For any two vector A and B - if A-B-A-B - the magnitude of C-A-B is

The correct option is B √(A^2+B^2+√(2)AB) |A.B|=|A×B| ⇒ AB cosθ= AB sin θ ⇒ tanθ= 1⇒ cosθ=1√(2) Now, |A+B|=√(A^2+B^2+2AB cosθ) #160; ...

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Standard XII

Mathematics

Cross Product of Two Vectors

For any two v...

Question

For any two vector $\to A$ and $\to B$ , if $\to A . \to B = | \to A \times \to B |$ , the magnitude of $\to C = \to A + \to B$ is

A + B

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\sqrt{A^{2} + B^{2} + \sqrt{2} A B}

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\sqrt{A^{2} + B^{2}}

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\sqrt{A^{2} + B^{2} + \frac{A B}{\sqrt{2}}}

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Solution

The correct option is B $\sqrt{A^{2} + B^{2} + \sqrt{2} A B}$
$| \to A . \to B | = | \to A \times \to B | \Rightarrow A B c o s θ = A B s i n θ$
$\Rightarrow t a n θ = 1 \Rightarrow c o s θ = \frac{1}{\sqrt{2}}$
Now, $| \to A + \to B | = \sqrt{A^{2} + B^{2} + 2 A B c o s θ}$ $= \sqrt{A^{2} + B^{2} + 2 A B . \frac{1}{\sqrt{2}}}$ $= \sqrt{A^{2} + B^{2} + \sqrt{2} A B}$

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Similar questions

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
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c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

The lowest form of $\frac{(a + b)}{a b} - \frac{2}{(a + b)}$ if ab=1.

Q. If AB = A and BA = B, where A and B are square matrices, then
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Find:
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(b) a² + b² − c² + 2ab = 4s (s − c)

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For any two vector →A and →B , if →A.→B=|→A×→B|, the magnitude of →C=→A+→B is

The correct option is B √A2+B2+√2AB|→A.→B|=|→A×→B|⇒ABcosθ=ABsinθ⇒tanθ=1⇒cosθ=1√2Now, |→A+→B|=√A2+B2+2ABcosθ =√A2+B2+2AB.1√2 =√A2+B2+√2AB

For any two vector $\to A$ and $\to B$ , if $\to A . \to B = | \to A \times \to B |$ , the magnitude of $\to C = \to A + \to B$ is