Let a- b- c- be three linearly independent vectors then -a- - 2-b- - c- 2-a- - b- - c- 4-a- - b- - 5-c-a- b- c- is

The correct option is C 0 [a⃗+2b⃗ c⃗ #160; #160; 2a⃗+b⃗+c⃗ #160; #160; 4a⃗ b⃗+5c⃗] 1 amp; 2 amp; 1 #160; 2 amp; 1 amp; 1 ...

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Multiplication with Vectors

Let a⃗, b⃗,...

Question

Let $\to a, \to b, \to c$ be three linearly independent vectors then
$\frac{[\to a + \to 2 b - \to c \to 2 a + \to b + \to c \to 4 a - \to b + \to 5 c]}{[\to a \to b \to c]}$ is

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Solution

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Similar questions

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, \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.

\to a =^i, \to b =^j

\to c =^i +^j

\to a

\to b

\to a, \to b

\to c

\to a

\to b

\to c

\to a, \to b

\to c

\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] = 5

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

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