Pi_1, Pi_2, Pi_3 are three planes which are r | vedclass

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(A) Let the origin be $O(0, 0, 0)$. The vertices of the rectangular parallelopiped are $O(0, 0, 0)$, $A(a, 0, 0)$, $E(0, b, 0)$, $D(0, 0, c)$, and oth

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$\cos ^{-1}\left(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\right)$

Vedclass · Answer

(A) Let the origin be $O(0, 0, 0)$. The vertices of the rectangular parallelopiped are $O(0, 0, 0)$, $A(a, 0, 0)$, $E(0, b, 0)$, $D(0, 0, c)$, and others. $d_1$ is a diagonal of the face in the $XY$-plane not passing through the origin. The vertices of this face are $(0, 0, 0), (a, 0, 0), (0, b, 0), (a, b, 0)$. The diagonal not passing through the origin is the line segment connecting $(a, 0, 0)$ and $(0, b, 0)$, i.e.,$AE$. The vector along $d_1$ is $\vec{v_1} = (0-a)\hat{i} + (b-0)\hat{j} + (0-0)\hat{k} = -a\hat{i} + b\hat{j}$. $d_2$ is a diagonal of the plane $\Pi_2$ (which is parallel to $ZX$ plane at distance $b$) coterminous with $d_1$. The plane $\Pi_2$ contains points $(0, b, 0), (a, b, 0), (0, b, c), (a, b, c)$. The diagonal coterminous with $d_1$ (which starts at $A(a, 0, 0)$) is $AD$, where $D$ is $(0, 0, c)$. Wait, looking at the provided image, $d_1$ is $AE$ and $d_2$ is $AD$. The vector $\vec{AE} = (0-a)\hat{i} + (b-0)\hat{j} + (0-0)\hat{k} = -a\hat{i} + b\hat{j}$. The vector $\vec{AD} = (0-a)\hat{i} + (0-0)\hat{j} + (c-0)\hat{k} = -a\hat{i} + c\hat{k}$. The angle $	heta$ between them is given by $\cos 	heta = \frac{\vec{AE} \cdot \vec{AD}}{|\vec{AE}| |\vec{AD}|}$. $\vec{AE} \cdot \vec{AD} = (-a)(-a) + (b)(0) + (0)(c) = a^2$. $|\vec{AE}| = \sqrt{(-a)^2 + b^2} = \sqrt{a^2+b^2}$. $|\vec{AD}| = \sqrt{(-a)^2 + c^2} = \sqrt{a^2+c^2}$. Thus, $\cos 	heta = \frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}$. Therefore, $	heta = \cos ^{-1}\left(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}ight)$.

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