Let v_1, v_2, v_3, v_4 be unit vectors in the | vedclass

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(C) Let the unit vectors be represented by their angles $	heta_1, 	heta_2, 	heta_3, 	heta_4$ where $	heta_1 \in (0, 90^{\circ})$,$	heta_2 \in (9

Vedclass · Accepted Answer

There exist $i, j$ with $1 \leq i < j \leq 4$ such that $v_i \cdot v_j < 0$

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(C) Let the unit vectors be represented by their angles $	heta_1, 	heta_2, 	heta_3, 	heta_4$ where $	heta_1 \in (0, 90^{\circ})$,$	heta_2 \in (90^{\circ}, 180^{\circ})$,$	heta_3 \in (180^{\circ}, 270^{\circ})$,and $	heta_4 \in (270^{\circ}, 360^{\circ})$.$(a)$ The sum $v_1 + v_2 + v_3 + v_4$ is not necessarily zero. For example,if the vectors are very close to the positive $X$-axis and positive $Y$-axis,the sum will not be zero.$(b)$ The sum $v_i + v_j$ is not necessarily in the first quadrant. For instance,if $v_1$ is near $90^{\circ}$ and $v_2$ is near $180^{\circ}$,their sum will have a negative $X$-component.$(c)$ Consider $v_1$ in the first quadrant and $v_3$ in the third quadrant. The angle between them is $|	heta_1 - 	heta_3|$. Since $	heta_1 \in (0, 90^{\circ})$ and $	heta_3 \in (180^{\circ}, 270^{\circ})$,the difference $	heta_3 - 	heta_1$ lies in $(90^{\circ}, 270^{\circ})$. The dot product $v_1 \cdot v_3 = \cos(	heta_1 - 	heta_3)$. Since the angle difference can be greater than $90^{\circ}$,the cosine can be negative. Specifically,if we choose $v_1$ near $45^{\circ}$ and $v_3$ near $225^{\circ}$,the angle is $180^{\circ}$,and the dot product is $\cos(180^{\circ}) = -1 $(d)$ This is not necessarily true for all pairs,as shown in $(c)$.

Let $v_1, v_2, v_3, v_4$ be unit vectors in the $XY$-plane,one each in the interior of the four quadrants. Which of the following statements is necessarily true?

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