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 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

Q. If

\to b, \to c, \to d

are non-coplanar vectors, then the vector

(\to a \times \to b) \times (\to c \times \to d) + (\to a \times \to c) \times (\to d \times \to b) + (\to a \times \to d) \times (\to b \times \to c)

is parallel to

Q. If

\to a, \to b, \to c, \to d

are non-coplanar vectors then the vector

(\to a \times \to b) \times (\to c \times \to d) + (\to a \times \to c) \times (\to d \times \to b) + (\to a \times \to d) \times (\to b \times \to c)

is parallel to:

Q. If

\to b, \to c, \to d

are non-coplanar vectors then the vector

(\to a \times \to b) \times (\to c \times \to d) + (\to a \times \to c) \times (\to d \times \to b) + (\to a \times \to d) \times (\to b \times \to c)

is parallel to

Q. If

\to a, \to b, \to c

and

\to d

are unit vectors such that

(\to a \times \to b) \cdot (\to c \times d) = 1

and

\to a \cdot \to c = \frac{1}{2}

, then

Q. If

\to a, \to b, \to c, \to d

are coplanar vectors then

{(\to a \times \to b) \times (\to c \times \to d)} \times (\to a - \to b) =

Join BYJU'S Learning Program

If →a,→b,→c, and →d, are the unit vectors such that (→a×→b).(→c×→d)=1 and →a.→c=12, then

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