The volume in cubic units of tetrahedron with vertices - -2k- - 3- - k- -4- -3- -6k-2-3- -2k- is equal to-

Let nbsp;A=ĵ +2k̂ ,B= 3î + k̂ ,C=4î +3ĵ +6k̂, D=2î+3ĵ +2k̂ Hence, position vectors of sides of tetrahedron are: AB = 3 î ĵ k̂ AC = 4 î +2ĵ +4k̂ AD = 2 î + ...

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Standard VIII

Mathematics

Polyhedrons

The volume in...

Question

The volume (in cubic units) of tetrahedron with vertices $^j + 2^k, 3^i +^k, 4^i + 3^j + 6^k, 2^i + 3^j + 2^k$ is equal to:

1

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36

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Solution

The correct option is C $6$
Let $A =^j + 2^k, B = 3^i +^k, C = 4^i + 3^j + 6^k, D = 2^i + 3^j + 2^k$
Hence, position vectors of sides of tetrahedron are:
$- - \to A B = 3^i -^j -^k$
$- - \to A C = 4^i + 2^j + 4^k$
$- - \to A D = 2^i + 2^j$
Volume of tetrahedron $= \frac{1}{6} ∣ ∣ ∣ [- - \to A B - - \to A C - - \to A D] ∣ ∣ ∣$
$= \frac{1}{6} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 1 & - 1 4 & 2 & 4 2 & 2 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$= \frac{1}{6} | 2 (- 4 + 2) - 2 (12 + 4) |$
$= \frac{1}{6} | - 4 - 32 |$
$= \frac{1}{6} \times 36 = 6$ cubic units.

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Polyhedrons

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The volume (in cubic units) of tetrahedron with vertices ^j+2^k,3^i+^k,4^i+3^j+6^k,2^i+3^j+2^k is equal to:

The volume (in cubic units) of tetrahedron with vertices $^j + 2^k, 3^i +^k, 4^i + 3^j + 6^k, 2^i + 3^j + 2^k$ is equal to: