Unit vectors a- b- c are coplanar- A unit vector d is perpendicular to them- If a-b- c-d- 16- 13- 13k- and the angle between a and b is 30- then c is equal to

A, nbsp; b, c are coplanar. nbsp; ⇒ [a nbsp; b nbsp; c ]=0 nbsp; also (a×b)× (c×d)= 16î 13ĵ+ 13k̂ Hence we have [a nbsp; b nbsp; d ]c [a ...

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

Unit vectors ...

Question

Unit vectors $\to a, \to b, \to c$ are coplanar. A unit vector $\to d$ is perpendicular to them. If $(\to a \times \to b) \times (\to c \times \to d) = \frac{1}{6}^i - \frac{1}{3}^j + \frac{1}{3}^k$ and the angle between $\to a$ and $\to b$ is $30^{\circ}$ then $\to c$ is equal to

\frac{^i - 2^j + 2^k}{3}

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\frac{2^i +^j -^k}{3}

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\frac{-^i + 2^j + 3^k}{3}

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\frac{-^i + 2^j +^k}{3}

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Solution

The correct option is A $\frac{^i - 2^j + 2^k}{3}$
$\to a, \to b, \to c$ are coplanar.
$\Rightarrow [\to a \to b \to c] = 0$ also
$(\to a \times \to b) \times (\to c \times \to d) = \frac{1}{6}^i - \frac{1}{3}^j + \frac{1}{3}^k$
Hence we have $[\to a \to b \to d] \to c - [\to a \to b \to c] \to d = \frac{1}{6}^i - \frac{1}{3}^j + \frac{1}{3}^k$
$\Rightarrow [\to a \to b \to d] \to c - \to 0 = \frac{1}{6}^i - \frac{1}{3}^j + \frac{1}{3}^k$
$\Rightarrow [\to a \to b \to d] = (\to a \times \to b) \cdot \to d$
$\Rightarrow [\to a \to b \to d] = | \to a \times \to b | | \to d | cos 0^{\circ}$ or $| \to a \times \to b | | \to d | cos 180^{\circ}$
$(∵ angle between them can be any of 0^{\circ} or 180^{\circ})$
$\Rightarrow [\to a \to b \to d] = \pm (1) | \to a | | \to b | sin 30^{\circ} | \to d |$
On solving, we have $[\to a \to b \to d] = \pm \frac{1}{2}$
Hence $\to c = \pm \frac{^i - 2^j + 2^k}{3}$

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Unit vectors →a,→b,→c are coplanar. A unit vector →d is perpendicular to them. If (→a×→b)×(→c×→d)=16^i−13^j+13^k and the angle between →a and →b is 30∘ then →c is equal to

Unit vectors $\to a, \to b, \to c$ are coplanar. A unit vector $\to d$ is perpendicular to them. If $(\to a \times \to b) \times (\to c \times \to d) = \frac{1}{6}^i - \frac{1}{3}^j + \frac{1}{3}^k$ and the angle between $\to a$ and $\to b$ is $30^{\circ}$ then $\to c$ is equal to