If a- and b- are two vectors such that a- -b-0 and a-b-0- then

We have, a⃗.b⃗=0 and a⃗×b⃗=0⃗ ⇒ |a⃗||b⃗|cos θ=0 and |a⃗||b⃗|sin θ =0 ⇒ |a⃗|=0 or |b⃗|=0 or cos θ =0 and |a⃗|=0|b⃗|=0 or sin θ =0 ...

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Angle between Two Vectors

If a⃗ and b⃗ ...

Question

If $\to a$ and $\to b$ are two vectors such that $\to a . \to b = 0$ and $\to a \times \to b = \to 0$ , then

\to a | | \to b

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\to a ⊥ \to b

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e i t h e r p a r a l l e l o r p e r p e d i c u l a r

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Solution

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\to a

\to b

\to a . \to b = 0

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

\to a, \to b, \to c

\to a . \to b = 0

\to a + \to b = \to c

\to a, \to b, \to c

\to a . \to b = \to a . \to c, \to a \times \to b = \to a \times \to c, \to a \neq 0

\to b = \to c

\to a, \to b, \to c

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = 2, | \to c | = 4

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

\to a . \to b + 3 \to b . \to c + 3 \to c . \to a

\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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