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

The correct option is D 2 HINT: Let r⃗=Xa⃗+Yb⃗+Z(a⃗×b⃗) r⃗.a⃗=2 0x3f7 4X+Y=2 r⃗.b⃗=8,gives X+Y=8 0x3f7=X= 2,Y=10 also by other given condition Z=2 0x3f7 r⃗ ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

a and b are n...

Question

$\to a$ 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 $\to b$ is $\frac{π}{3}$ . 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=

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Solution

The correct option is B 2
HINT: Let $\to r = X \to a + Y \to b + Z (\to a \times \to b)$
$\to r . \to a = 2$
$Ϸ 4 X + Y = 2$
$\to r . \to b = 8, g i v e s X + Y = 8$
$Ϸ = X = - 2, Y = 10$
also by other given condition Z=2
$Ϸ \to r + 2 \to a - 10 \to b = 2 (\to a \times \to b)$ .Hence I=2

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

The correct option is B 2 HINT: Let →r=X→a+Y→b+Z(→a×→b) →r.→a=2 Ϸ 4X+Y=2 →r.→b=8,gives X+Y=8 Ϸ=X=−2,Y=10 also by other given condition Z=2 Ϸ →r+2→a−10→b=2(→a×→b).Hence I=2

The correct option is B 2
HINT: Let $\to r = X \to a + Y \to b + Z (\to a \times \to b)$
$\to r . \to a = 2$
$Ϸ 4 X + Y = 2$
$\to r . \to b = 8, g i v e s X + Y = 8$
$Ϸ = X = - 2, Y = 10$
also by other given condition Z=2
$Ϸ \to r + 2 \to a - 10 \to b = 2 (\to a \times \to b)$ .Hence I=2