Let a vector a→ has a magnitude 9. Let a vector b→be such that for every x,y∈R×R-0,0, the vector xa→+yb→ is perpendicular to the vector 6ya→-18xb→. Then the value of a→×b→ is equal to:

a⇀=9 &xa⇀+yb⇀·6ya⇀-18×b⇀=0 ⇒6xya2-18x2a·b+6y2a·b-18xyb2=0 ⇒6xya⇀2-3b2+a·by2-3x2=0 This should hold ∀x, y∈R×R ∴a⇀2=3b2 & a·b=0 Now a×b2=a2b2- a·b2 =a2·a23 ∴a×b=a23=813=273

Let a vector a→ has a magnitude 9. Let a vector b→be such that for every x,y∈R×R-0,0, the vector xa→+yb→ is perpendicular to the vector 6ya→-18xb→. Then the value of a→×b→ is equal to:

a⇀=9 &xa⇀+yb⇀·6ya⇀-18×b⇀=0 ⇒6xya2-18x2a·b+6y2a·b-18xyb2=0 ⇒6xya⇀2-3b2+a·by2-3x2=0 This should hold ∀x, y∈R×R ∴a⇀2=3b2 & a·b=0 Now a×b2=a2b2- a·b2 =a2·a23 ∴a×b=a23=813=273

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Let a vector $\vec{a}$ has a magnitude 9. Let a 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
$(6 y \vec{a} - 18 x \vec{b}) .$ Then the value of $|\vec{a} \times \vec{b}|$ is equal to:

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$9 \sqrt{3}$

$27 \sqrt{3}$

answer is B.

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Detailed Solution

$|\overset{⇀}{a}| = 9 & (x \overset{⇀}{a} + y \overset{⇀}{b}) \cdot (6 y \overset{⇀}{a} - 18 \times \overset{⇀}{b}) = 0 \Rightarrow 6 x y {|a|}^{2} - 18 x^{2} (a \cdot b) + 6 y^{2} (a \cdot b) - 18 x y {|b|}^{2} = 0 \Rightarrow 6 x y ({|\overset{⇀}{a}|}^{2} - 3 {|b|}^{2}) + (a \cdot b) (y^{2} - 3 x^{2}) = 0 T h i s s h o u l d h o l d \forall x, y \in R \times R ∴ {|\overset{⇀}{a}|}^{2} = 3 {|b|}^{2} & (a \cdot b) = 0 N o w {|a \times b|}^{2} = {|a|}^{2} {|b|}^{2} - {(a \cdot b)}^{2} = {|a|}^{2} \cdot \frac{{|a|}^{2}}{3} ∴ |a \times b| = \frac{{|a|}^{2}}{\sqrt{3}} = \frac{81}{\sqrt{3}} = 27 \sqrt{3}$