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 b-b-2-a-b-a-b-2-b-b-a-b-a-b-2-b-b-2-a-b-a-b-None of these-

The correct option is A b⃗/|b⃗|^2+a⃗×b⃗/|a⃗×b⃗|^2 Since a ⊥ b, ∴ a, b, a× b are linerly independent ∴v⃗ can be expressed uniquely in terms of a⃗,b⃗ and (a⃗×b ...

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Byju's Answer

Standard VII

Mathematics

Change in the Value of a Number by Moving Decimal

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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Similar questions

Q. Show that

| \to a | \to b + | \to b | \to a

is perpendicular to

| \to a | \to b - | \to b | \to a

, for any two nonzero vectors

\to a

and

\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. If

\to a and \to b

are two vectors such that

| \to a + \to b | = | \to b |,

then prove that

(\to a + 2 \to b)

is perpendicular to

\to a .

$I f \to u = \to a - \to b a n d \to v = \to a + \to b a n d | \to a | = | \to b | = 2, t h e n | \to u \times \to v | =$

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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

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