Let b and c be two non-collinear vectors- If a is a vector such that a-b-c-4 and a-b-c-x2-2x-6b-sin yc- then x-y lies on the line

We know that, a×(b×c)=(a·c)b (a·b)c Given, a×(b×c)=(x^2 2x+6)b+(sin y)c ⇒ (a·c)b (a·b)c=(x^2 2x+6)b+(sin y)c On comparing, we get a·c=x^2 2x+6, a·b= sin y ∴a·( ...

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Byju's Answer

Standard XIII

Mathematics

Applications of Dot Product

Let b and c b...

Question

Let $\to b$ and $\to c$ be two non-collinear vectors. If $\to a$ is a vector such that $\to a \cdot (\to b + \to c) = 4$ and $a \times (\to b \times \to c) = (x^{2} - 2 x + 6) \to b + (sin y) \to c,$ 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 $x = 1$
We know that, $\to a \times (\to b \times \to c) = (\to a \cdot \to c) \to b - (\to a \cdot \to b) \to c$
Given, $\to a \times (\to b \times \to c) = (x^{2} - 2 x + 6) \to b + (sin y) \to c$
$\Rightarrow (\to a \cdot \to c) \to b - (\to a \cdot \to b) \to c = (x^{2} - 2 x + 6) \to b + (sin y) \to c$
On comparing, we get
$\to a \cdot \to c = x^{2} - 2 x + 6, \to a \cdot \to b = - sin y$
$∴ \to a \cdot (\to b + \to c) = 4 \Rightarrow x^{2} - 2 x + 2 = sin y$
$x^{2} - 2 x + 2 = (x - 1)^{2} + 1 \geq 1$ and $sin y \leq 1$
$∴$ Both sides are equal only for $x = 1$

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

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