Statement I: For any three vectors $\vec{a}, \vec{b}$ and $\vec{c}$\ $(\vec{a}-\vec{b}) \cdot(\vec{

Question

Statement I: For any three vectors $\vec{a}, \vec{b}$ and $\vec{c}$\ $(\vec{a}-\vec{b}) \cdot(\vec{b}-\vec{c}) 	imes(\vec{c}-\vec{a})=2 \vec{a} \cdot \vec{b} 	imes \vec{c}$.\ Statement II: Let $\vec{A}, \vec{B}, \vec{C}$ be unit vectors. Suppose that $\vec{A} \cdot \vec{B}=\vec{A} \cdot \vec{C}=0$ and the angle between $\vec{B}$ and $\vec{C}=\frac{\pi}{6}$. Then $\vec{A}= \pm 2(\vec{B} 	imes \vec{C})$\

Professor PC Thomas Classes · Answer

False- For $(\vec{a}-\vec{b}) \cdot(\vec{b}-\vec{c}) 	imes(\vec{c}-\vec{a})=0$
True- As $\vec{A} \cdot \vec{B}=0=\vec{A} \cdot \vec{C}, \vec{A}$ is perpendicular to both $\vec{B}$ and $\vec{C}$\ $\Rightarrow \vec{A}=\lambda(\vec{B} 	imes \vec{C}), \vec{B} 	imes \vec{C}=|\vec{B}||\vec{C}| \sin \frac{\pi}{6} \hat{n}$\ = (1) (1) $\left(\frac{1}{2}ight) \hat{n}$\ $|\vec{B} 	imes \vec{C}|=\frac{1}{2}|\hat{n}|=\frac{1}{2}$\ $	herefore \quad \vec{A}=\lambda\left(\frac{1}{2} \hat{n}ight) 	herefore|\vec{A}|=|\lambda| \cdot \frac{1}{2}$ But $|\vec{A}|=1$\ $	herefore \quad|\lambda|=2 \quad 	herefore \lambda= \pm 2 \quad 	herefore \vec{A}= \pm 2(\vec{B} 	imes \vec{C})$\ Hence the statement is true.\