If - a- - 2- b- - 3 and - 2a- - b- - 5- then - 2a- - b- equals-

The correct option is C 5 We have | a| = 2 | b| = 3 And, | 2a #160; b| = 5 We have to find the value of | 2a #160; + b| Now | ...

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Addition of Vectors

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If $| \to a | = 2, ∣ ∣ \to b ∣ ∣ = 3$ and $∣ ∣ 2 \to a - \to b ∣ ∣ = 5,$ then $∣ ∣ 2 \to a + \to b ∣ ∣$ equals:

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(\to A - \to B) \times (\to A + \to B)

2 (\to A \times \to B)

(\to a - \to b) \times (\to a + \to b) = 2 (\to a \times \to b)

\to a

\to b

| \to a + \to b | = \sqrt{3}

(2 \to a - 5 \to b) . (3 \to a + \to b) =

\to a

\to b

(\to a - \to b) \times [(\to b + \to a) \times (2 \to a + \to b)] = \to a + \to b

θ

\to a

\to b

\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}

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