If a- - b- - c- b-d- - 0 and c- - d- - 0 then d- a-d-d-2- is equal to

d⃗× ( a⃗×d⃗ )d⃗^2 =d⃗× [ ( b⃗+c⃗ )×d⃗ ]d⃗^2 =d⃗× [ b⃗×d⃗ +c⃗×d⃗ ]d⃗^2 =d⃗× [ 0 +c⃗×d⃗ ]d⃗^2 = ( d⃗·d⃗ )c⃗ ( d⃗·c⃗ )d⃗d⃗^2 #160; ...

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

Complex Numbers

If a⃗ = b⃗ ...

Question

If $\to a = \to b + \to c, \to b \times \to d = 0$ and $\to c . \to d = 0$ then $\frac{\to d \times (\to a \times \to d)}{^{2}}$ is equal to

Open in App

Solution

Suggest Corrections

Similar questions

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

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

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

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

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

k

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

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

\to b, \to c, \to d

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

\to b, \to c, \to d

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

Join BYJU'S Learning Program