Vector a- b- c- are three non-zero vectors such that a-b- -c-0 Let a - b - c - b - c - a -4- x 2 b -4 x cos 2- a- where a and b are non-collinear vectors and x -0-0-10 - then number of different vales of - will be

A× (b×c)+b×(c×a)= (4+x^2 nbsp; )b ( nbsp;4xcos^2 θ )a ⇒ (a. c )b (a.b)c+(b.a)c (b.c)a=(4+x^2)b (4x cos^2 θ )a So b.c=4x cos^2 θ amp; a.c=4+x^2 But b.c ...

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Byju's Answer

Standard XII

Physics

Potential Energy of a Dipole in Uniform Electric Field

Vector a⃗, b⃗...

Question

Vector $\to a, \to b, \to c$ are three non-zero vectors such that $(\to a - \to b) . \to c = 0.$
Let $\to a \times (\to b \times \to c) + \to b \times (\to c \times \to a) = (4 + x^{2}) \to b - (4 x {cos}^{2} θ) \to a,$ where $\to a$ and $\to b$ are non-collinear vectors and $x > 0, 0 < θ < 10,$ then number of different vales of $θ$ will be

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Solution

$\to a \times (\to b \times \to c) + \to b \times (\to c \times \to a) = (4 + x^{2}) \to b - (4 x {cos}^{2} θ) \to a$
$\Rightarrow (\to a . \to c) \to b - (\to a . \to b) \to c + (\to b . \to a) \to c - (\to b . \to c) \to a = (4 + x^{2}) \to b - (4 x {cos}^{2} θ) \to a$
So $\to b . \to c = 4 x {cos}^{2} θ$ & $\to a . \to c = 4 + x^{2}$
But
$\to b . \to c = \to a . \to c \Rightarrow 4 x {cos}^{2} θ = 4 + x^{2}$
$x^{2} - 4 x {cos}^{2} θ + 4 = 0$
$D = 16 {cos}^{4} θ - 16 = 16 ({cos}^{4} θ - 1)$
$D$ should be greater than or equal to $0$
$∴ {cos}^{4} θ = 1 {cos}^{2} θ = 1 \Rightarrow {sin}^{2} θ = 0$
$θ = π, 2 π, 3 π$

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