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Work Done as Dot Product

Two vectors a...

Question

Two vectors are necessarily mutually perpendicular when

\to A + \to B = \to 0

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\to A - \to B = \to 0

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\to A \times \to B = \to 0

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\to A . \to B = 0

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Solution

The correct option is D $\to A . \to B = 0$
$\to A \cdot \to B = 0$

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Similar questions

\to a

\to b

\to a . \to b = 0

\to a \times \to b = \to 0

$\to a \neq \to 0, \to b \neq \to 0, \to a \times \to b = \to 0, \to c \times \to b = \to 0 \Rightarrow \to a \times \to c =$

(\to a + \to b) \cdot (\to a + \to b) = {| \to a |}^{2} + {∣ ∣ \to b ∣ ∣}^{2}

\to a \to b

\to a \neq \to 0, \to b \neq \to 0

| \to a | = 2, ∣ ∣ \to b ∣ ∣ = 3

\to a, \to b

\to 0, \to a + \to b, \to a - \to b

\to a, \to b, \to c

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

(\to a, \to b) = \frac{π}{3}

| \to a \times \to b | + | \to b \times \to c | + | \to c \times \to a | =

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