SATHEE: Vectors 3 Question 24

Vectors 3 Question 24

24. If $| \begin{array}{lll} a & a^{2} & 1 + a^{3} b & b^{2} & 1 + b^{3} c & c^{2} & 1 + c^{3} \end{array} | = 0$ and the vectors $\vec{A} = (1, a, a^{2}), \vec{B} = (1, b, b^{2}), \vec{C} (1, c, c^{2})$ are non-coplanar, then the product a b c=. .

(1985, 2M)

Show Answer

Answer:

Correct Answer: 24. $(- 1)$

Solution:

Since, $| \begin{array}{lll} a & a^{2} & 1 + a^{3} b & b^{2} & 1 + b^{3} c & c^{2} & 1 + c^{3} \end{array} | = 0$ $\Rightarrow | \begin{array}{lll} a & a^{2} & 1 b & b^{2} & 1 c & c^{2} & 1 \end{array} | + | \begin{array}{ccc} a & a^{2} & a^{3} b & b^{2} & b^{3} c & c^{2} & c^{3} \end{array} | = 0$

$\Rightarrow (1 + a b c) | \begin{array}{lll} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{array} | = 0$

$\Rightarrow$ Either $(1 + a b c) = 0$ or $| \begin{array}{lll} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{array} | = 0$

But $(1, a, a^{2}), (1, b, b^{2}), (1, c, c^{2})$ are non-coplanar.

$\begin{array}{ll} \Rightarrow & | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | \neq 0 \\ ∴ & a b c = - 1 \end{array}$

Vectors 3 Question 24

24. If $| \begin{array}{lll} a & a^{2} & 1 + a^{3} b & b^{2} & 1 + b^{3} c & c^{2} & 1 + c^{3} \end{array} | = 0$ and the vectors $\vec{A} = (1, a, a^{2}), \vec{B} = (1, b, b^{2}), \vec{C} (1, c, c^{2})$ are non-coplanar, then the product a b c=. .

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Vectors 3 Question 24

24. If |aa21+a3 bb21+b3 cc21+c3|=0 and the vectors A→=(1,a,a2),B→=(1,b,b2),C→(1,c,c2) are non-coplanar, then the product a b c=. .

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

24. If $| \begin{array}{lll} a & a^{2} & 1 + a^{3} b & b^{2} & 1 + b^{3} c & c^{2} & 1 + c^{3} \end{array} | = 0$ and the vectors $\vec{A} = (1, a, a^{2}), \vec{B} = (1, b, b^{2}), \vec{C} (1, c, c^{2})$ are non-coplanar, then the product a b c=. .