Which of the following is not always true?
$\left( A \right)$ ${{\left| \vec{a}+\vec{b} \right|}^{2}}={{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}$ if $\vec{a}$ and $\vec{b}$ are perpendicular to each other
$\left( B \right)$ $\left| \vec{a}+\lambda \vec{b} \right|\ge \left| {\vec{a}} \right|$ for all $\lambda \in R$ if $\vec{a}$ and $\vec{b}$ are perpendicular to each other
$\left( C \right)$ $\left| \vec{a}+\vec{b} \right|+\left| \vec{a}-\vec{b} \right|=2\left( {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}} \right)$
$\left( D \right)$ $\left| \vec{a}+\lambda \vec{b} \right|\ge \left| {\vec{a}} \right|$ for all $\lambda \in R$ if $\vec{a}$ is parallel to $\vec{b}$

Hint: In this question we have been told to find which one of the given options is not true. we will solve this question by considering the given options and solving them by expanding them as per their given condition. We will then check whether the condition holds true or not. The expression which does not hold its condition always true will be the required answer.

Complete step by step solution:
Consider the first option, we have the condition as:
${{\left| \vec{a}+\vec{b} \right|}^{2}}={{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}$ if $\vec{a}$ and $\vec{b}$ are perpendicular to each other.
Now consider the left-hand side of the expression:
$\Rightarrow {{\left| \vec{a}+\vec{b} \right|}^{2}}$
On squaring and expanding the terms, we get:
$\Rightarrow {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+2\vec{a}\cdot \vec{b}$
Now since $\vec{a}$ and $\vec{b}$ are perpendicular to each other we have $\vec{a}\cdot \vec{b}=0$ therefore, on substituting, we get:
$\Rightarrow {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+2\left( 0 \right)$
On simplifying, we get:
$\Rightarrow {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}$, which is the right-hand side therefore, this condition holds true.
Consider the second option, we have the condition as:
$\left| \vec{a}+\lambda \vec{b} \right|\ge \left| {\vec{a}} \right|$ for all $\lambda \in R$ if $\vec{a}$ and $\vec{b}$ āre perpendicular to each other.
Now consider $\vec{a}=\hat{i}$ and $\vec{b}=\hat{j}$
We have the expression as:
$\Rightarrow \left| \vec{a}+\lambda \vec{b} \right|\ge \left| {\vec{a}} \right|$
On substituting the values, we get:
$\Rightarrow \left| \hat{i}+\lambda \hat{b} \right|\ge \left| {\hat{i}} \right|$
On squaring both sides, we get:
$\Rightarrow {{\left| \hat{i}+\lambda \hat{b} \right|}^{2}}\ge {{\left| {\hat{i}} \right|}^{2}}$
Now since both are unit vectors, we they have value of $1$ therefore, we get:
$\Rightarrow {{\left| 1+\lambda \right|}^{2}}\ge {{\left| 1 \right|}^{2}}$
On removing the modulus, we get:
$\Rightarrow \sqrt{{{1}^{2}}+{{\lambda }^{2}}}=\sqrt{{{1}^{2}}}$
Since ${{\lambda }^{2}}$ is positive and is added in the left-hand side, the value of the left-hand side will always be greater than $1$ therefore, this condition is true.
Consider the third option, we have the condition as:
${{\left| \vec{a}+\vec{b} \right|}^{2}}+{{\left| \vec{a}-\vec{b} \right|}^{2}}=2\left( {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}} \right)$
Consider the left-hand side of the expression.
$\Rightarrow {{\left| \vec{a}+\vec{b} \right|}^{2}}+{{\left| \vec{a}-\vec{b} \right|}^{2}}$
On expanding, we get:
$\Rightarrow {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+2\vec{a}\cdot \vec{b}+{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}-2\vec{a}\cdot \vec{b}$
On cancelling the terms, we get:
$\Rightarrow {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}$
On adding the same terms, we get:
$\Rightarrow 2\times {{\left| {\vec{a}} \right|}^{2}}+2\times {{\left| {\vec{b}} \right|}^{2}}$
On taking $2$ common, we get:
$\Rightarrow 2\left( {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}} \right)$, which is the right-hand side therefore, this condition is also always true.
Consider the fourth option, we have the condition as:
$\left| \vec{a}+\lambda \vec{b} \right|\ge \left| {\vec{a}} \right|$ for all $\lambda \in R$ if $\vec{a}$ is parallel to $\vec{b}$
Now since $\vec{a}$ is parallel to $\vec{b}$ the values of both the vectors are the same. Consider the value of vectors as $\hat{i}$. On substituting, we get:
$\Rightarrow \left| \hat{i}+\lambda \hat{i} \right|\ge \left| {\hat{i}} \right|$
On removing the modulus, we get:
$\Rightarrow \sqrt{{{\left( 1+\lambda \right)}^{2}}}\ge \sqrt{{{1}^{2}}}$
Now if lambda has negative value such as $\lambda =-1$, we get:
$\Rightarrow \sqrt{{{\left( 1-1 \right)}^{2}}}\ge \sqrt{{{1}^{2}}}=\sqrt{{{0}^{2}}}\ge \sqrt{{{1}^{2}}}$
On simplifying, we get:
$\Rightarrow \sqrt{0}\ge \sqrt{1}$, which is not valid.

So, the correct answer is “Option D”.

Note: It is to be remembered that vector addition and multiplication is different. In this question we used the dot product. It is to be remembered that to remove the modulus signs both the sides are to be squared but since this changes the expression the square root is to be taken to get the expression with the same value.