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

Hint: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.