Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

If  \begin{vmatrix} a & a^{2} & 1+a^{3}\\ b&b^{2} & 1+b^{3}\\ c &c^{2} & 1+c^{3} \end{vmatrix}=0  and vectors  (1,\vec{a},\vec{a}^{2}),(1,\vec{b},\vec{b}^{2}),and\; (1,\vec{c},\vec{c}\, ^{2}) are non­coplanar, then the product abc equals

  • Option 1)

    –1

  • Option 2)

    1

  • Option 3)

    0

  • Option 4)

    2

 

Answers (1)

best_answer

As we have learned

Summation of two determinants -

-

 

 \begin{vmatrix} a & a^2& (1+a)(a^2-a+1) \\ b & b^2 & (1+b)(b^2-b+1) \\ c&c^2 & (1+c)(c^2-c+1) \end{vmatrix} = 0

 

\begin{vmatrix} a & a^2& (1+a)a^2 \\ b & b^2 & (1+b)b^2 \\ c&c^2 & (1+c)c^2\end{vmatrix} + \begin{vmatrix} a & a^2& -(1+a)a \\ b & b^2 & -(1+b) b\\ c&c^2 & -(1+c) c \end{vmatrix}+ \begin{vmatrix} a & a^2& (1+a) \\ b & b^2 & (1+b) \\ c&c^2 & (1+c) \end{vmatrix} = 0 

 

\begin{vmatrix} a & a^2& a^3 \\ b & b^2 & b^3 \\ c&c^2 & c^3\end{vmatrix} +0+ \begin{vmatrix} a & a^2&1 \\ b & b^2 &1 \\ c&c^2 &1 \end{vmatrix} = 0

(abc + 1 )\begin{vmatrix} 1 & a&a^2 \\ 1& b &b^2 \\ 1 &c &c^2 \end{vmatrix} = 0 \\

\Rightarrow abc = -1

 

 

 

 

 

 

 


Option 1)

–1

Option 2)

1

Option 3)

0

Option 4)

2

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question