Question

find the volume of tetrahedron whose vertices are A(3, 7, 4), B(5, -2, 3), C(-4, 5, 6) and D(1 , 2, 3)​

Answer 1

AnswEr :

⋆ Reference of Image in Attachment :

Given Vertices of Tetrahedron are :

$\bold{Vertices} \begin{cases}\sf{A=(3, 7, 4)}\\ \sf{B=(5, -2, 3)}\\ \sf{C=(-4, 5, 6)}\\ \sf{D=(1, 2, 3)}\end{cases}$

• Finding the Vector of Sides :

$\sf\vec{AB} = \vec{B} - \vec{A} \\ \: \: \: \: \: \: \: = \sf(5 - 3)\hat{\imath} + ( - 2 - 7)\hat{\jmath} + (3 - 4)\hat{k}\\ \: \: \: \: \: \: \: = 2\hat{\imath} - 9\hat{\jmath} - \hat{k}$

$\sf\vec{AC} = \vec{C} - \vec{A} \\ \: \: \: \: \: \: \: = \sf(-4 - 3)\hat{\imath} + (5 - 7)\hat{\jmath} + (6 - 4)\hat{k}\\ \: \: \: \: \: \: \: = - 7\hat{\imath} - 2\hat{\jmath} + 2 \hat{k}$

$\sf\vec{AD} = \vec{D} - \vec{A} \\ \: \: \: \: \: \: \: = \sf(1 - 3)\hat{\imath} + (2 - 7)\hat{\jmath} + (3 - 4)\hat{k}\\ \: \: \: \: \: \: \: = - 2\hat{\imath} - 5\hat{\jmath} - \hat{k}$

• Now we will find the Volume :

$\leadsto \rm Volume = \dfrac{1}{6} |\vec{AB} \: \vec{AC} \: \vec{AD}|$

$\leadsto \rm Volume = \dfrac{1}{6}\left|\begin{array}{c c c}2 & - 9 & - 1\\- 7 & - 2 & 2\\- 2 & - 5 & - 1\end{array}\right|$

$\leadsto \rm Volume = \dfrac{1}{6}|2{(-2*-1) - (2*-5)} - (-9){(-7*-1) - (2*-2)} + (-1){(-7*-5) - (-2*-2)}|$

$\leadsto \rm Volume = \dfrac{1}{6}|2(2 + 10) + 9(7 + 4) - 1(35 - 4)|$

$\leadsto \rm Volume = \dfrac{1}{6}|2(12)+9(11)-1(31)|$

$\leadsto \rm Volume = \dfrac{1}{6}|24+99-31|$

$\leadsto \rm Volume = \dfrac{1}{\cancel6}\times\cancel{92}$

$\leadsto\boxed{\sf Volume =15.33 \: {unit}^{3} }$

⠀

∴ Volume of Tetrahedron is 15.33 unit³.

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• The Method to find any Determinant :

$\left|\begin{array}{c c c}a & b & c\\d & e & f\\g & h & i\end{array}\right|$

⋆ |A| = a(ei − fh) − b(di − fg) + c(dh − eg)

Answer 2

Question :---

• we have to find volume of tetrahedron whose vertices are A(3, 7, 4), B(5, -2, 3), C(-4, 5, 6) and D(1 , 2, 3) ???

Formula used :----

• The volume of Tetrahedron is equal to the 1/6 of the absolute value of the triplet product ..

Given vertices :---

• A(3, 7, 4)
• B(5, -2, 3)
• C(-4, 5, 6)
• D(1 , 2, 3)

AB = B-A = (5-3)i + (-2-7)j + (3-4)k = 2i -9j -k

AC = C-A = (-4-3)i + (5-7)j + (6-4)k = -7i -2j +2k

AD = D-A = (1-3)i + (2-7)j + (3-4)k = -2i -5j -1 k

Now Volume = 1/6 ∆ [ AB AC AD ]

$Volume = \frac{1}{6} [ 2( 2- ( - 10)) + 9(7 - ( - 4)) - 1(35 - 4)) ] \\ \\ Volume = \frac{1}{6} [2 \times 12 + 9 \times 11 - 31] \\ \\ Volume \: = \frac{1}{6} [24 + 99 - 31] \\ \\Volume = \frac{1}{6}[ \: 92] \: \\ \\ Volume = \frac{46}{3} \: units^{3}$

So volume of tetrahedran will be (46/3) cubic units ..

(Hope it helps you)