If b and c are non-zero and non-collinear vectors- a- b-c-a-bb-4-2x-sin yb-x2-1c and c-ca-c- Then-

A× (b× nbsp;c)+(a·b)b=(4 2x sin y)b+(x^2 1)c (a·c)b (a·b)c+(a·b)b =(4 2x sin y)b+(x^2 1)c Now, (c·c)a=c. Therefore, nbsp; (c·c)(a·c)=(c·c) or a·c=1 ⇒ 1+a ...

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Standard XIII

Mathematics

Vector Triple Product

If b and c ar...

Question

If $\to b$ and $\to c$ are non-zero and non-collinear vectors, $\to a \times (\to b \times \to c) + (\to a \cdot \to b) \to b$ $= (4 - 2 x - sin y) \to b + (x^{2} - 1) \to c$ and $(\to c \cdot \to c) \to a = \to c$ . Then,

x = 1

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x = - 1

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y = (4 n + 1) \frac{π}{2}, n \in I

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y = (2 n + 1) \frac{π}{2}, n \in I

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Solution

The correct option is C $y = (4 n + 1) \frac{π}{2}, n \in I$
$\to a \times (\to b \times \to c) + (\to a \cdot \to b) \to b$ $= (4 - 2 x - sin y) \to b + (x^{2} - 1) \to c$

$(\to a \cdot \to c) \to b - (\to a \cdot \to b) \to c + (\to a \cdot \to b) \to b$
$= (4 - 2 x - sin y) \to b + (x^{2} - 1) \to c$

Now, $(\to c \cdot \to c) \to a = \to c$ . Therefore,
$(\to c \cdot \to c) (\to a \cdot \to c) = (\to c \cdot \to c)$ or $\to a \cdot \to c = 1$

$\Rightarrow 1 + \to a \cdot \to b = 4 - 2 x - sin y$ and $x^{2} - 1 = - (\to a \cdot \to b)$
$1 = 4 - 2 x - sin y + x^{2} - 1$
$sin y = x^{2} - 2 x + 2 = (x - 1)^{2} + 1$
But $sin y \leq 1 \Rightarrow x = 1, sin y = 1$
$\Rightarrow y = (4 n + 1) \frac{π}{2}, n \in I$

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If →b and →c are non-zero and non-collinear vectors, →a×(→b×→c)+(→a⋅→b)→b=(4−2x−siny)→b+(x2−1)→c and (→c⋅→c)→a=→c. Then,

If $\to b$ and $\to c$ are non-zero and non-collinear vectors, $\to a \times (\to b \times \to c) + (\to a \cdot \to b) \to b$ $= (4 - 2 x - sin y) \to b + (x^{2} - 1) \to c$ and $(\to c \cdot \to c) \to a = \to c$ . Then,