Question

If $x,y$ and $z$ are three unit vectors in a three dimensional space, then the minimum value of ${\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2}$is
A. $\dfrac{3}{2}$
B. $3$
C. $3\sqrt 3$
D. $6$

Answer 1

Here we use the knowledge of unit vectors that they always have magnitude equal to one. Using the formula ${\left| {\overrightarrow a } \right|^2} = (\overrightarrow a ).(\overrightarrow a )$ we expand each term which is in square form. Then using the expansion of ${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$ we find the value of $2(\overrightarrow a \overrightarrow b + \overrightarrow b \overrightarrow c + \overrightarrow c \overrightarrow a )$and use it for finding the minimum value of the given vectors.

Formula used: We have the formula ${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2(\overrightarrow a \overrightarrow b + \overrightarrow b \overrightarrow c + \overrightarrow c \overrightarrow a )$

Complete step-by-step answer:
We have three unit vectors $x,y$ and $z$. Since, we know unit vectors have magnitude 1
$\left| {\hat x} \right| = 1,\left| {\hat y} \right| = 1,\left| {\hat z} \right| = 1$
We have to find the value of ${\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2}$.
We solve each term separately.
First we solve ${\left| {\hat x + \hat y} \right|^2}$
Using the formula ${\left| {\overrightarrow a } \right|^2} = (\overrightarrow a ).(\overrightarrow a )$ we can write
$\Rightarrow {\left| {\hat x + \hat y} \right|^2} = (\hat x + \hat y).(\hat x + \hat y)$
Multiplying the terms in RHS of the equation
$\Rightarrow {\left| {\hat x + \hat y} \right|^2} = \hat x.\hat x + \hat x.\hat y + \hat y.\hat x + \hat y.\hat y$
Using the formula ${\left| {\overrightarrow a } \right|^2} = (\overrightarrow a ).(\overrightarrow a )$ again
$\Rightarrow {\left| {\hat x + \hat y} \right|^2} = {\left| {\hat x} \right|^2} + \hat x.\hat y + \hat x.\hat y + {\left| {\hat y} \right|^2}$
Since $\left| {\hat x} \right| = 1,\left| {\hat y} \right| = 1,\left| {\hat z} \right| = 1$
$\Rightarrow {\left| {\hat x + \hat y} \right|^2} = 1 + 2\hat x.\hat y + 1$
$\Rightarrow {\left| {\hat x + \hat y} \right|^2} = 2 + 2\hat x.\hat y …….. (1)$
Now we solve ${\left| {\hat y + \hat z} \right|^2}$
Using the formula ${\left| {\overrightarrow a } \right|^2} = (\overrightarrow a ).(\overrightarrow a )$we can write
$\Rightarrow {\left| {\hat y + \hat z} \right|^2} = (\hat y + \hat z).(\hat y + \hat z)$
Multiplying the terms in RHS of the equation
$\Rightarrow {\left| {\hat y + \hat z} \right|^2} = \hat y\hat y + \hat y.\hat z + \hat z.\hat y + \hat z.\hat z$
Using the formula ${\left| {\overrightarrow a } \right|^2} = (\overrightarrow a ).(\overrightarrow a )$ again
$\Rightarrow {\left| {\hat y + \hat z} \right|^2} = {\left| {\hat y} \right|^2} + \hat y.\hat z + \hat y.\hat z + {\left| {\hat z} \right|^2}$
Since $\left| {\hat x} \right| = 1,\left| {\hat y} \right| = 1,\left| {\hat z} \right| = 1$
$\Rightarrow {\left| {\hat y + \hat z} \right|^2} = 1 + 2\hat y.\hat z + 1$
$\Rightarrow {\left| {\hat y + \hat z} \right|^2} = 2 + 2\hat y.\hat z .......… (2)$
Now we solve ${\left| {\hat z + \hat x} \right|^2}$
Using the formula ${\left| {\overrightarrow a } \right|^2} = (\overrightarrow a ).(\overrightarrow a )$we can write
$\Rightarrow {\left| {\hat z + \hat x} \right|^2} = (\hat z + \hat x).(\hat z + \hat x)$
Multiplying the terms in RHS of the equation
$\Rightarrow {\left| {\hat z + \hat x} \right|^2} = \hat z.\hat z + \hat z.\hat x + \hat x.\hat z + \hat x.\hat x$
Using the formula ${\left| {\overrightarrow a } \right|^2} = (\overrightarrow a ).(\overrightarrow a )$again
$\Rightarrow {\left| {\hat z + \hat x} \right|^2} = {\left| {\hat z} \right|^2} + \hat z.\hat x + \hat z.\hat x + {\left| {\hat x} \right|^2}$
Since $\left| {\hat x} \right| = 1,\left| {\hat y} \right| = 1,\left| {\hat z} \right| = 1$
$\Rightarrow {\left| {\hat z + \hat x} \right|^2} = 1 + 2\hat z.\hat x + 1$
$\Rightarrow {\left| {\hat z + \hat x} \right|^2} = 2 + 2\hat z.\hat x………….… (3)$
Now we substitute values from equations (1), (2) and (3) in the sum of terms

$$\Rightarrow {\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2} = 2 + 2\hat x.\hat y + 2 + 2\hat y.\hat z + 2 + 2\hat z.\hat x \\\ \Rightarrow {\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2} = 6 + 2\hat x.\hat y + 2\hat y.\hat z + 2\hat z.\hat x \\\$$

Take 2 common from the last three terms
$\Rightarrow {\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2} = 6 + 2(\hat x.\hat y + \hat y.\hat z + \hat z.\hat x) …….. (4)$
Now we know the expansion ${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2(\overrightarrow a \overrightarrow b + \overrightarrow b \overrightarrow c + \overrightarrow c \overrightarrow a )$.
Substitute the values of $\overrightarrow a = \hat x,\overrightarrow b = \hat y,\overrightarrow c = \hat z$.
$\Rightarrow {\left| {\hat x + \hat y + \hat z} \right|^2} = {\left| {\hat x} \right|^2} + {\left| {\hat y} \right|^2} + {\left| {\hat z} \right|^2} + 2(\hat x\hat y + \hat y\hat z + \hat z\hat x)$
Substitute the values of $\left| {\hat x} \right| = 1,\left| {\hat y} \right| = 1,\left| {\hat z} \right| = 1$

$$\Rightarrow {\left| {\hat x + \hat y + \hat z} \right|^2} = 1 + 1 + 1 + 2(\hat x\hat y + \hat y\hat z + \hat z\hat x) \\\ \Rightarrow {\left| {\hat x + \hat y + \hat z} \right|^2} = 3 + 2(\hat x\hat y + \hat y\hat z + \hat z\hat x) \\\$$

We can see that ${\left| {\hat x + \hat y + \hat z} \right|^2} > 0$ as the terms on RHS are adding.
$\therefore 3 + 2(\hat x\hat y + \hat y\hat z + \hat z\hat x) > 0$
Shifting the value of constant to one side of the equation we get
$\Rightarrow 2(\hat x\hat y + \hat y\hat z + \hat z\hat x) > - 3…………. (5)$
Therefore we can use equation (5) in equation (4)
$\Rightarrow {\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2} = 6 + 2(\hat x.\hat y + \hat y.\hat z + \hat z.\hat x)$
Substitute $\Rightarrow 2(\hat x\hat y + \hat y\hat z + \hat z\hat x) > - 3$

$$\Rightarrow {\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2} > 6 + ( - 3) \\\ \Rightarrow {\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2} > 3 \\\$$

Therefore the minimum value of ${\left| {\hat x + \hat y} \right|^2} + {\left| {\hat y + \hat z} \right|^2} + {\left| {\hat z + \hat x} \right|^2}$ is 3.

So, the correct answer is “Option B”.

Note: Students many times make mistake of writing the vectors multiplied in the bracket $2(\hat x\hat y + \hat y\hat z + \hat z\hat x)$ as $2(1 + 1 + 1) = 2 \times 3 = 6$ which is wrong because we don’t know the direction of the vectors and direction plays very important role in vectors.