Adding Vectors by Components — Question 4

Question 4

A linear combination is not an unweighted sum

Let $\vec A=(2\hat\imath-3\hat\jmath)$ m and $\vec B=(-\hat\imath+4\hat\jmath)$ m. Define $\vec R=2\vec A-3\vec B$. The coefficients 2 and 3 are dimensionless.

1. Find R and its magnitude.

2. Solve the equation for B in terms of A and R, then recover the original B.

3. A student computes $R_y=2(-3)-3(-4)$. Identify the error and explain how to avoid it.

Question 4 - Solution

A linear combination is not an unweighted sum

(a) Distribute each coefficient to both components.

$$\vec R=2(2,-3)-3(-1,4)=\boxed{(7,-18)\,\mathrm m},\quad R=\sqrt{373}\,\mathrm m$$ $2\vec A=(4,-6)$ m and $-3\vec B=(3,-12)$ m. Thus $\vec R=(7\hat\imath-18\hat\jmath)$ m and $R=\sqrt{373}\approx19.313$ m. The negative multiplier reverses B before tripling its length.

(b) Invert the whole vector equation.

$$\vec B=\frac{2\vec A-\vec R}{3}=\frac{(4,-6)-(7,-18)}3=(-1,4)\,\mathrm m$$ $3\vec B=2\vec A-\vec R$, so $\vec B=(2\vec A-\vec R)/3=[(4,-6)-(7,-18)]/3=(-1,4)$ m. This substitution independently checks both components of the forward calculation.

(c) Do not reverse a sign twice.

$$R_y=2(-3)-3(+4)=-18\,\mathrm m\ne2(-3)-3(-4)=6\,\mathrm m$$ The given vertical component is $B_y=+4$ m. The subtraction is already represented by the coefficient $-3$, so the term is $-3(+4)=-12$ m. Replacing $B_y$ by $-4$ inside that same subtraction reverses it twice. Either scale the original B by $-3$ directly or form $-\vec B$ and then add three copies.