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

\to u + \to v + \to w = \to 0

, where

\to 0

is the null vector

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^i +^j +^k

is normal to the plane containing

\to u, \to v

and

\to w

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The angle between any two of the vectors

\to u, \to v

and

\to w

120^{\circ}

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\to u \times \to v = \to v \times \to w = \to w \times \to u

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Solution

The correct options are
A $\to u + \to v + \to w = \to 0$ , where $\to 0$ is the null vector
B The angle between any two of the vectors $\to u, \to v$ and $\to w$ is $120^{\circ}$
C $^i +^j +^k$ is normal to the plane containing $\to u, \to v$ and $\to w$
D $\to u \times \to v = \to v \times \to w = \to w \times \to u$
Any three vectors $\to a, \to b$ and $\to c$ are co-planar only if the value of $\to a . (\to b \times \to c)$ or $[\to a \to b \to c] = 0$

$\Rightarrow \to a . (\to b \times \to c)$ is also written as $[\to a \to b \to c]$

Now the given vector are $\to u, \to v$ and $\to w$ , which are given as:

$\to u = x^i + y^j + z^k$

$\to v = y^i + z^j + x^k$

$\to w = z^i + x^j + z y^k$

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

The value of $(\to v \times \to w)$ $= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k y & z & x z & x & y \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow \to v \times \to w =$ $^i (y z - x^{2}) -^j (y^{2} - x z) +^k (x y - z^{2})$

Now the value of $\to u . (\to v \times \to w) =$ $(x^i + y^j + z^k) . (^i (y z - x^{2}) -^j (y^{2} - x z) +^k (x y - z^{2}))$

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

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

Or $3 x y z - x^{3} - y^{3} - z^{3} = 0$

$\to x^{3} + y^{3} + z^{3} = 3 x y z$

We know that from the algebraic identities that $a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 x y z$

and $a^{3} + b^{3} + c^{3} = 3 x y z$ , if and only if the value of $a + b + c = 0$

Using this Identity we can write that,

$∵ x^{3} + y^{3} + z^{3} = 3 x y z$

$∴ x + y + z = 0$ ..... $(1)$

Now If we calculate the value of $\to u + \to v + \to w$ , we get,

$\Rightarrow \to u + \to v + \to w =$ $(x^i + y^j + z^k)$ $+ (y^i + z^j + x^k)$ $+ (z^i + x^j + z y^k)$

$\Rightarrow \to u + \to v + \to w =$ $(x + y + z) (^i +^j +^k)$

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

Hence $\Rightarrow \to u + \to v + \to w = \to 0$

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

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

We also know that $\Rightarrow \to u + \to v + \to w = \to 0$

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 $\Rightarrow \to u + \to v + \to w = \to 0$ ..... $(2)$

Taking cross-product of eq. $(2)$ with $\to v$ , we get,

$\Rightarrow \to u \times \to v + \to v \times \to v + \to w \times \to v = 0$

$\Rightarrow \to u \times \to v = \to v \times \to w$ $∵ (\to a \times \to b = - \to b \times \to a)$ ...... $(3)$

Similarly, we can also find that $\Rightarrow \to v \times \to w = \to w \times \to u$ .... $(4)$

If we take the cross-product of Eq. $(2)$ with vector $\to w$

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

$\to u \times \to v = \to v \times \to w = \to w \times \to u$

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

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