Let a and b be two vectors such that -a-3- -b-6 and -a-b-7- Then the value of 3a-2b-2a-5b-4a-b is

(3a 2b) · (2a+5b 4a×b) =6|a|^2+15a·b 3a· (4a×b) 4a·b 10|b|^2+2b· (4a×b) Now 3a· (4a×b)=0 and nbsp;2b· (4a×b)=0 because nbsp;a and nbsp;b will be pe ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

Let a and b b...

Question

Let $\to a$ and $\to b$ be two vectors such that $| \to a | = 3$ , $| \to b | = 6$ and $| \to a + \to b | = 7$ . Then the value of $(3 \to a - 2 \to b) . (2 \to a + 5 \to b - 4 \to a \times \to b)$ is

- 306

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- 295

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- 284

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- 290

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Solution

The correct option is C $- 284$
$(3 \to a - 2 \to b) \cdot (2 \to a + 5 \to b - 4 \to a \times \to b)$
$= 6 | \to a |^{2} + 15 \to a \cdot \to b - 3 \to a \cdot (4 \to a \times \to b) - 4 \to a \cdot \to b - 10 | \to b |^{2} + 2 \to b \cdot (4 \to a \times \to b)$

Now $3 \to a \cdot (4 \to a \times \to b) = 0$ and $2 \to b \cdot (4 \to a \times \to b) = 0$ because $\to a$ and $\to b$ will be perpendicular to $4 \to a \times \to b$

So the equation reduces to
$(3 \to a - 2 \to b) \cdot (2 \to a + 5 \to b - 4 \to a \times \to b) = 6 | \to a |^{2} + 15 \to a \cdot \to b - 4 \to a \cdot \to b - 10 | \to b |^{2}$
$= 6 (3)^{2} + 11 \to a \cdot \to b - 10 (6)^{2}$
$= 11 \to a \cdot \to b - 306 \dots (1)$

Now $| \to a + \to b | = 7$
$\Rightarrow | \to a + \to b |^{2} = 7^{2}$
$\Rightarrow | \to a |^{2} + 2 \to a \cdot \to b + | \to b |^{2} = 7^{2}$
$\Rightarrow 3^{2} + 2 \to a \cdot \to b + 6^{2} = 7^{2}$
$\Rightarrow \to a \cdot \to b = 2$
Putting this value in equation $(1),$ we get
$(3 \to a - 2 \to b) \cdot (2 \to a + 5 \to b - 4 \to a \times \to b) = 11 \times 2 - 306 = - 284$

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Similar questions

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Let →a and →b be two vectors such that |→a|=3, |→b|=6 and |→a+→b|=7. Then the value of (3→a−2→b).(2→a+5→b−4→a×→b) is

Let $\to a$ and $\to b$ be two vectors such that $| \to a | = 3$ , $| \to b | = 6$ and $| \to a + \to b | = 7$ . Then the value of $(3 \to a - 2 \to b) . (2 \to a + 5 \to b - 4 \to a \times \to b)$ is