設 $\left|\overset{\large\rightharpoonup}{v}\right| = k > 0$，則根據題意： $$\left|\overset{\large\rightharpoonup}{u}\right| = 2k, \ \left|2\overset{\large\rightharpoonup}{u} + 3\overset{\large\rightharpoonup}{v}\right| = 2k$$ 將長度等式兩邊平方，得： $$\left|2\overset{\large\rightharpoonup}{u} + 3\overset{\large\rightharpoonup}{v}\right|^2 = (2k)^2$$ $$4\left|\overset{\large\rightharpoonup}{u}\right|^2 + 12(\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v}) + 9\left|\overset{\large\rightharpoonup}{v}\right|^2 = 4k^2$$ 將 $\left|\overset{\large\rightharpoonup}{u}\right| = 2k$ 與 $\left|\overset{\large\rightharpoonup}{v}\right| = k$ 代入，得： $$4(2k)^2 + 12(\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v}) + 9k^2 = 4k^2$$ $$16k^2 + 12(\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v}) + 9k^2 = 4k^2$$ $$25k^2 + 12(\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v}) = 4k^2 \implies 12(\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v}) = -21k^2 \implies \overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} = -\dfrac{7}{4}k^2$$ 又向量內積定義為： $$\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} = \left|\overset{\large\rightharpoonup}{u}\right|\left|\overset{\large\rightharpoonup}{v}\right|\cos\theta = (2k)(k)\cos\theta = 2k^2\cos\theta$$ 因此： $$2k^2\cos\theta = -\dfrac{7}{4}k^2 \implies \cos\theta = -\dfrac{7}{8}$$