Three vectors a-b- c- are such that a-b-3a-c- - Also -a-b-1- -c-13- If the angle between b- and c- is 600- then

The correct option is A b⃗=3c⃗+a⃗ a⃗× (b⃗ 3c⃗)=0 which show b⃗ 3c⃗=λa⃗ |b⃗ 3c⃗|=λ |a⃗| √(1+99 6b⃗.c⃗)=λ √(2 1)=λ λ =1 ...

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Applications of Cross Product

Three vectors...

Question

Three vectors $\to a, \to b, \to c$ are such that $\to a \times \to b = 3 (\to a \times \to c)$ . Also $| \to a | = | \to b | = 1, | \to c | = \frac{1}{3}$ . If the angle between $\to b$ and $\to c$ is $60^{0}$ , then

\to b = 3 \to c + \to a

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\to b = 3 \to a - \to c

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\to b = 3 \to c + 2 \to a

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\to b = 3 \to c - 2 \to a

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Solution

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Similar questions

\to a, \to b, \to c

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}

\to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

\to a, \to b, \to c

\to a \neq 0

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

| \to b \times \to c | = \sqrt{15}

\to b - 2 \to c = λ \to a

λ

(\to b \times \to c) \times (\to c \times \to a) = 3 \to c

[\to b \times \to c

\to c \times \to a

\to a \times \to b] =

\to a, \to b, \to c

| \to a | = 3

| \to b | = 1

| \to c | = 2

| \to b \times \to c | = \sqrt{3}

\to b - 3 \to c = λ \to a

[λ]

[.]

\to a, \to b

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

\to c = 2 (\to a \times \to b) - 3 \to b

\to b

\to c

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