MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Scalar Triple Product

[ a⃗,b⃗, c⃗ ]...

Question

$[\to a, \to b, \to c] \neq [\to b, \to c, \to a] \neq [\to c, \to a, \to b]$

A

True

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

False

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B False
Let
$\to a = a_{1}^i + a_{2}^j + a_{3}^k$
$\to b = b_{1}^i + b_{2}^j + b_{3}^k$
$\to c = c_{1}^i + c_{2}^j + c_{3}^k$
Then $[\to a, \to b, \to c] = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$
$[\to b, \to c, \to a] = ∣ ∣ ∣ ∣ \begin{matrix} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} a_{1} & a_{2} & a_{3} \end{matrix} ∣ ∣ ∣ ∣ = - ∣ ∣ ∣ ∣ \begin{matrix} b_{1} & b_{2} & b_{3} a_{1} & a_{2} & a_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = [\to a, \to b, \to c]$
$[\to c, \to a, \to b] = ∣ ∣ ∣ ∣ \begin{matrix} c_{1} & c_{2} & c_{3} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ ∣ = - ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} c_{1} & c_{2} & c_{3} b_{1} & b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = [\to a, \to b, \to c]$
So $[\to a, \to b, \to c] = [\to b, \to c, \to a] = [\to c, \to a, \to b]$
In the above determinants, we have interchanged the rows twice. First when we interchange the rows then the determinant is multiplied by (-1). Then when we again interchange two rows its again multiplied by -1.
So value of scalar triple product does not change when this cyclic order is maintained.

Suggest Corrections

thumbs-up

0

Similar questions

(\to a + \to b) \times (\to a - \to b) + (\to b + \to c) \times (\to b - \to c) + (\to c + \to a) \times (\to c - \to a) =

\to A = \to b \times \to c, \to B = \to c \times \to a, \to C = \to a \times \to b

, then the vectors

\to A \times (\to B \times \to C)

\to B \times (\to C \times \to A)

\to C \times (\to A \times \to B)

[\to a + 2 \to b - \to c, \to a - \to b, \to a - \to b - \to c] =

[\to a, \to b, \to c] \neq [\to b, \to c, \to a] \neq [\to c, \to a, \to b]

Q. The value of

[\to a - \to b

\to b - \to c

\to c - \to a] + [\to a \times \to b

\to b \times \to c

\to c \times \to a]

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Scalar Triple Product

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Scalar Triple Product

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon