Which of the following is not always true?

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(D) If $a$ and $b$ are perpendicular to each other, then $a \cdot b = 0$. Now consider, $|a+b|^{2} = (a+b) \cdot (a+b) = |a|^{2} + |b|^{2} + 2(a \cdot

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$|a+\lambda b| \geq |a|$ for all $\lambda \in R$ if $a$ is parallel to $b$

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(D) If $a$ and $b$ are perpendicular to each other, then $a \cdot b = 0$. Now consider, $|a+b|^{2} = (a+b) \cdot (a+b) = |a|^{2} + |b|^{2} + 2(a \cdot b) = |a|^{2} + |b|^{2}$. So, option $(a)$ is always true. $(b)$ If $a$ and $b$ are perpendicular to each other, then $a \cdot b = 0$. Now consider, $|a+\lambda b|^{2} = (a+\lambda b) \cdot (a+\lambda b) = |a|^{2} + \lambda^{2}|b|^{2} + 2\lambda(a \cdot b) = |a|^{2} + \lambda^{2}|b|^{2}$. Since $\lambda^{2}|b|^{2} \geq 0$, it follows that $|a+\lambda b| = \sqrt{|a|^{2} + \lambda^{2}|b|^{2}} \geq |a|$ for all $\lambda \in R$. So, option $(b)$ is always true. $(c)$ Consider, $|a+b|^{2} + |a-b|^{2} = (a+b) \cdot (a+b) + (a-b) \cdot (a-b) = (|a|^{2} + |b|^{2} + 2a \cdot b) + (|a|^{2} + |b|^{2} - 2a \cdot b) = 2(|a|^{2} + |b|^{2})$. So, option $(c)$ is always true. $(d)$ Consider $a = -b$ and $b 
eq 0$. Then, $|a+\lambda b| = |-b + \lambda b| = |\lambda - 1||b|$. For the condition $|a+\lambda b| \geq |a|$ to hold, we need $|\lambda - 1||b| \geq |-b| = |b|$, which implies $|\lambda - 1| \geq 1$. This is not true for all $\lambda \in R$ (e.g.,if $\lambda = 0.5$, then $|0.5 - 1| = 0.5$, which is not $\geq 1$). Hence, option $(d)$ is not always true.

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

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