Let hat{i}, hat{j} and hat{k} be the unit vec | vedclass

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(A) Given $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$,$\vec{b} = \hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.

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$B, C, D$

Vedclass · Answer

(A) Given $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$,$\vec{b} = \hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.From the matrix equation,we get:$b_2c_3 - b_3c_2 = c_1 - 3$ ... $(1)$$c_3 - b_3c_1 = 1 - c_2$ ... $(2)$$c_2 - b_2c_1 = 1 + c_3$ ... $(3)$These equations represent the cross product $\vec{b} 	imes \vec{c} = \vec{c} - \vec{a}$.Taking the dot product with $\vec{b}$ on both sides: $\vec{b} \cdot (\vec{b} 	imes \vec{c}) = \vec{b} \cdot (\vec{c} - \vec{a}) \implies 0 = \vec{b} \cdot \vec{c} - \vec{b} \cdot \vec{a}$. Since $\vec{a} \cdot \vec{b} = 0$,we get $\vec{b} \cdot \vec{c} = 0$. Thus,$(B)$ is true.Taking the dot product with $\vec{c}$ on both sides: $\vec{c} \cdot (\vec{b} 	imes \vec{c}) = \vec{c} \cdot (\vec{c} - \vec{a}) \implies 0 = |\vec{c}|^2 - \vec{c} \cdot \vec{a}$. So $\vec{a} \cdot \vec{c} = |\vec{c}|^2 
eq 0$. Thus,$(A)$ is false.From $\vec{b} 	imes \vec{c} = \vec{c} - \vec{a}$,squaring both sides: $|\vec{b} 	imes \vec{c}|^2 = |\vec{c} - \vec{a}|^2 \implies |\vec{b}|^2|\vec{c}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c}$.Since $\vec{a} \cdot \vec{c} = |\vec{c}|^2$,we have $|\vec{b}|^2|\vec{c}|^2 = |\vec{c}|^2 + 11 - 2|\vec{c}|^2 = 11 - |\vec{c}|^2$.$|\vec{c}|^2(1 + |\vec{b}|^2) = 11 \implies |\vec{c}|^2 = \frac{11}{1 + |\vec{b}|^2} \leq 11$ (since $|\vec{b}|^2 \geq 1$),so $|\vec{c}| \leq \sqrt{11}$. Thus,$(D)$ is true.Since $\vec{a} \cdot \vec{b} = 3 + b_2 - b_3 = 0 \implies b_3 - b_2 = 3$. Squaring: $b_3^2 + b_2^2 - 2b_2b_3 = 9$. Since $b_2b_3 > 0$,$b_3^2 + b_2^2 = 9 + 2b_2b_3 > 9$. Thus $|\vec{b}|^2 = 1 + b_2^2 + b_3^2 > 10$,so $|\vec{b}| > \sqrt{10}$. Thus,$(C)$ is true.Therefore,$(B), (C), (D)$ are true.

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