Coordinate planes and the planes pi_1, pi_2, | vedclass

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(A) The vertices of the rectangular parallelepiped are $O(0,0,0)$,$A(a,0,0)$,$B(0,b,0)$,$C(0,0,c)$,$F(a,b,0)$,$D(a,0,c)$,$E(0,b,c)$,and $G(a,b,c)$.Bas

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$\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}$

Vedclass · Answer

(A) The vertices of the rectangular parallelepiped are $O(0,0,0)$,$A(a,0,0)$,$B(0,b,0)$,$C(0,0,c)$,$F(a,b,0)$,$D(a,0,c)$,$E(0,b,c)$,and $G(a,b,c)$.Based on the provided figure,$d_1$ is the diagonal $CG$ on the plane parallel to the $XY$-plane (at $z=c$),connecting $C(0,0,c)$ and $G(a,b,c)$. Thus,the vector $\vec{d}_1 = \vec{G} - \vec{C} = (a-0)\hat{i} + (b-0)\hat{j} + (c-c)\hat{k} = a\hat{i} + b\hat{j}$.$d_2$ is the diagonal $GB$ on the plane $\pi_2$ (parallel to $ZX$-plane at $y=b$),connecting $G(a,b,c)$ and $B(0,b,0)$. Thus,the vector $\vec{d}_2 = \vec{B} - \vec{G} = (0-a)\hat{i} + (b-b)\hat{j} + (0-c)\hat{k} = -a\hat{i} - c\hat{k}$.However,considering the direction of the vectors as shown in the figure,we take $\vec{d}_1 = a\hat{i} + b\hat{j}$ and $\vec{d}_2 = a\hat{i} + c\hat{k}$.The angle $	heta$ between $\vec{d}_1$ and $\vec{d}_2$ is given by $\cos 	heta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1| |\vec{d}_2|}$.$\vec{d}_1 \cdot \vec{d}_2 = (a\hat{i} + b\hat{j}) \cdot (a\hat{i} + c\hat{k}) = a^2$.$|\vec{d}_1| = \sqrt{a^2 + b^2}$ and $|\vec{d}_2| = \sqrt{a^2 + c^2}$.Therefore,$\cos 	heta = \frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}$.

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