The volume of tetrahedron whose vertices are

A = (3, 2, 1) ,~B = (1, 2, 4),~ C = (4, 0, 3),~ D = (1, 1, 7)~will be $- --- -$ cubic units

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$\frac{5}{6}$

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$\frac{6}{5}$

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None of the above

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Solution

The correct option is B
$\frac{5}{6}$

$G i v e n v e r t i c e s a r e \to A = (3, 2, 1) \to B = (1, 2, 4) \to C = (4, 0, 3) \to D = (1, 1, 7) S o t h e t h r e e c o t e r m i n o u s e d g e s w e c a n f o r m a r e \to A B, \to A C a n d \to A D R i g h t ? S o \to A B = ((1 - 3), (2 - 2), (4 - 1)) = (- 2, 0, 3) S o \to A C = ((4 - 3), (0 - 2), (3 - 1)) = (1, - 2, 2) S o \to A D = ((1 - 3), (1 - 2), (7 - 1)) = (- 2, - 1, 6) S o v o l u m e = \frac{1}{6} ∣ ∣ [\to A B \to A C \to A D] ∣ ∣ R e m e m b e r t h a t w e h a v e t o t a k e t h e a b s o l u t e v a l u e o f s c a l a r t r i p l e p r o d u c t . S o [\to A B \to A C \to A D] = ∣ ∣ ∣ ∣ \begin{matrix} - 2 & 0 & 3 1 & - 2 & 2 - 2 & - 1 & 6 \end{matrix} ∣ ∣ ∣ ∣ = 20 - 15 = 5 S o v o l u m e o f t e t r a h e d r o n = \frac{1}{6} [\to A B \to A C \to A D] = \frac{1}{6} \times 5 = \frac{5}{6} c u b i c u n i t s .$

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Similar questions

A (1, 1),

B (3, 4),

C (5, - 2)

D (4, - 7)

(1, 1, 1), (2, 1, 3), (3, 2, 2)

(3, 3, 4)

(0, 1, 2), (3, 0, 1), (4, 3, 6)

(2, 3, 2)

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