If the non zero vectors a- and b- are perpendicular to each other then the solution of the equation r-a-b- is

The correct option is A r⃗=xa⃗+1/a⃗.a⃗ ( a⃗×b⃗ ) Given: ab [a.b=0] Since, a⃗.b⃗=abcos90^∘ Since, a,b,a×b are non coplanar; r=xa+y ...

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Algebra of Limits

If the non ze...

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If the non zero vectors $\to a$ and $\to b$ are perpendicular to each other then the solution of the equation $\to r \times \to a = \to b$ is

\to r = x \to a + \frac{1}{\to a . \to a} (\to a \times \to b)

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\to r = x \to b - \frac{1}{\to b . \to b} (\to a \times \to b)

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\to r = x (\to a \times \to b)

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none of these

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\to a, \to b, \to c

\to r

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

\to R

\to A

\to A

\to R

\to A

\to R = (\frac{\to R . \to A}{\to A . \to A}) \to A - (\frac{1}{\to A . \to A}) (\to A \times (\to A \times \to R))

\to a, \to b, \to c

\to a + \to b, \to b + \to c, \to c + \to a

\to r \times \to a = \to b, \to r . \to a = 1,

\to r = \to a + \to a \times \to b .

\to r \times \to a = \to b

\to r = \to a \times \to b

\to a, \to b, \to c

\to r

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

\to r \times \to a = \to b \times \to a; \to r \times \to b = \to a \times \to b; \to a \neq 0, \to b \neq 0, \to a \neq λ \to b; \to a

\to b

\to r =

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