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(A) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$. Using the vector triple product formula: $a 	imes (b 	imes c) = (a \cdot c)b - (

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$. Using the vector triple product formula: $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$. Given $a 	imes (b 	imes c) = \frac{1}{2}b$,we have: $(a \cdot c)b - (a \cdot b)c = \frac{1}{2}b$. Since $b$ and $c$ are generally non-collinear,we compare the coefficients: $a \cdot c = \frac{1}{2}$ and $a \cdot b = 0$. For $a \cdot c = \frac{1}{2}$,we have $|a||c| \cos 	heta_2 = \frac{1}{2} \Rightarrow (1)(1) \cos 	heta_2 = \frac{1}{2} \Rightarrow 	heta_2 = 60^{\circ}$. For $a \cdot b = 0$,we have $|a||b| \cos 	heta_1 = 0 \Rightarrow (1)(1) \cos 	heta_1 = 0 \Rightarrow 	heta_1 = 90^{\circ}$. Therefore,$	heta_1 + 	heta_2 = 90^{\circ} + 60^{\circ} = 150^{\circ}$.

Let $a, b, c$ be three unit vectors such that $a \times(b \times c)=\frac{1}{2} b$. If the angle between $a$ and $b$ is $\theta_1$ and the angle between $a$ and $c$ is $\theta_2$,then $\theta_1+\theta_2$ is equal to (in $^{\circ}$)

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