Question

If the vectors \[a = i + aj + {a^2}k,b = i + bj + {b^2}k\] and \[c = i + cj + {c^2}k\] are three non-coplanar vectors and \[\left| \begin{gathered}
a&{a^2}&1+{a^3} \\
b&{b^2}&1+{b^3}\\
c&{c^2}&1+{c^3}\\
\end{gathered} \right|\] $= 0$
then the value of \[abc\] is
A. \[0\]
B. \[1\]
C. \[2\]
D. \[ - 1\]

Answer 1

Hint:In the given question, we are given the three vectors that are non-coplanar. If any three vectors lie in the same plane, then the scalar product of the three vectors is equal to zero. Hence, the triple scalar product of the three vectors must not be equal to zero. Hence, we will be solving the equation involving the determinant given to us keeping in mind that the scalar triple product of the three given vectors cannot be zero.

Complete step by step answer:
Vectors that are part of the same plane, in this way, are coplanar vectors . In contrast, vectors that belong to different planes are called non-coplanar vectors. To find out if the vectors are coplanar or not coplanar, it is possible to use the operation known as mixed product or triple dot product . If the result of the mixed product is different from 0 , the vectors are non-coplanar(they are in different planes). Following the same reasoning, we can affirm that when the result of the triple scalar product is equal to 0 , the vectors in question are coplanar (they are in the same plane). Since \[a,b,c\] are non-coplanar vectors, therefore
\[\left| \begin{gathered}
1&a&{a^2} \\
1&b&{b^2} \\
1&c&{c^2} \\
\end{gathered} \right|\]

\[\vartriangle \ne 0\]
Consider,
\[\left| \begin{gathered}
1&{a^2}&1+{a^3} \\
1&{b^2}&1+{b^3}\\
1&{c^2}&1+{c^3}\\
\end{gathered} \right|= 0\]
This can be written as
\[\left| \begin{gathered}
a&{a^2}&1+{a^3} \\
b&{b^2}&1+{b^3}\\
c&{c^2}&1+{c^3}\\
\end{gathered} \right|\] = \[\left| \begin{gathered}
a&{a^2}&1 \\
b&{b^2}&1\\
c&{c^2}&1\\
\end{gathered} \right|\] + \[\left| \begin{gathered}
a&{a^2}&{a^3} \\
b&{b^2}&{b^3}\\
c&{c^2}&{c^3}\\
\end{gathered} \right|\]

\[\left| \begin{gathered}
a&{a^2}&1+{a^3} \\
b&{b^2}&1+{b^3}\\
c&{c^2}&1+{c^3}\\
\end{gathered} \right|\] = \[\left| \begin{gathered}
a&{a^2}&1 \\
b&{b^2}&1\\
c&{c^2}&1\\
\end{gathered} \right|\] + \[\left| \begin{gathered}
a&{a^2}&{a^3} \\
b&{b^2}&{b^3}\\
c&{c^2}&{c^3}\\
\end{gathered} \right|\]
Taking a common from first row, b common from second row, and c common from third row, we get,

\[\left| \begin{gathered}
a&{a^2}&1+{a^3} \\
b&{b^2}&1+{b^3}\\
c&{c^2}&1+{c^3}\\
\end{gathered} \right|\] = \[\left| \begin{gathered}
a&{a^2}&1 \\
b&{b^2}&1\\
c&{c^2}&1\\
\end{gathered} \right|\] + \[abc\left| \begin{gathered}
1&a&{a^2} \\
1&b&{b^2}\\
1&c&{c^2}\\
\end{gathered} \right|\]
Taking determinant common from the terms, we get,
\[\vartriangle (1 + abc) = 0\]
Therefore \[abc = - 1\]

Therefore option D is the correct answer.

Note:Vector is a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity’s magnitude. Although a vector has magnitude and direction, it does not have a position. Vectors that are part of the same plane, in this way, are coplanar vectors. In contrast, vectors that belong to different planes are called non-coplanar vectors.