What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?

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Payal Gupta | Contributor-Level 10

2 months ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t e a c h e d g e o f t h e c u b o i d i s 2 u n i t s . \\ ∴ C o o r d i n a t e s o f t h e v e r t i c e s a r e \\ A (2, 0, 0), B (2, 2, 0), C (0, 2, 0), D (0, 2, 2), E (0, 0, 2), F (2, 0, 2), G (2, 2, 2) a n d O (0, 0, 0) . \end{array}$

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Similar Questions for you

Prove that the points $(0, 1, 7)$ , $(2, 1, 9)$ , and $(6, 5, 13)$ are collinear. Find the ratio in which the first point divides the join of the other two.

Payal Gupta

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e g i v e n points a r e A (0, - 1, - 7), B (2, 1, - 9), C (6, 5, - 13) \\ A B = \sqrt[]{{(2 - 0)}^{2} + {(1 + 1)}^{2} + {(- 9 + 7)}^{2}} = \sqrt[]{4 + 4 + 4} = \sqrt[]{12} = 2 \sqrt[]{3} \\ B C = \sqrt[]{{(6 - 2)}^{2} + {(5 - 1)}^{2} + {(- 13 + 9)}^{2}} = \sqrt[]{16 + 16 + 16} = \sqrt[]{48} = 4 \sqrt[]{3} \\ A C = \sqrt[]{{(6 - 0)}^{2} + {(5 + 1)}^{2} + {(- 13 + 7)}^{2}} = \sqrt[]{36 + 36 + 36} = \sqrt[]{108} = 6 \sqrt[]{3} \\ 2 \sqrt[]{3} + 4 \sqrt[]{3} = 6 \sqrt[]{3} \\ i . e ., A B + B C = A C \\ ∴ A B : A C = 2 \sqrt[]{3} : 6 \sqrt[]{3} = 1 : 3 \\ H e n c e, point A d i v i d e s B a n d C i n 1 : 3 e x t e r n a l l y . \end{array}$

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What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?

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