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 $\lambda ϵR$ 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 $\lambda ϵR$ 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 $\lambda ϵR$ if $\overrightarrow{a}$ is parallel to $\overrightarrow{b}$

Let vector

$\overrightarrow{a}=\hat{i}$

$\overrightarrow{b}=\hat{j}$

${|\overrightarrow{a}+\overrightarrow{b}|}^2=\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+2\overrightarrow{a}\cdot \overrightarrow{b}$

Here a and are perpendicular So

$\overrightarrow{a}\cdot \overrightarrow{b}=0$

${|\overrightarrow{a}+\overrightarrow{b}|}^2=\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|$

Valid statement

$|\overrightarrow{a}+\lambda \overrightarrow{b}|\ge \left|\overrightarrow{a}\right|$

$|\hat{i}+\lambda \hat{j}|\ge \left|\hat{i}\right|$

$\sqrt{1^2+{\lambda }^2}\ge \sqrt{1^2}$

$\sqrt{1+{\lambda }^2}\ge \sqrt{1}$

It is also always true

if vectors are parallel then $b=\hat{i}$

$|\overrightarrow{a}+\lambda \overrightarrow{b}|\ge \left|\overrightarrow{a}\right|$

$|\hat{i}+\lambda \hat{i}|\ge \left|\hat{i}\right|$

$\sqrt{(1+\lambda {)}^2}\ge \sqrt{1^2}$

$\sqrt{1}\ge \sqrt{1}$ if lambda has zero value

but if lambda has -1 value then

$|\hat{i}-\hat{i}|\ge \left|\hat{i}\right|$

$0\ge 1$

It is not true