Question

Let a ⃗ = (3, -1, 0) and b ⃗=(1/2,3/2,1). Find the vector c ⃗satisfying a ⃗×c ⃗=4b ⃗ and a ⃗ .( c) ⃗=1.

Answer 1

Answer on Question #44079 – Math - Vector Calculus

Let $a^{\prime} = (3, -1, 0)$ and $b^{\prime} = (1/2, 3/2, 1)$. Find the vector $c^{\prime}$ satisfying $a^{\prime} \times c^{\prime} = 4b^{\prime}$ and $a^{\prime} \cdot (c)^{\prime} = 1$.

Solution

$$(\vec{a} \times \vec{c}) = 4\vec{b}$$ $$\vec{a} \times (\vec{a} \times \vec{c}) = 4\vec{a} \times \vec{b}.$$

Thus

$$\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} (\vec{a} \cdot \vec{c}) - \vec{c} (\vec{a} \cdot \vec{a}) = 4\vec{a} \times \vec{b}.$$

But $(\vec{a} \cdot \vec{c}) = 1$. So

$$\vec{c} = \frac{\vec{a} - 4\vec{a} \times \vec{b}}{(\vec{a} \cdot \vec{a})}.$$ $$(\vec{a} \cdot \vec{a}) = 3^2 + (-1)^2 + 0^2 = 10.$$ $$\vec{a} \times \vec{b} = \left( (-1) \cdot 1 - 0 \cdot \frac{3}{2}; \quad 0 \cdot \frac{1}{2} - 3 \cdot 1; \quad 3 \cdot \frac{3}{2} - (-1) \cdot \frac{1}{2} \right) = (-1, -3, 5).$$ $$\vec{c} = \frac{(3, -1, 0) - 4(-1, -3, 5)}{10} = \left( \frac{7}{10}, \frac{11}{10}, -2 \right).$$

Answer: $\left( \frac{7}{10}, \frac{11}{10}, -2 \right)$.

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