Mathematics

Problem

Let

a_1, a_2, \dots, a_n

n

(

n \geq 10

) natural numbers in AP such that volume of parallelopiped formed by vectors

\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

\vec{b} = a_2\hat{i} + a_3\hat{j} + a_1\hat{k}

and

\vec{c} = a_3\hat{i} + a_1\hat{j} + a_2\hat{k}

is 54 then

a_6

equals: (A) 2 (B) 5 (C) 7 (D) 10 (A) 2 (B) 5 (C) 7 (D) 10

Intuitive explanation

The volume of a parallelepiped is the absolute value of the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ , which equals the determinant of the matrix formed by the three vectors. Since the $a_i$ form an arithmetic progression, you can write each term as $a_i = a_1 + (i-1)d$ for some common difference $d$ . The key insight is that the determinant has a special structure due to the cyclic permutation of coordinates—it will simplify dramatically when you substitute the AP formula, leaving you with an equation in $a_1$ and $d$ that you can solve using the volume constraint of 54.

Intuitive explanation

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