Question

Which of the following is not always true ?

• A. $|\overrightarrow{a}+\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2$ if $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular to each other
• B. $|\overrightarrow{a}+\lambda \overrightarrow{b}|\ge |\overrightarrow{a}|$ for all $\lambda \in R$ if $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular to each other
• C. $|\overrightarrow{a}+\overrightarrow{b}{|}^2+|\overrightarrow{a}-\overrightarrow{b}{|}^2=2(|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2)$
• D. $|\overrightarrow{a}+\lambda \overrightarrow{b}|\ge |\overrightarrow{a}|$ for all $\lambda \in R$ if $\overrightarrow{a}$ is parallel to $\overrightarrow{b}$

Answer 1

The correct option is D $|\overrightarrow{a}+\lambda \overrightarrow{b}|\ge |\overrightarrow{a}|$ for all $\lambda \in R$ if $\overrightarrow{a}$ is parallel to $\overrightarrow{b}$

$|\overrightarrow{a}+\lambda \overrightarrow{b}{|}^2\ge |\overrightarrow{a}{|}^2$ for all $\lambda \in R$ if $\overrightarrow{a}$ is parallel to $\overrightarrow{b}$ does not hold true for each value of $\lambda \in R$

If $\overrightarrow{b}=\overrightarrow{a}\ne \overrightarrow{0}$ and $\lambda =-1,$ then $|\overrightarrow{a}+\lambda \overrightarrow{b}|=0$, which is less than $|\overrightarrow{a}|$.