If d is a unit vector such that d-b-c-c-a-a-b then - d - a b - c - d -b c-a - d-c a-b - is equal to

The correct option is A | [ a b⃗ c ] | d⃗.a⃗ = λ[ a⃗ amp; b⃗ amp; c⃗ ] d⃗.b⃗ = μ[ c⃗ amp; nbsp;a⃗ amp; b⃗ ] = μ[ a⃗ amp; ...

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

Area Method to Find Condition for Co-Linearity

If d is a u...

Question

If $\to d$ is a unit vector such that $\to d = λ \to b \times \to c + μ \to c \times \to a + ν \to a \times \to b$ then $∣ ∣ ∣ (\to d \cdot \to a) (\to b \times \to c) + (\to d \cdot \to b) (\to c \times \to a) + (\to d \cdot \to c) (\to a \times \to b) ∣ ∣ ∣$ is equal to

∣ ∣ [\begin{matrix} \to a & \to b & \to c \end{matrix}] ∣ ∣

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

1

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

3 ∣ ∣ [\begin{matrix} \to a & \to b & \to c \end{matrix}] ∣ ∣

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

None of these

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

Open in App

Solution

Suggest Corrections

Similar questions

\to a, \to b, \to c

\to d

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

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

(\to a - \to d) \times (\to b - \to c) = \to 0

(\to a \times \to d) \times (\to b \times \to c) = \to 0

\to a, \to b, \to c, \to d

\to a = \to b + \to c

\to b \times \to d = \to 0

\to c \cdot \to d = 0

\frac{\to d \times (\to a \times \to d)}{| \to d |^{2}}

\to a, \to b, \to c, \to d

[\to a \times \to b \to a \times \to c \to d]

\to a, \to b, \to c

\to d

\to a + \to b + \to c

\to d = x \to b \times \to c + y \to c \times \to a + z \to a \times \to b

[\to a \times \to b \to b \times \to c \to c \times \to a] = λ [\to a \to b \to c]^{2}

λ

Join BYJU'S Learning Program