If the vectors a- and b- are perpendicular to each other- then a vector v- in terms of a- and b- satisfying the equations v- a- -0- v- b-1 and -v- a- b-1 is

Since a ⊥ b, ∴ a, b, a× b are linerly independent ∴v⃗ can be expressed uniquely in terms of a⃗,b⃗ and (a⃗×b⃗) Let v⃗ = xa⃗+yb⃗+z(a⃗×b⃗) Given: a⃗.b⃗=0, v⃗.a⃗= ...

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Byju's Answer

Standard XIII

Mathematics

Linear Dependence and Independence of Vectors

If the vector...

Question

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

$\frac{\to b}{| \to b |^{2}} + \frac{\to a \times \to b}{| \to a \times \to b |^{2}}$

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$\frac{\to b}{| \to b |} + \frac{\to a \times \to b}{| \to a \times \to b |^{2}}$

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$\frac{\to b}{| \to b |^{2}} + \frac{\to a \times \to b}{| \to a \times \to b |}$

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None of these

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Solution

The correct option is A
$\frac{\to b}{| \to b |^{2}} + \frac{\to a \times \to b}{| \to a \times \to b |^{2}}$

Since $a ⊥ b, ∴ a, b, a \times b$ are linerly independent
$∴ \to v$ can be expressed uniquely in terms of $\to a, \to b a n d (\to a \times \to b)$
Let $\to v = x \to a + y \to b + z (\to a \times \to b)$
Given: $\to a . \to b = 0, \to v . \to a = 0, \to v . \to b = 1, [\to v \to a \to b] = 1 \Rightarrow \to v . \to a = x | \to a |^{2} [∵ \to a . \to b = 0, \to a . \to a \times \to b = 0] \Rightarrow 0 = x a^{2} \Rightarrow x = 0 \to v . \to b = y b^{2} \Rightarrow 1 = y b^{2} ∴ y = \frac{1}{b^{2}} \to v . \to a \times \to b = z | \to a \times \to b |^{2} \Rightarrow 1 = z | \to a \times \to b |^{2} \Rightarrow z = \frac{1}{| \to a \times \to b |^{2}} H e n c e, \to v = \frac{1}{| \to b |^{2}} \to b + \frac{1}{| \to a \times \to b |^{2}} (\to a \times b)$

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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