- a-b- -c-a- - b-c- -TrueFalse

The correct option is A TrueLet a⃗ = a_1i + a_2j + a_3k b⃗ = b_1i + b_2j + b_3k c⃗ = c_1i + c_2j + c_3k (a⃗×b⃗).c⃗→ Isn’t it scalar triple product of a⃗, b⃗ an ...

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Byju's Answer

Standard IX

Mathematics

Scalar Triple Product

( a⃗×b⃗ ).c⃗=...

Question

$(\to a \times \to b) . \to c = \to a (\to b \times \to c)$

True

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False

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Solution

The correct option is A True
Let
$\to a = a_{1} ˆ i + a_{2} ˆ j + a_{3} ˆ k$
$\to b = b_{1} ˆ i + b_{2} ˆ j + b_{3} ˆ k$
$\to c = c_{1} ˆ i + c_{2} ˆ j + c_{3} ˆ k$
$(\to a \times \to b) . \to c \to$ Isn’t it scalar triple product of $\to a, \to b a n d \to c$ ?
= $= ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$
Now $\to a . (\to b \times \to c)$ can be written as $(\to b \times \to c) . \to a$ (Dot product is commutative).
Now $(\to b \times \to c) . \to a$ is again a scalar triple product given by $∣ ∣ ∣ ∣ \begin{matrix} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} a_{1} & a_{2} & a_{3} \end{matrix} ∣ ∣ ∣ ∣$
Now $∣ ∣ ∣ ∣ \begin{matrix} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} a_{1} & a_{2} & a_{3} \end{matrix} ∣ ∣ ∣ ∣$ $= ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ why?
Because in the second determinant we have just done following.
$R_{2}$ $\leftrightarrow$ $R_{3}$ and then $R_{1}$ $\leftrightarrow$ $R_{2}$ which does not change the determinant value.
In first row operation it multiplies the determinant by -1. Second row operation again multiplies it by -1. Thus it becomes equal to the first determinant.

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Similar questions

(\to a \times \to b) . \to c = \to a (\to b \times \to c)

Q. If

\to a, \to b, \to c

are any three vectors such that

(\to a + \to b) . \to c = (\to a - \to b) . \to c = 0

then

(\to a \times \to b) \times \to c

Q. If

[\to a \to b \to c] = 2

, then

\frac{\to a . (\to b \times \to c)}{(\to c \times \to a) . \to b} + \frac{\to b . (\to c \times \to a)}{(\to a \times \to b) . \to c} + \frac{\to c . (\to a \times \to b)}{(\to b \times \to c) . \to a} =

(\to a \times \to b) . (\to c \times \to d) + (\to b . \to c) (\to a . \to d) =

Q. ABCD is parallelogram. If

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

then show that

\to A B = \to a, + \to b

and

\to B D = \to b, - \to a

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(→a×→b).→c=→a(→b×→c)

$(\to a \times \to b) . \to c = \to a (\to b \times \to c)$