If -a- b- c-k- where k- 0- then -b-c-c-a- c-a-a-b- a-b-b-c-

The correct option is C k^4 We have (b̅×c̅) × (c̅×a̅) = [a̅ #160;b̅ c̅] c̅ Similarly, #160; (c̅×a̅) × (a̅×b̅) = [ #160;c̅ #160;a̅ b̅ ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Nth Term of GP

If [a⃗ b⃗ c...

Question

If $[\to a \to b \to c] = \to k$ , where $k \neq 0$ , then $[(\to b \times \to c) \times (\to c \times \to a) (\to c \times \to a) \times (\to a \times \to b) (\to a \times \to b) \times (\to b \times \to c)] =$

k

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

k^{2}

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

k^{3}

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

k^{4}

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

Open in App

Solution

Suggest Corrections

Similar questions

\to a, \to b, \to c

\to a + \to b + \to c = \to 0,

\to a \times \to b = \to b \times \to c = \to c \times \to a,

[\to a \to b \to c] = 0.

\to a, \to b, \to c

[\to a \to b \to c] = 4

[\to a \times \to b

\to b \times \to c

\to c \times \to a] = ?

\to a, \to b, \to c

\to a, \to b, \to c

\to a \times \to b = \to b \times \to c = \to c \times \to a

[\to a \to b \to c] = 0.

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

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

(\to a . \to b) \neq 0

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

Join BYJU'S Learning Program