Let a vector a→ has magnitude 8. Let the vector b→ be such that for every (x,y)∈R×R−{(0,0)} the vector xa→+yb→ is perpendicular to the vector (3ya→−6xb→) then the value of |a→×b→| is equal to

Given |a→|=8(xa→+yb→).(3ya→−6xb→)=0 ⇒3xy(|a→|2−2|b→|2)+3(a→.b→)(y2−2x2)=0 ⇒ ∴ |a→|2=2|b→|2=0 and a→ . b→=0Now |a→ × b→|2=|a→|2|b→|2 sin2θ=|a→ |2.(|a→|22).1=|a→|42 ⇒|a→×b→|=|a→|22=642=322

Let a vector a→ has magnitude 8. Let the vector b→ be such that for every (x,y)∈R×R−{(0,0)} the vector xa→+yb→ is perpendicular to the vector (3ya→−6xb→) then the value of |a→×b→| is equal to

Given |a→|=8(xa→+yb→).(3ya→−6xb→)=0 ⇒3xy(|a→|2−2|b→|2)+3(a→.b→)(y2−2x2)=0 ⇒ ∴ |a→|2=2|b→|2=0 and a→ . b→=0Now |a→ × b→|2=|a→|2|b→|2 sin2θ=|a→ |2.(|a→|22).1=|a→|42 ⇒|a→×b→|=|a→|22=642=322

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Let a vector $\vec{a}$ has magnitude 8. Let the vector $\vec{b}$ be such that for every $(x, y) \in R \times R - {(0, 0)}$ the vector $x \vec{a} + y \vec{b}$ is perpendicular to the vector $(3 y \vec{a} - 6 x \vec{b})$ then the value of $| \vec{a} \times \vec{b} |$ is equal to

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$64 \sqrt{2}$

$32 \sqrt{2}$

$16 \sqrt{2}$

$8 \sqrt{2}$

answer is B.

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Detailed Solution

Given $| \vec{a} | = 8$
$(x \vec{a} + y \vec{b}) . (3 y \vec{a} - 6 x \vec{b}) = 0$
$\Rightarrow 3 x y ({| \vec{a} |}^{2} - 2 {| \vec{b} |}^{2}) + 3 (\vec{a} . \vec{b}) (y^{2} - 2 x^{2}) = 0$
$\Rightarrow ∴ {| \vec{a} |}^{2} = 2 {| \vec{b} |}^{2} = 0$ and $\vec{a} . \vec{b} = 0$
Now ${| \vec{a} \times \vec{b} |}^{2} = {| \vec{a} |}^{2} {| \vec{b} |}^{2} \sin^{2} θ$
$= {| \vec{a} |}^{2} . (\frac{{| \vec{a} |}^{2}}{2}) .1 = \frac{{| \vec{a} |}^{4}}{2}$
$\Rightarrow | \vec{a} \times \vec{b} | = \frac{{| \vec{a} |}^{2}}{2} = \frac{64}{\sqrt{2}} = 32 \sqrt{2}$