If \overrightarrow a ,\overrightarrow b ,\overrightarrow c and \overrightarrow d are unit vectors such that (\overrightarrow a \times \overrightarrow b )\,.\,(\overrightarrow c \times \overrightarrow d ) = 1 and \overrightarrow a \,.\,\overrightarrow c = 1 \over 2 , then

\overrightarrow b \,,\overrightarrow d are non-parallel

\overrightarrow a \,,\,\overrightarrow b ,\overrightarrow c are non-coplanar

\overrightarrow b \,,\,\overrightarrow c ,\overrightarrow d are non-coplanar

\overrightarrow a ,\overrightarrow d parallel and \overrightarrow b ,\overrightarrow c are parallel

Vector Algebra PYQ — JEE Advanced 2009 · Shift 1 · Q24 with Solution

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Correct answer: C

We use the identity

(\vec a\times \vec b)\cdot(\vec c\times \vec d)= (\vec a\cdot \vec c)(\vec b\cdot \vec d)- (\vec a\cdot \vec d)(\vec b\cdot \vec c).

Given

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

and

\vec a\cdot \vec c=\frac12.

Since all vectors are unit vectors, let

\vec b\cdot \vec d=x,\qquad \vec a\cdot \vec d=y,\qquad \vec b\cdot \vec c=z,

with $x,y,z\in[-1,1]$ .

Then

1=\frac12 x-yz.

So,

yz=\frac12 x-1.

Now use bounds. Since $x\le 1$ , we get

\frac12 x-1\le \frac12-1=-\frac12.

Hence

yz\le -\frac12.

But also $|yz|\le 1$ .

For the equality

1=\frac12 x-yz

to hold, the term $\frac12 x$ is at most $\frac12$ , so we must have

-yz\ge \frac12.

In particular, $yz<0$ , so $y$ and $z$ have opposite signs.

A sharper way is to use magnitude inequality:

|(\vec a\times \vec b)\cdot(\vec c\times \vec d)|\le |\vec a\times \vec b|\,|\vec c\times \vec d|\le 1.

But the given value is exactly $1$ , so equality must hold everywhere. Therefore:

$|\vec a\times \vec b|=1$ and $|\vec c\times \vec d|=1$ ,
and $\vec a\times \vec b$ is parallel to $\vec c\times \vec d$ .

Now, for unit vectors,

|\vec a\times \vec b|=1 \implies \sin\theta_{ab}=1 \implies \vec a\perp \vec b,

and similarly,

|\vec c\times \vec d|=1 \implies \vec c\perp \vec d.

Thus,

\vec a\cdot \vec b=0,\qquad \vec c\cdot \vec d=0.

Also, since $\vec a\times \vec b$ is parallel to $\vec c\times \vec d$ , the planes of $(\vec a,\vec b)$ and $(\vec c,\vec d)$ are parallel; in fact both pairs lie in the same oriented plane structure.

Since $\vec a\cdot \vec c=\frac12$ , the angle between $\vec a$ and $\vec c$ is

\cos^{-1}\left(\frac12\right)=60^\circ.

Now in the plane perpendicular to the common cross-product direction, each of $\vec b$ and $\vec d$ is obtained by rotating $\vec a$ and $\vec c$ respectively by $90^\circ$ (with the same orientation, because the scalar triple product is positive and equals $1$ ).

Therefore the angle between $\vec b$ and $\vec d$ is also $60^\circ$ , so

\vec b\cdot \vec d=\frac12.

Hence $\vec b$ and $\vec d$ are certainly not parallel.

So option C is true.

Check the other options:

A: $\vec a,\vec b,\vec c$ are non-coplanar. Since $\vec a,\vec b$ span a plane and $\vec c$ is in the same plane as forced by the equality case above, they are coplanar. So A is false.
B: $\vec b,\vec c,\vec d$ are non-coplanar. Similarly, these vectors are coplanar. So B is false.
D: $\vec a,\vec d$ parallel and $\vec b,\vec c$ parallel. If $\vec a\parallel \vec d$ and $\vec b\parallel \vec c$ , then
$\vec a\cdot \vec c = \pm (\vec d\cdot \vec b),$
and with orthogonality structure this does not fit the given value and equality condition. In fact from the planar picture, $\vec d$ is perpendicular to $\vec c$ , not parallel to $\vec a$ in general. So D is false.

Therefore the only correct option is

\boxed{\text{C}}.

More from Vector Algebra

View written solutionFree

Correct answer: C

We use the identity

(\vec a\times \vec b)\cdot(\vec c\times \vec d)= (\vec a\cdot \vec c)(\vec b\cdot \vec d)- (\vec a\cdot \vec d)(\vec b\cdot \vec c).

Given

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

and

\vec a\cdot \vec c=\frac12.

Since all vectors are unit vectors, let

\vec b\cdot \vec d=x,\qquad \vec a\cdot \vec d=y,\qquad \vec b\cdot \vec c=z,

with $x,y,z\in[-1,1]$ .

Then

1=\frac12 x-yz.

So,

yz=\frac12 x-1.

Now use bounds. Since $x\le 1$ , we get

\frac12 x-1\le \frac12-1=-\frac12.

Hence

yz\le -\frac12.

But also $|yz|\le 1$ .

For the equality

1=\frac12 x-yz

to hold, the term $\frac12 x$ is at most $\frac12$ , so we must have

-yz\ge \frac12.

In particular, $yz<0$ , so $y$ and $z$ have opposite signs.

A sharper way is to use magnitude inequality:

|(\vec a\times \vec b)\cdot(\vec c\times \vec d)|\le |\vec a\times \vec b|\,|\vec c\times \vec d|\le 1.

But the given value is exactly $1$ , so equality must hold everywhere. Therefore:

$|\vec a\times \vec b|=1$ and $|\vec c\times \vec d|=1$ ,
and $\vec a\times \vec b$ is parallel to $\vec c\times \vec d$ .

Now, for unit vectors,

|\vec a\times \vec b|=1 \implies \sin\theta_{ab}=1 \implies \vec a\perp \vec b,

and similarly,

|\vec c\times \vec d|=1 \implies \vec c\perp \vec d.

Thus,

\vec a\cdot \vec b=0,\qquad \vec c\cdot \vec d=0.

Since $\vec a\cdot \vec c=\frac12$ , the angle between $\vec a$ and $\vec c$ is

\cos^{-1}\left(\frac12\right)=60^\circ.

Therefore the angle between $\vec b$ and $\vec d$ is also $60^\circ$ , so

\vec b\cdot \vec d=\frac12.

Hence $\vec b$ and $\vec d$ are certainly not parallel.

So option C is true.

Check the other options:

A: $\vec a,\vec b,\vec c$ are non-coplanar. Since $\vec a,\vec b$ span a plane and $\vec c$ is in the same plane as forced by the equality case above, they are coplanar. So A is false.
B: $\vec b,\vec c,\vec d$ are non-coplanar. Similarly, these vectors are coplanar. So B is false.
D: $\vec a,\vec d$ parallel and $\vec b,\vec c$ parallel. If $\vec a\parallel \vec d$ and $\vec b\parallel \vec c$ , then
$\vec a\cdot \vec c = \pm (\vec d\cdot \vec b),$
and with orthogonality structure this does not fit the given value and equality condition. In fact from the planar picture, $\vec d$ is perpendicular to $\vec c$ , not parallel to $\vec a$ in general. So D is false.

Therefore the only correct option is

\boxed{\text{C}}.

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