Vector x- is Vectors x-y- and z- each of magnitude -2- make an angle of 60- with each other-x- y-z- -a-y- z-x- -b- and x-y-c-

Given that |x⃗|=|y⃗|=|z⃗|=√(2) and they are inclined at an angle of 60^∘ with each other. ∴ nbsp;x⃗.y⃗=y⃗.z⃗=z⃗.x⃗=√(2).√(2). cos 60^∘ nbsp;=1 ⇒ nbsp;x⃗× ...

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Byju's Answer

Standard IX

Mathematics

Linear Dependence and Independence of Vectors

Vector x⃗ is ...

Question

Vector $\to x$ is Vectors $\to x, \to y$ and $\to z,$ each of magnitude $\sqrt{2},$ make an angle of $60^{\circ}$ with each other.
$\to x \times (\to y \times \to z) = \to a, \to y \times (\to z \times \to x) = \to b$ and $\to x \times \to y = \to c .$

\frac{1}{2} [(\to a - \to b) \times \to c + (\to a + \to b)]

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\frac{1}{2} [(\to a + \to b) \times \to c + (\to a - \to b)]

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\frac{1}{2} [- (\to a + \to b) \times \to c + (\to a + \to b)]

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\frac{1}{2} [(\to a + \to b) \times \to c - (\to a + \to b)]

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Solution

The correct option is D $\frac{1}{2} [(\to a + \to b) \times \to c - (\to a + \to b)]$
given that $| \to x | = | \to y | = | \to z | = \sqrt{2}$ and they are inclined at an angle of $60^{\circ}$ with each other.
$∴ \to x . \to y = \to y . \to z = \to z . \to x = \sqrt{2} . \sqrt{2} . c o s 60^{\circ} = 1$
$\Rightarrow \to x \times (\to y \times \to z) = \to a$
or $(\to x . \to z) \to y - (\to x . \to y) \to z = \to a$ or $\to y - \to z = \to a (i)$
Similarly, $\to y \times (\to z \times \to x) = \to b \Rightarrow \to z - \to x = \to b (i i)$
$\to y = \to a + \to z, \to x = \to z - \to b$ [from (i) and (ii) ] (iii)
Now, $\to x \times \to y = \to c \Rightarrow (\to z - \to b) \times (\to z + \to a) = \to c$
or $\to z \times \to a - \to b \times \to z - \to b \times \to a = \to c$
or $\to z \times (\to a + \to b) = \to c + (\to b \times \to a) . . . . (i v)$
or $(\to a + \to b) \times {\to z + (\to a + \to b)} = (\to a + \to b) \times \to c + (\to a + \to b) \times (\to b \times \to a)$
or $(\to a + \to b)^{2} \to z - {(\to a + \to b) . \to z} (\to a + \to b) =$
$(\to a + \to b) \times \to c + | \to a |^{2} \to b - | \to b |^{2} \to a + (\to a . \to b) (\to b - \to a) . . . . (v)$
Now, $(i) \Rightarrow | \to a |^{2} = | \to y - \to z |^{2} = 2 + 2 - 2 = 2$
Similarly, $(i i) \Rightarrow | \to b |^{2} = 2$ Also (i) and (ii)
$\Rightarrow \to a + \to b = \to y - \to x \Rightarrow | \to a + \to b |^{2} = 2 . . . (v i)$
Also $(\to a + \to b) . \to z = (\to y - \to z) . \to z = \to y . \to z - \to x . \to z = 1 - 1 = 0$ and
$\to a . \to b = (\to y - \to z) . (\to z - \to x) = \to y . \to z - \to x . \to y - | \to z |^{2} + \to x . \to z = - 1$
Thus from (v), we have
$2 \to z = (\to a + \to b) \times \to c + 2 (\to b - \to a) - (\to b - \to a)$
or $\to z = (1 / 2) [(\to a + \to b) \times \to c + \to b - \to a]$
$∴ \to y = \to a + \to z = (1 / 2) [(\to a + \to b) \times \to c + \to b + \to a]$ and
$\to x = \to z - \to b = (1 / 2) [(\to a + \to b) \times \to c - (\to a + \to b)]$

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Vector →x is Vectors →x,→y and →z, each of magnitude √2, make an angle of 60∘ with each other. →x×(→y×→z)=→a,→y×(→z×→x)=→b and →x×→y=→c.

Vector $\to x$ is Vectors $\to x, \to y$ and $\to z,$ each of magnitude $\sqrt{2},$ make an angle of $60^{\circ}$ with each other.
$\to x \times (\to y \times \to z) = \to a, \to y \times (\to z \times \to x) = \to b$ and $\to x \times \to y = \to c .$