If -a- a- -b- -b and a-b-q- then the value of -a-b-

The vectors $\to a$ and $\to b$ have magnitudes 2 and 2 $\sqrt{2}$ respectively. It is found that $\to a$ . $\to b$ = | $\to a$ $\times$ $\to b$ |. Then the value of | $\frac{\to a + \to b}{\to a - \to b}$ | will be

Q. If

\to a \times (\to a \times \to b) = \to b \times (\to b \times \to c)

and

(\to a . \to b) \neq 0

, then

[\to a \to b \to c] =

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If |→a|=a, |→b| =b and →a.→b=q, then the value of |(→a×→b)|=

The correct option is D √a2b2−q2→a.→b=|a||b|cosθabcosθ=qcosθ=qab ...(i)Now |→a×→b| =absinθ=ab(√1−cos2θ)=ab(√1−q2(ab)2)=√a2b2−q2

If $| \to a | = a$ , $| \to b |$ $= b$ and $\to a . \to b = q$ , then the value of $| (\to a \times \to b) | =$

The correct option is D $\sqrt{a^{2} b^{2} - q^{2}}$
$\to a . \to b = | a | | b | cos θ$
$a b cos θ = q$
$cos θ = \frac{q}{a b}$ ...(i)
Now $| \to a \times \to b |$ $= a b sin θ$
$= a b (\sqrt{1 - {cos}^{2} θ})$
$= a b (\sqrt{1 - \frac{q^{2}}{(a b)^{2}}})$
$= \sqrt{a^{2} b^{2} - q^{2}}$