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# If $\overrightarrow a=\hat i+\hat j+p\hat k\:\:and\:\:\overrightarrow b=\hat i+\hat j+\hat k$, then $|\overrightarrow a+\overrightarrow b|=|\overrightarrow a|+|\overrightarrow b|$ holds for ?

$\begin{array}{1 1} All \;real\;p \\ No\;real\;p \\ p=1 \\ p=-1 \end{array}$

$\overrightarrow a +\overrightarrow b=(\hat i+\hat j+p\hat k)+(\hat i+\hat j+\hat k)=2\hat i+2\hat j+(p+1)\hat k$
$|\overrightarrow a+\overrightarrow b|=\sqrt {4+4+(p+1)^2}=\sqrt {p^2+2p+9}$
$|\overrightarrow a|=\sqrt {2+p^2}\:\:and\:\:|\overrightarrow b|=\sqrt {3}$
Given: $|\overrightarrow a+\overrightarrow b|=|\overrightarrow a|+|\overrightarrow b|$
$\Rightarrow\:\sqrt {p^2+2p+9}=\sqrt {2+p^2}+\sqrt 3$
Squaring both the sides, $p^2+2p+9=2+p^2+3+2\sqrt 3.\sqrt {2+p^2}$
$\Rightarrow\:2p+4=2\sqrt 3.\sqrt {2+p^2}$
$\Rightarrow\:p+2=\sqrt 3 \sqrt {2+p^2}$
Again squaring both the sides $p^2+4p+4=3p^2+6$
or $2p^2-4p+2=0$
$\Rightarrow\:p=1$