If a- b- c- and d- are the unit vectors such that a-b- -c-d-1 and a-c-1-2- thenA- a- b- d- are non-coplanarB- b - d are non-parallelC- a - d are parallel andb - c are parallelD- a - b - c are non-coplanar

The correct option is B b⃗, d⃗ are non parallel Let angle between a⃗ and b⃗ be θ_1, c⃗ and d⃗ be θ_2 and a⃗×b⃗ and a⃗×b⃗ be θ. Since, (a⃗×b⃗).(c⃗×d⃗)=1 ⇒sin θ ...

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Byju's Answer

Standard IX

Mathematics

Scalar Triple Product

If a⃗, b⃗, c⃗...

Question

If $\to a, \to b, \to c$ , and $\to d,$ are the unit vectors such that $(\to a \times \to b) . (\to c \times \to d) = 1$ and $\to a . \to c = \frac{1}{2},$ then

$\to a, \to b, \to c$ are non-coplanar

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$\to a, \to b, \to d$ are non-coplanar

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$\to b, \to d$ are non-parallel

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$\to a, \to d$ are parallel and
$\to b, \to c$ are parallel

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Solution

The correct option is C
$\to b, \to d$ are non-parallel

Let angle between $\to a$ and $\to b$ be $θ_{1}$ , $\to c$ and $\to d$ be $θ_{2}$ and $\to a \times \to b$ and $\to a \times \to b$ be $θ$ .
Since, $(\to a \times \to b) . (\to c \times \to d) = 1$
$\Rightarrow sin θ_{1} . sin θ_{2} . cos θ = 1 \Rightarrow θ_{1} = 90^{\circ}, θ_{2} = 90^{\circ}, θ = 0^{\circ} \Rightarrow \to a ⊥ \to b, \to c ⊥ \to d, (\to a \times \to b) | | (\to c \times \to d) So, \to a \times \to b = k (\to c \times \to d) \Rightarrow (\to a \times \to b) . \to c = k (\to c \times \to d) . \to d and (\to a \times \to b) . \to d = k (\to c \times \to d) . \to d \Rightarrow [\to a \to b \to c] = 0 and [\to a \to b \to d] = 0$
$\Rightarrow \to a, \to b$ , $\to c$ and $\to a$ , $\to b$ , $\to d$ are coplanar vectors, so options (a) and (b) are incorrect.
$L e t \to b | | \to d \Rightarrow \to b = \pm \to d As (\to a \times \to b) . (\to c \times \to d) = 1 \Rightarrow (\to a \times \to b) . (\to c \times \to b) = \pm 1 \Rightarrow [\to a \times \to b \to c \to b] = \pm 1 \Rightarrow [\to c \to b \to a \times \to b] = \pm 1 \Rightarrow \to c . [\to b \times (\to a \times \to b)] = \pm 1 \Rightarrow \to c . [\to a - (\to b . \to a) \to b] = \pm 1 \Rightarrow \to c . \to a = \pm 1 [∵ \to a . \to b = 0]$
Which is a contradiction, so option (c) is correct.
Let option (d) be correct.
$\Rightarrow \to d = \pm \to a and \to c = \pm \to b As (\to a \times \to b) . (\to c \times \to d) = \pm 1 (\to a \times \to b) . (\to b \times \to a) = \pm 1$
which is a contradiction, so option (d) is incorrect. Alternatively, options (c) and (d) may be observed from the given figure.

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