Problem 79 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0\) and vectors \(\left(1, a, a^{2}\right)\left(1, b, b^{2}\right)\) and \(\left(1, c, c^{2}\right)\) are non-coplanar, then the product \(a b c\) equals [2003] (A) 2 (B) \(-1\) (C) 1 (D) 0

Step-by-Step Solution

Answer

The product \(abc\) equals 0.

Step 1: Understand the Problem

We need to find the value of the product \(abc\) given a determinant equals zero and vectors are non-coplanar. This involves understanding a special property of the determinant and conditions for non-coplanarity of vectors.

Step 2: Analyze the Determinant Condition

Given the determinant \(\left|\begin{array}{ccc}a & a^2 & 1+a^3 \ b & b^2 & 1+b^3 \ c & c^2 & 1+c^3\end{array}\right| = 0\), this suggests that the rows (or columns) are linearly dependent.

Step 3: Recall the Condition for Coplanarity

For vectors \((1, a, a^2)\), \((1, b, b^2)\), and \((1, c, c^2)\) to be non-coplanar, their determinant must be non-zero. This corresponds to the determinant of a matrix formed by these vectors.

Step 4: Relate Determinant to Vandermonde Determinant

For a generalized Vandermonde determinant of form \(\left|\begin{array}{ccc}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}\right|\), the value is \((b-a)(c-a)(c-b)\). Non-zero value implies \(b eq a\), \(c eq a\), and \(c eq b\).

Step 5: Use the Information Given to Solve for abc

Originally given that the overall determinant evaluates to zero, and given the non-coplanarity, consider the interchangeable roles of \( (1+a^3), (1+b^3), (1+c^3) \) terms to be zero by balancing the two conditions. Since \(a = b = c\) addresses both conditions making each \(determinant \equiv 0\), consider what satisfies this: \(abc = 0\) would work because all equal leads to a special solution matching coplanarity constraint.

Step 6: Determine the Solution

The conditions satisfy \(abc = 0\), suggesting this results in zeroing the determinant due to shared roots invalidating vector uniqueness for non-coplanarity or else leading back to linear dependence simplifications where potential solutions revert cycle to zero.

Key Concepts

Non-coplanar vectors

Vectors are non-coplanar if they do not lie on the same plane. In three-dimensional space, for three vectors to be non-coplanar, they must not be linearly dependent.
This means there cannot be any scalar multiples that can sum up to zero among these vectors. If vectors are non-coplanar, the volume of the parallelepiped they form is non-zero. This is directly related to their determinant.
If the determinant of a set of vectors, say (1, a, a^2), (1, b, b^2), and (1, c, c^2), is non-zero, they are non-coplanar, confirming their independence in space. In simple terms:

• Non-zero determinant = Non-coplanar vectors
• Zero determinant = Coplanar vectors

Hence, the determinant gives insight into whether vectors span across space or lie flat on a plane.

Vandermonde Determinant

The Vandermonde determinant is a special type of determinant used primarily in polynomial equations. The formula for a 3x3 Vandermonde determinant e.g., \[ \begin{vmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{vmatrix} \] is derived as \((b-a)(c-a)(c-b)\).
This product results from differences of terms and determines whether each variable is unique.
When the Vandermonde determinant is zero, it means some values are equal, indicating that at least two vectors become linearly dependent in terms of their coefficients in polynomial expression leading to coplanarity.

• Non-zero: distinct, unique terms in polynomial
• Zero: repeated terms, causing loss of independence

Linearly dependent rows

A matrix's rows (or columns) are linearly dependent if you can express one row as a linear combination of the others. When dealing with determinants, linear dependence is crucial because it results in a determinant value of zero.
Consider three rows of a matrix in terms of vectors, if they are linearly dependent, these vectors lie in the same plane, which corresponds to coplanarity in the spatial context. For the matrix given by \[ \begin{pmatrix} a & a^2 & 1+a^3 \ b & b^2 & 1+b^3 \ c & c^2 & 1+c^3 \end{pmatrix} \], if their determinant is zero, it implies linear dependency.

• At least one row is a combination of others
• Determinant equals zero = linear dependency

In this context, breaking down the reasons why the given determinant equals zero helps elucidate how conditions such as shared values in coefficients can harm vector independence, revealing why linear dependency is pivotal in matrix evaluation.