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Unknown component found from the given volume of a tetrahedron or parallelepiped

This type of question, asked 6 times: 11 May 2023 S1, 9 May 2024 S1, 10 May 2024 S2, 11 May 2024 S1, 13 Apr 2026 S1, 18 Apr 2026 S2. All 6 questions below, oldest first.

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  1. Question 1

    11 May 2023 S1

    141
    If the volume of tetrahedron whose vertices are A≡(1,−6,10)A\equiv(1,-6,10), B≡(−1,−3,7)B\equiv(-1,-3,7), C≡(5,−1,k)C\equiv(5,-1,k) and D≡(7,−4,7)D\equiv(7,-4,7) is 11 cu. units, then the value of k is

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  2. Question 2

    9 May 2024 S1

    147
    The volume of parallelepiped, whose coterminous edges are given by u⃗=i^+j^+λk^\vec{u}=\hat{i}+\hat{j}+\lambda\hat{k}, v⃗=i^+j^+3k^\vec{v}=\hat{i}+\hat{j}+3\hat{k}, w⃗=2i^+j^+k^\vec{w}=2\hat{i}+\hat{j}+\hat{k} is 1 cu units. If θ\theta is the angle between u⃗\vec{u} and w⃗\vec{w}, then the value of cos⁡θ\cos\theta is

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  3. Question 3

    10 May 2024 S2

    101
    If the volume of tetrahedron, whose vertices are with position vectors i^−6j^+10k^, −i^−3j^+7k^\hat{i}-6\hat{j}+10\hat{k},\,-\hat{i}-3\hat{j}+7\hat{k}, 5i^−j^+λk^5\hat{i}-\hat{j}+\lambda\hat{k} and 7i^−4j^+7k^7\hat{i}-4\hat{j}+7\hat{k} is 11 cubic units, then value of λ\lambda is

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  4. Question 4

    11 May 2024 S1

    114
    If the volume of the parallelepiped is 158 cu. units whose coterminous edges are given by the vectors a⃗=(i^+j^+nk^)\vec{a} = (\hat{i}+\hat{j}+n\hat{k}), b⃗=2i^+4j^−nk^\vec{b} = 2\hat{i}+4\hat{j}-n\hat{k} and c⃗=i^+nj^+3k^\vec{c} = \hat{i}+n\hat{j}+3\hat{k}, where n≥0n \geq 0, then the value of n is

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  5. Question 5

    13 Apr 2026 S1

    87
    The volume of a tetrahedron with vertices 5i^−j^+k^,7i^−4j^+pk^,i^−6j^+10k^5\hat{i} - \hat{j} + \hat{k}, 7\hat{i} - 4\hat{j} + p\hat{k}, \hat{i} - 6\hat{j} + 10\hat{k} and −i^−3j^+7k^-\hat{i} - 3\hat{j} + 7\hat{k} is 11 cubic units, then one of the values of p is

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  6. Question 6

    18 Apr 2026 S2

    89
    If the volume of the tetrahedron whose coterminous edges are given by the vectors aˉ=−2i^+3j^−3k^,bˉ=4i^+5j^+(λ−10)k^,cˉ=6i^+2j^−3k^\bar{a} = -2\hat{i} + 3\hat{j} - 3\hat{k}, \bar{b} = 4\hat{i} + 5\hat{j} + (\lambda - 10)\hat{k}, \bar{c} = 6\hat{i} + 2\hat{j} - 3\hat{k} is 11 cubic units, then the sum of the possible values of λ\lambda is …\ldots

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