Let b and c are non-collinear vectors- If a is a vector such that a - b - c -4 and a - b - c - x 2-2 x -6 b -sin y c- then x - y lies onthe lineA- x-y-0B- x-1C- x-y-0 D- y -

The correct option is B x = 1a⃗× (b⃗×c⃗)=(x^2 2x+6)b⃗+(sin y)c⃗ ⇒ (a⃗.c⃗)b⃗ (a⃗.b⃗)c⃗=(x^2 2x+6)b⃗+(sin y)c⃗ a⃗.c⃗=x^2 2x+6,a⃗.b⃗= sin y Given, a⃗.(b⃗+c⃗)=4 ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

Let b and c a...

Question

Let $\to b$ and $\to c$ are non-collinear vectors. If $\to a$ is a vector such that $\to a . (\to b + \to c) = 4$ and $\to a \times (\to b \times \to c) = (x^{2} - 2 x + 6) \to b + (s i n y) \to c$ , then (x, y) lies on the line

x + y = 0

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x - y = 0

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x = 1

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y = π

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Solution

The correct option is C x = 1
$\to a \times (\to b \times \to c) = (x^{2} - 2 x + 6) \to b + (s i n y) \to c \Rightarrow (\to a . \to c) \to b - (\to a . \to b) \to c = (x^{2} - 2 x + 6) \to b + (s i n y) \to c \to a . \to c = x^{2} - 2 x + 6, \to a . \to b = - s i n y$
Given, $\to a . (\to b + \to c) = 4 \Rightarrow - s i n y + x^{2} - 2 x + 6 = 4 \Rightarrow x^{2} - 2 x + 2 = s i n y \Rightarrow (x - 1)^{2} + 1 = s i n y ∴ x = 1 a n d s i n y = 1$
$∴$ (x,y) lies on the line x =1

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Let →b and →c are non-collinear vectors. If →a is a vector such that →a.(→b+→c)=4 and →a×(→b×→c)=(x2−2x+6)→b+(sin y)→c, then (x, y) lies on the line

Let $\to b$ and $\to c$ are non-collinear vectors. If $\to a$ is a vector such that $\to a . (\to b + \to c) = 4$ and $\to a \times (\to b \times \to c) = (x^{2} - 2 x + 6) \to b + (s i n y) \to c$ , then (x, y) lies on the line