A and b are non zero non-collinerar vectors such that - a -2- a - b -1 and angle between a and b is -3 - If r is any vector satisfying r - a -2- r - b -8- r -2 a -10 b - a - b -6 and r -2 a -10 b - a - b then -A-B- 2 C- 4-3D- 3

The correct option is B 2Let r⃗= x a⃗+ y b⃗+z(a⃗×b⃗) r⃗. a⃗=2 ⇒ 4x+y=2 r⃗.b⃗=8, gives x+y=8 ⇒ x= 2, y=10 also by other given condition z = 2 ⇒r⃗+2 a⃗ ...

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Byju's Answer

Standard XII

Mathematics

Monotonically Increasing Functions

a and b are n...

Question

$¯ a a n d ¯ b$ are non zero non-collinerar vectors such that $| \to a | = 2, \to a . \to b = 1$ and angle between $\to a a n d \to b i s \frac{π}{3} . I f \to r$ is any vector satisfying $\to r . \to a = 2, \to r . \to b = 8, (\to r + 2 \to a - 10 \to b) . (\to a \times \to b) = 6 a n d \to r + 2 \to a - 10 \to b = λ (\to a \times \to b) t h e n λ =$

\frac{1}{2}

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\frac{4}{\sqrt{3}}

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Solution

The correct option is B 2
$L e t \to r = x \to a + y \to b + z (\to a \times \to b)$
$\to r . \to a = 2 \Rightarrow 4 x + y = 2$
$\to r . \to b = 8, g i v e s x + y = 8$
$\Rightarrow x = - 2, y = 10$ also by other given condition z = 2
$\Rightarrow \to r + 2 \to a - 10 \to b = 2 (\to a \times \to b) . H e n c e λ = 2$

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Similar questions

\to a

and

\to b

are non-zero non-collinear vectors such that |

\to a

|=2,

\to a . \to b

=1 and angle between

\to a

and

\to b

\frac{π}{3}

. If

\to r

any vector satisfying

\to r

\to a = 2, \to r . \to b

= 8,

(\to r + 2 \to a - 10) . (\to a \times \to b)

= 6 and is equal to

\to r + 2 \to a - 10 \to b = | (\to a \times \to b) |

= then I=

\to a

and

\to b

are two vectors such that

| \to a | = 1, | \to b | = 4

and

\to a \cdot \to b = 2

. If

\to c = (2 \to a \times \to b) - 3 \to b

, then the angle between

\to b

and

\to c

\to a

and

\to b

are two vectors such that

| \to a | = 1, | \to b | = 4, | \to c |^{2} = 192

and

\to a . \to b = 2

. If

\to c = (2 \to a \times \to b) - 3 \to b

, then the angle between

\to b

and

\to c

\to a

and

\to b

are two vectors such that

| \to a | = 1, | \to b | = 4, | \to c |^{2} = 192

and

\to a . \to b = 2

. If

\to c = (2 \to a \times \to b) - 3 \to b

, then the angle between

\to b

and

\to c

Q. Let

\to a

and

\to b

be two vectors such that

| \to a | = 1,

| \to b | = 4

and

\to a \cdot \to b = 2.

\to c = (2 \to a \times \to b) - 3 \to b,

then the angle between

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¯a and ¯b are non zero non-collinerar vectors such that |→a|=2, →a.→b=1 and angle between →a and →b is π3. If →r is any vector satisfying →r.→a=2,→r.→b=8, (→r+2→a−10→b).(→a×→b)=6 and →r+2→a−10→b=λ(→a×→b) then λ=

The correct option is B 2Let →r=x→a+y→b+z(→a×→b) →r.→a=2 ⇒4x+y=2 →r.→b=8, gives x+y=8 ⇒x=−2, y=10 also by other given condition z = 2 ⇒→r+2→a−10→b=2(→a×→b). Hence λ=2