If a-b-c- are three unit vectors- each is inclined at an angle -3 with the other- then scalar triple product - a-b-c- - equals

The correct option is B 1√(2) Given a⃗, b⃗, c⃗ are unit vectors |a⃗|=|b⃗|=|c⃗|=1 Now, a⃗. b⃗=|a⃗||b⃗|cosπ3 ⇒ #160;a⃗ . #160;b⃗=1/2 ...

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If a⃗,b⃗,c⃗...

Question

If $\to a, \to b, \to c$ are three unit vectors, each is inclined at an angle $\frac{π}{3}$ with the other, then scalar triple product $[\to a \to b \to c]$ equals

\frac{1}{2}

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\frac{1}{\sqrt{2}}

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\frac{1}{\sqrt{3}}

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\frac{1}{4}

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Solution

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Similar questions

\to a, \to b, \to c

\frac{π}{3}

| \to a | = 1, | \to b | = 2, | \to c | = 3

2 \to a + 3 \to b - 5 \to c

4 \to a - 6 \to b + 10 \to c

\to a, \to b, \to c

\frac{π}{3}

\to a, \to b, \to c

\to a, \to b, \to c

\frac{π}{3}

\to a = a_{1}^i + a_{2}^j + a_{3}^k; \to b = b_{1}^i + b_{2}^j + b_{3}^k; \to c = c_{1}^i + c_{2}^j + c_{3}^k

A

3 \times 3

a d j (a d j A) = ⎡ ⎢ ⎣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎦

| A |^{4} =

∣ \to a ∣= 1, ∣ \to b ∣= 2

∣ \to c ∣= 3

\to a, \to b, \to c

∣ \to d ∣= 4

∣ \frac{[\to d \to b \to c] \to a + [\to d \to c \to a] \to b + [\to d \to a \to b] \to c}{[\to a \to b \to c]} ∣

\to a, \to b

\to c

[2 \to a - \to b 2 \to b - \to c 2 \to c - \to a]

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