If 3a+2b=5c and 8a−7b=4c, Are a,b,c Collinear and Is |a|>|b|?

A standard vector algebra problem, not tied to a specific exam paper.

Problem

If $3\vec{a}+2\vec{b}=5\vec{c}$ and $8\vec{a}-7\vec{b}=4\vec{c}$, determine whether the following statements are true: 1. $|\vec{a}| > |\vec{b}|$ 2. $\vec{a}$, $\vec{b}$, and $\vec{c}$ are collinear vectors.

Key insight. Two equations both expressing $\vec{c}$ in terms of $\vec{a}$ and $\vec{b}$ can be combined by eliminating $\vec{c}$ entirely — what’s left is a direct relationship between $\vec{a}$ and $\vec{b}$ alone. If that relationship says one is a scalar multiple of the other, both magnitude comparison and collinearity fall out immediately.

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Approach

Solve both given equations for $\vec{c}$, then set the two expressions equal to eliminate $\vec{c}$. This produces a direct scalar relationship between $\vec{a}$ and $\vec{b}$, showing they are parallel. Since $\vec{c}$ is itself built from $\vec{a}$ and $\vec{b}$, it inherits the same direction, making all three collinear — and the scalar relationship also settles the magnitude comparison.

Solution

Step 1 — Express c from each equation

$$\text{From } 3\vec{a}+2\vec{b}=5\vec{c}: \qquad \vec{c} = \frac{3\vec{a}+2\vec{b}}{5}$$

$$\text{From } 8\vec{a}-7\vec{b}=4\vec{c}: \qquad \vec{c} = \frac{8\vec{a}-7\vec{b}}{4}$$

Step 2 — Eliminate c by equating

$$\frac{3\vec{a}+2\vec{b}}{5} = \frac{8\vec{a}-7\vec{b}}{4}$$

Cross-multiplying:

$$4(3\vec{a}+2\vec{b}) = 5(8\vec{a}-7\vec{b})$$

$$12\vec{a}+8\vec{b} = 40\vec{a}-35\vec{b}$$

Step 3 — Solve for the relationship between a and b

$$8\vec{b}+35\vec{b} = 40\vec{a}-12\vec{a} \implies 43\vec{b} = 28\vec{a} \implies \vec{a} = \frac{43}{28}\vec{b}$$

Since $\vec{a}$ is a scalar multiple of $\vec{b}$, they are parallel (collinear) vectors.

Step 4 — Find c in terms of b too

Substituting $\vec{a}=\dfrac{43}{28}\vec{b}$ into $\vec{c}=\dfrac{3\vec{a}+2\vec{b}}{5}$:

$$\vec{c} = \frac{3\left(\frac{43}{28}\vec{b}\right)+2\vec{b}}{5} = \frac{\frac{129}{28}\vec{b}+\frac{56}{28}\vec{b}}{5} = \frac{\frac{185}{28}\vec{b}}{5} = \frac{37}{28}\vec{b}$$

Since $\vec{c}$ is also a scalar multiple of $\vec{b}$, all three vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ point along the same direction — they are collinear.

Step 5 — Compare magnitudes of a and b

Since $\vec{a}=\dfrac{43}{28}\vec{b}$ and $\dfrac{43}{28}>1$:

$$|\vec{a}| = \frac{43}{28}|\vec{b}| > |\vec{b}|$$

Answer

Common mistakes

• Trying to check collinearity by computing $\vec{a}\times\vec{b}$ without first eliminating $\vec{c}$. Since $\vec{a}$ and $\vec{c}$ (and $\vec{b}$ and $\vec{c}$) aren’t directly related in the original equations, eliminating $\vec{c}$ first is what reveals the direct proportionality between $\vec{a}$ and $\vec{b}$.
• Stopping after finding $\vec{a}$ in terms of $\vec{b}$ and assuming collinearity of all three without checking $\vec{c}$. The collinearity of $\vec{a}$ and $\vec{b}$ alone doesn’t automatically prove $\vec{c}$ shares that direction — it needs to be confirmed by substituting back, even though it turns out to hold here.

Practise next

• If $2\vec{p}+3\vec{q}=4\vec{r}$ and $5\vec{p}-\vec{q}=2\vec{r}$, determine whether $\vec{p}$, $\vec{q}$, $\vec{r}$ are collinear, and compare $|\vec{p}|$ and $|\vec{q}|$, using the same elimination method.