Unit vector perpendicular to both A and B is A- A - B - A - B -A-B-A-B- cos-A - B - A - B - sin-A - B - A - B - sin-

The correct option is C A×B|A| |B| sinθLet A×B= C C will be perpendicular to both nbsp; nbsp;Aand nbsp;B. A×B= |A| |B| sinθ ĉ A×B|A| |B| sinθ=ĉ nbsp; ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

Unit vector p...

Question

Unit vector perpendicular to both $\to A$ and $\to B$ is

\frac{\to A \times \to B}{| \to A | | \to B |}

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\frac{\to A \cdot \to B}{| \to A | | \to B | cos θ}

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\frac{\to A \times \to B}{| \to A | | \to B | sin θ}

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\frac{\to A \cdot \to B}{| \to A | | \to B | sin θ}

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Solution

The correct option is C $\frac{\to A \times \to B}{| \to A | | \to B | sin θ}$
Let $\to A \times \to B = \to C$
$\to C$ will be perpendicular to both $\to A and \to B$ .
$\to A \times \to B = | \to A | | \to B | sin θ^c$
$\frac{\to A \times \to B}{| \to A | | \to B | sin θ} =^c$

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Similar questions

Q. Let

\to a = 3^i + 2^j + 2^k, b =^i + 2^j - 2^k

. Then a unit vector perpendicular to both

\to a - \to b

and

\to a + \to b

is :

Q. Projection of

\to A

along the vector

\to B

If the vectors $\to a$ and $\to b$ are perpendicular to each other, then a vector $\to v$ in terms of $\to a$ and $\to b$ satisfying the equations
$\to v . \to a = 0, \to v . \to b = 1 a n d [\to v \to a \to b] = 1$ is

Q. Which of the following expression is equal to

cos θ

, where

θ

is angle between

\to A and \to B

Q. Which of the following is the unit vector perpendicular to

\to A and \to B

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Unit vector perpendicular to both→A and →B is

The correct option is C →A×→B|→A| |→B| sinθLet →A×→B=→C →C will be perpendicular to both →Aand →B. →A×→B=|→A| |→B| sinθ ^c →A×→B|→A| |→B| sinθ=^c

Unit vector perpendicular to both $\to A$ and $\to B$ is

The correct option is C $\frac{\to A \times \to B}{| \to A | | \to B | sin θ}$
Let $\to A \times \to B = \to C$
$\to C$ will be perpendicular to both $\to A and \to B$ .
$\to A \times \to B = | \to A | | \to B | sin θ^c$
$\frac{\to A \times \to B}{| \to A | | \to B | sin θ} =^c$