Let a-b and c be three non-coplanar vectors and d-sin x a-b - cos y b-c-2 c-a be a non-zero vector- which is perpendicular to a-b-c- If the minimum value of x2-y2 is equal to - 24- then the value of - is

Given, nbsp;d=sin x (a×b) + cos y (b×c)+2 (c×a) ⇒d·a=cos y[a b c]= d· (b+c) (∵d· (a+b+c)=0) ⇒cos y= d·(b+c)[a b c] ⋯(1) Similarly, sin x= d· (b+a)[a b ...

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Byju's Answer

Standard XII

Mathematics

Scalar Triple Product

Let a,b and c...

Question

Let $\to a, \to b$ and $\to c$ be three non-coplanar vectors and $\to d = sin x (\to a \times \to b) + cos y (\to b \times \to c) + 2 (\to c \times \to a)$ be a non-zero vector, which is perpendicular to $\to a + \to b + \to c .$ If the minimum value of $x^{2} + y^{2}$ is equal to $\frac{λ π^{2}}{4},$ then the value of $λ$ is

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Solution

Given, $\to d = sin x (\to a \times \to b) + cos y (\to b \times \to c) + 2 (\to c \times \to a)$
$\Rightarrow \to d \cdot \to a = cos y [\to a \to b \to c] = - \to d \cdot (\to b + \to c)$
$(∵ \to d \cdot (\to a + \to b + \to c) = 0)$
$\Rightarrow cos y = \frac{- \to d \cdot (\to b + \to c)}{[\to a \to b \to c]} \dots (1)$
Similarly, $sin x = \frac{- \to d \cdot (\to b + \to a)}{[\to a \to b \to c]} \dots (2)$
and $2 = \frac{- \to d \cdot (\to c + \to a)}{[\to a \to b \to c]} \dots (3)$

Adding equation $(1), (2)$ and $(3),$ we get
$sin x + cos y + 2 = 0$
$\Rightarrow sin x + cos y = - 2$
which is possible, only when
$sin x = - 1$ and $cos y = - 1$
$\Rightarrow x = (4 n - 1) \frac{π}{2}$ and $y = (2 n + 1) π$
Since we want minimum value of $x^{2} + y^{2},$
so $x = - \frac{π}{2}$ and $y = π$
$\Rightarrow x^{2} + y^{2} = \frac{5 π^{2}}{4}$
$∴ λ = 5$

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Let →a,→b and →c be three non-coplanar vectors and →d=sinx(→a×→b)+cosy(→b×→c)+2(→c×→a) be a non-zero vector, which is perpendicular to →a+→b+→c. If the minimum value of x2+y2 is equal to λπ24, then the value of λ is