vec a + t\vec b$ and rewrite the given equation. Since $\vec a$ and $\vec b$ are non-parallel, if $s\vec a + t\vec b = s'\vec a + t'\vec b$, then we must have $s=s'$ and $t=t'$.

Answer 2

HINT

We have $\vec u=t\vec a + s\vec b$ that is

$$2\vec u=2t\vec a + 2s\vec b=\left(t+s\frac{\sqrt 2}2\right) \vec{b} + 4\vec{a}$$

thus

• $2t=4$

• $2s=t+s\frac{\sqrt 2}2$

from which we can find $t$ and $s$.

Answer 3

HINT

Note that $a \cdot b = b \cdot a = \cos (\pi/4) = \sqrt{2}/2$. If we dot your equation with $a$ we get $$ (a\cdot u)(b \cdot a) + 4 = 2 u \cdot a $$ so you can solve for $u \cdot a$. Similarly you can dot with $b$ and solve for $u \cdot b$....

Answer 4

If you use $\vec{a}$ and $\vec{b}$ as a basis and write $u=x\vec{a}+y\vec{b}$ then you have $(x+\frac{y}{\sqrt{2}})\vec{b}+4\vec{a}=2x\vec{a}+2y\vec{b}$. So $x=2$ and $y$ is immediately found too.