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If the vector...

Question

If the vectors $\to a, \to b$ and $\to c$ are represented by the sides $B C, C A$ and $A B$ , of a triangle respectively, then

A

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

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B

\to a \times \to b = \to b \times \to c = \to c \times \to a

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C

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

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D

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

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Solution

The correct option is C $\to a \times \to b = \to b \times \to c = \to c \times \to a$
here cross product gives us area
SO area of triangle be same wheather you calculate it from $\to a \times \to b, \to b \times \to c, \to c \times \to a$

$\to a \times \to b = \to b \times \to c = \to c \times \to a$
Area is a positive quantity so the sum of area cannit be zero
but dot product gives us projection So projection will different from different vector on different vector in triangle

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Similar questions

\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 | = 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 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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