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If \hat{x},\hat{y},and\: \hat{z} are three unit vectors in 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

  • Option 1)

    \frac{3}{2}

  • Option 2)

    3

  • Option 3)

    3\sqrt{3}

  • Option 4)

    6

 

Answers (1)

best_answer

As we have learned

Unit vector -

A vector of unit magnitude in direction of a vector \vec{a} is called unit vector along \widehat{a}.

- wherein

It is denoted by \widehat{a}.

 

 

Scalar Product of two vectors -

\vec{a}.\vec{b}=\vec{b}.\vec{a}

\vec{a}.\vec{a}=a^2=\left | \vec{a} \right |^2

- wherein

Dot product is commutative for \theta =0, where \theta  is the angle between the vectors \vec{a}\:and\:\vec{b}

 

 |\hat x + \hat y|^2 +|\hat y + \hat z|^2 +|\hat z + \hat x|^2 \\ 2+ 2 \hat x \hat y + 2+ 2 \hat y \hat z+2+ 2 \hat z \hat x\\ = 6+2 ( \hat x \hat y+ \hat y \hat z+ \hat z \hat x)\\ = 6+2 \left [ \frac{(\hat x + \hat y + \hat z -\hat x - \hat y - \hat z )}{2} \right ] \\

= 6+ [ (\hat x + \hat y + \hat z)^2- (\hat x^2 + \hat y^2 + \hat z^2)]

 = 6+ [ (\hat x + \hat y + \hat z)^2- (1+1+1)]

= 3+ (\hat x + \hat y + \hat z)^2

\geq 3

Minimum value 3 

 

 

 

 

 


Option 1)

\frac{3}{2}

Option 2)

3

Option 3)

3\sqrt{3}

Option 4)

6

Posted by

Himanshu

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