AB and BC are diagonals of adjacent faces of | vedclass

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(A) Let the dimensions of the rectangular box be $2a, 2b, 2c$ along the $x, y, z$ axes respectively. The center is at the origin $(0, 0, 0)$.The verti

Vedclass · Accepted Answer

$-1$

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(A) Let the dimensions of the rectangular box be $2a, 2b, 2c$ along the $x, y, z$ axes respectively. The center is at the origin $(0, 0, 0)$.The vertices can be represented as $A(-a, b, c)$,$B(a, b, -c)$,and $C(-a, -b, -c)$.The vectors from the origin are $\vec{OB} = a\hat{i} + b\hat{j} - c\hat{k}$,$\vec{OC} = -a\hat{i} - b\hat{j} - c\hat{k}$,and $\vec{OA} = -a\hat{i} + b\hat{j} + c\hat{k}$.The magnitudes are $|OB| = |OC| = |OA| = \sqrt{a^2 + b^2 + c^2}$.Using the dot product formula $\cos \alpha = \frac{\vec{OB} \cdot \vec{OC}}{|OB||OC|}$:$\cos \alpha = \frac{(a)(-a) + (b)(-b) + (-c)(-c)}{a^2 + b^2 + c^2} = \frac{-a^2 - b^2 + c^2}{a^2 + b^2 + c^2}$.Similarly,$\cos \beta = \frac{\vec{OC} \cdot \vec{OA}}{|OC||OA|} = \frac{(-a)(-a) + (-b)(b) + (-c)(c)}{a^2 + b^2 + c^2} = \frac{a^2 - b^2 - c^2}{a^2 + b^2 + c^2}$.And $\cos \gamma = \frac{\vec{OA} \cdot \vec{OB}}{|OA||OB|} = \frac{(-a)(a) + (b)(b) + (c)(-c)}{a^2 + b^2 + c^2} = \frac{-a^2 + b^2 - c^2}{a^2 + b^2 + c^2}$.Summing these values:$\cos \alpha + \cos \beta + \cos \gamma = \frac{(-a^2 - b^2 + c^2) + (a^2 - b^2 - c^2) + (-a^2 + b^2 - c^2)}{a^2 + b^2 + c^2} = \frac{-a^2 - b^2 - c^2}{a^2 + b^2 + c^2} = -1$.

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