If a- - b- -c- are vectors of lengths 2-3-4- If a- is orthogonal to b-c- -b- is orthogonal to c-a- and c- is orthogonal to a-b- then the modulus of a-b-c- is

The correct option is B √(29) Given Orthogonal condition a·(b+c)=0 #160; —–(1) b·(c+a)=0 #160; —–(2) c·(a+b)=0 #160; —–(3)Now, ...

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Standard Formulae - 2

If a⃗ , b⃗ ...

Question

If $\to a, \to b, \to c$ are vectors of lengths $2, 3, 4$ . If $\to a$ is orthogonal to $\to b + \to c, \to b$ is orthogonal to $\to c + \to a$ and $\to c$ is orthogonal to $\to a + \to b$ , then the modulus of $\to a + \to b + \to c$ is

\sqrt{29}

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\sqrt{39}

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\sqrt{45}

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2 \sqrt{19}

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Solution

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∣ ∣ \to a + \to b + \to c ∣ ∣

\to a, \to b, \to c

{\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}

\to A = \to b \times \to c, \to B = \to c \times \to a, \to C = \to a \times \to b

\to A \times (\to B \times \to C)

\to B \times (\to C \times \to A)

\to C \times (\to A \times \to B)

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