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

The correct options are C x=1 D y=π2 a×(b×c)=(a.c).b (a.b).c ∴a.c=x^2 2x+6= sin y a.(b+c)=4 ⇒ sin y+x^2 2x+6=4 ⇒x^2 2x+2=sin y ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Let b and ...

Question

Let $\to b$ and $\to c$ be non collinear vectors.If $\to a$ is a vector such that $\to a . (\to b + \to c) = 4$ and $\to a \times (\to b \times \to c) = (x^{2} - 2 x + 6) \to b + sin \to c$ 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 = \frac{π}{2}

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Solution

The correct options are
C $x = 1$
D $y = \frac{π}{2}$
$\to a \times (\to b \times \to c) = (\to a . \to c) . \to b - (\to a . \to b) . \to c$
$∴ \to a . \to c = x^{2} - 2 x + 6 = - sin y$
$\to a . (\to b + \to c) = 4 \Rightarrow - sin y + x^{2} - 2 x + 6 = 4$
$\Rightarrow x^{2} - 2 x + 2 = sin y$
$\Rightarrow {(x - 1)}^{2} + 1 = sin y$
Leftside $\geq 1$ , right side $\leq 1$
$∴$ they are equal if ${(x - 1)}^{2} + 1 = sin y = 1$
$∴ y = \frac{π}{2}, x = 1$

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

The correct options are C x=1 D y=π2→a×(→b×→c)=(→a.→c).→b−(→a.→b).→c∴→a.→c=x2−2x+6=−siny→a.(→b+→c)=4⇒−siny+x2−2x+6=4⇒x2−2x+2=siny⇒(x−1)2+1=sinyLeftside ≥1, right side ≤1∴ they are equal if (x−1)2+1=siny=1∴y=π2,x=1

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