1. **Problem statement:** Given vectors $a$, $b$, and $c$ such that $a \neq 0$ and $a \times 3b = 2a \times c$, show that $3b - 2c = \lambda a$ for some scalar $\lambda$. 2. **Recall the vector cross product property:** For any vectors $u$, $v$, and $w$, if $u \times v = u \times w$ and $u \neq 0$, then $v - w$ is parallel to $u$. This means there exists a scalar $\lambda$ such that $v - w = \lambda u$. 3. **Apply the property:** Given $$a \times 3b = 2a \times c,$$ we can rewrite as $$a \times 3b - 2a \times c = 0,$$ which is $$a \times (3b) - a \times (2c) = a \times (3b - 2c) = 0.$$ 4. Since $a \times (3b - 2c) = 0$ and $a \neq 0$, the vector $3b - 2c$ must be parallel to $a$. Therefore, $$3b - 2c = \lambda a,$$ where $\lambda$ is a scalar constant. --- 1. **Problem statement:** Given that $a$ and $c$ are unit vectors, $|b| = 4$, and the angle between $b$ and $c$ is $60^\circ$, find the two possible values of $\lambda$ from the equation $$3b - 2c = \lambda a.$$ 2. **Take the dot product with itself:** $$|3b - 2c|^2 = |\lambda a|^2 = \lambda^2 |a|^2 = \lambda^2,$$ since $a$ is a unit vector. 3. **Calculate $|3b - 2c|^2$:** $$|3b - 2c|^2 = (3b - 2c) \cdot (3b - 2c) = 9|b|^2 - 12 b \cdot c + 4|c|^2.$$ 4. **Substitute known values:** - $|b| = 4$ so $|b|^2 = 16$. - $|c| = 1$ so $|c|^2 = 1$. - $b \cdot c = |b||c| \cos 60^\circ = 4 \times 1 \times \frac{1}{2} = 2$. Thus, $$|3b - 2c|^2 = 9 \times 16 - 12 \times 2 + 4 \times 1 = 144 - 24 + 4 = 124.$$ 5. **Therefore,** $$\lambda^2 = 124,$$ so $$\lambda = \pm \sqrt{124} = \pm 2\sqrt{31}.$$