Adding Vectors by Components — Question 9

Question 9

Solve for two vector weights

Dimensionless vectors are $\vec u=(1,2)$, $\vec v=(3,-1)$, and $\vec R=(8,2)$. Seek real numbers a and b such that $a\vec u+b\vec v=\vec R$.

1. Set up and solve the two component equations.

2. Verify the vector equation by substitution. Why is the solution unique?

3. If v is replaced by $(2,4)$, can any a,b produce the same R? What changes if the target is also replaced by $(5,10)$?

Question 9 - Solution

Solve for two vector weights

(a) Solve both equations together.

$$a+3b=8,\quad2a-b=2\Rightarrow7a=14\Rightarrow\boxed{a=b=2}$$ The equations are $a+3b=8$ and $2a-b=2$. The second gives $b=2a-2$; substituting yields $7a=14$. Thus $a=2$ and $b=2$.

(b) Check the result and independence.

$$2(1,2)+2(3,-1)=(8,2),\qquad\det\begin{pmatrix}1&3\\2&-1\end{pmatrix}=-7\ne0$$ $2(1,2)+2(3,-1)=(2,4)+(6,-2)=(8,2)$. The directions are not parallel; the coefficient determinant is $1(-1)-3(2)=-7\ne0$, ensuring uniqueness. Equivalently, elimination fixes one coefficient and then the other.

(c) Identify the loss of an independent direction.

$$\vec v=2\vec u\Rightarrow\vec R=(a+2b)(1,2);\quad2\ne2(8),\qquad a+2b=5\text{ for }(5,10)$$ With $\vec v=2\vec u$, every combination equals $(a+2b)(1,2)$ and must satisfy $R_y=2R_x$. The target $(8,2)$ does not, so no solution exists. The target $(5,10)$ does: any pair satisfying $a+2b=5$ works, giving infinitely many solutions.