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Unit Vectors

If | a⃗ | =...

Question

If $| \to a | = 5, | \to b | = 4,$ and $| \to c | = 3$ , then what will be the value of $| \to a, \to b + \to b \cdot \to c + \to c \cdot \to a | ?$ Given that $\to a + \to b + ¯ ¯ c = \to 0$

A

25

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B

50

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C

- 25

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D

- 50

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Solution

The correct option is A $25$
$\begin{matrix} ∣ ∣ \to a . \to b + \to b . \to c + \to c . \to a ∣ ∣ = | 16 + 0 + 9 | = 25 H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

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| \to a | = 3, ∣ ∣ \to b ∣ ∣ = 1, | \to c | = 4

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

, then the value of

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

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

| \to a | = 3, | \to b | = 5

| \to c | = 7

.

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

\to a, \to b, \to c

are unit vectors such that

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

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a =

\to a, \to b, \to c

are unit vectors such that

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

, find the value of

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

\to a, \to b, \to c

are three non-zero, non-coplanar vectors and

{\to b}_{1} = \to b - \frac{\to b \cdot \to a}{| \to a |^{2}} \to a,

{\to b}_{2} = \to b + \frac{\to b \cdot \to a}{| \to a |^{2}} \to a

{\to c}_{1} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1},

{\to c}_{2} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a - \frac{{\to b}_{1} \cdot \to c}{| {\to b}_{1} |^{2}} {\to b}_{1},

{\to c}_{3} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1}, {\to c}_{4} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a - \frac{\to b \cdot \to c}{| \to b |^{2}} {\to b}_{1}

, then the set of orthogonal vectors is

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