Left Right Triple 686D25

1. **State the problem:** We need to determine if the ordered triple of vectors $(\mathbf{c}, \mathbf{a}, \mathbf{b})$ forms a left hand triple, right hand triple, or neither. 2. **Recall the rule:** The orientation of three vectors $(\mathbf{u}, \mathbf{v}, \mathbf{w})$ can be determined by the sign of the scalar triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$. - If $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) > 0$, the triple is right-handed. - If $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) < 0$, the triple is left-handed. - If $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 0$, the vectors are coplanar and form neither. 3. **Calculate $\mathbf{a} \times \mathbf{b}$:** $$\mathbf{a} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix}-6 \\ -2 \\ 4\end{pmatrix}$$ $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ -6 & -2 & 4 \end{vmatrix} = \mathbf{i}(2 \cdot 4 - 3 \cdot (-2)) - \mathbf{j}(1 \cdot 4 - 3 \cdot (-6)) + \mathbf{k}(1 \cdot (-2) - 2 \cdot (-6))$$ $$= \mathbf{i}(8 + 6) - \mathbf{j}(4 + 18) + \mathbf{k}(-2 + 12) = 14\mathbf{i} - 22\mathbf{j} + 10\mathbf{k}$$ 4. **Calculate scalar triple product $\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})$:** $$\mathbf{c} = \begin{pmatrix}-4 \\ 3 \\ -5\end{pmatrix}$$ $$\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) = (-4)(14) + 3(-22) + (-5)(10) = -56 - 66 - 50 = -172$$ 5. **Interpret the result:** Since $\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) = -172 < 0$, the triple $(\mathbf{c}, \mathbf{a}, \mathbf{b})$ forms a left hand triple. **Final answer:** (c, a, b) forms a left hand triple.