Question

Let $\hat{i},\hat{j},\hat{k}$ be three mutually or orthogonal unit vectors and let $x,y,z$ be three distinct real numbers. If the vectors $\overrightarrow{u}=x\hat{i}+y\hat{j}+z\hat{k},\overrightarrow{v}=y\hat{i}+z\hat{j}+x\hat{k}$ and $\overrightarrow{w}=z\hat{i}+x\hat{j}+y\hat{k}$ are coplanar then

• A. $\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\overrightarrow{0}$, where $\overrightarrow{0}$ is the null vector
• B. $\hat{i}+\hat{j}+\hat{k}$ is normal to the plane containing $\overrightarrow{u},\overrightarrow{v}$ and $\overrightarrow{w}$
• C. The angle between any two of the vectors $\overrightarrow{u},\overrightarrow{v}$ and $\overrightarrow{w}$ is ${120}^{\circ }$
• D. $\overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{v}\times \overrightarrow{w}=\overrightarrow{w}\times \overrightarrow{u}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\overrightarrow{0}$, where $\overrightarrow{0}$ is the null vector
B The angle between any two of the vectors $\overrightarrow{u},\overrightarrow{v}$ and $\overrightarrow{w}$ is ${120}^{\circ }$
C $\hat{i}+\hat{j}+\hat{k}$ is normal to the plane containing $\overrightarrow{u},\overrightarrow{v}$ and $\overrightarrow{w}$
D $\overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{v}\times \overrightarrow{w}=\overrightarrow{w}\times \overrightarrow{u}$

Any three vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are co-planar only if the value of $\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})$ or $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

$\Rightarrow \overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})$ is also written as $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$

Now the given vector are $\overrightarrow{u},\overrightarrow{v}$ and $\overrightarrow{w}$, which are given as:

$\overrightarrow{u}=x\hat{i}+y\hat{j}+z\hat{k}$

$\overrightarrow{v}=y\hat{i}+z\hat{j}+x\hat{k}$

$\overrightarrow{w}=z\hat{i}+x\hat{j}+zy\hat{k}$

So if these vectors are given co-planar then value of $\overrightarrow{u}.(\overrightarrow{v}\times \overrightarrow{w})$ or $\left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]=0$

The value of $(\overrightarrow{v}\times \overrightarrow{w})$ $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ y& z& x\\ z& x& y\end{array}\right|$

$\Rightarrow \overrightarrow{v}\times \overrightarrow{w}=$ $\hat{i}(yz-x^2)-\hat{j}(y^2-xz)+\hat{k}(xy-z^2)$

Now the value of $\overrightarrow{u}.(\overrightarrow{v}\times \overrightarrow{w})=$ $(x\hat{i}+y\hat{j}+z\hat{k}).\left(\hat{i}\right(yz-x^2)-\hat{j}(y^2-xz)+\hat{k}(xy-z^2\left)\right)$

$\Rightarrow \overrightarrow{u}.(\overrightarrow{v}\times \overrightarrow{w})=3xyz-x^3-y^3-z^3$

As the vector are co-planar so $\overrightarrow{u}.(\overrightarrow{v}\times \overrightarrow{w})=0$

Or $3xyz-x^3-y^3-z^3=0$

$\to x^3+y^3+z^3=3xyz$

We know that from the algebraic identities that $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3xyz$

and $a^3+b^3+c^3=3xyz$, if and only if the value of $a+b+c=0$

Using this Identity we can write that,

$\because x^3+y^3+z^3=3xyz$

$\therefore x+y+z=0$ .....$\left(1\right)$

Now If we calculate the value of $\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}$, we get,

$\Rightarrow \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=$ $(x\hat{i}+y\hat{j}+z\hat{k})$ $+(y\hat{i}+z\hat{j}+x\hat{k})$ $+(z\hat{i}+x\hat{j}+zy\hat{k})$

$\Rightarrow \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=$ $(x+y+z)(\hat{i}+\hat{j}+\hat{k})$

From equation $\left(1\right)$, we know that for given co-planar vectors $x+y+z=0$

Hence $\Rightarrow \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\overrightarrow{0}$

Now we find the values of $\overrightarrow{v}.(\hat{i}+\hat{j}+\hat{k})$ or $\overrightarrow{v}.(\hat{i}+\hat{j}+\hat{k})$ or $\overrightarrow{w}.(\hat{i}+\hat{j}+\hat{k})$, We always get $(x+y+z)$

And from above calculations we know that $x+y+z=0$, hence the vector $(\hat{i}+\hat{j}+\hat{k})$ is normal to the plane containing the given three vectors.

We also know that $\Rightarrow \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\overrightarrow{0}$

If the sum of three co-planar vectors is zero or a null vector, then the angle between any two them is always equal to $120°$.

As $\Rightarrow \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\overrightarrow{0}$ .....$\left(2\right)$

Taking cross-product of eq. $\left(2\right)$ with $\overrightarrow{v}$, we get,

$\Rightarrow \overrightarrow{u}\times \overrightarrow{v}+\overrightarrow{v}\times \overrightarrow{v}+\overrightarrow{w}\times \overrightarrow{v}=0$

$\Rightarrow \overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{v}\times \overrightarrow{w}$ $\because (\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a})$ ......$\left(3\right)$

Similarly, we can also find that $\Rightarrow \overrightarrow{v}\times \overrightarrow{w}=\overrightarrow{w}\times \overrightarrow{u}$ ....$\left(4\right)$

If we take the cross-product of Eq. $\left(2\right)$ with vector $\overrightarrow{w}$

Hence from equation $3)$ and $\left(4\right)$, we can state that,

$\overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{v}\times \overrightarrow{w}=\overrightarrow{w}\times \overrightarrow{u}$

From above calculations and results, we say that all options are correct.