Question

P and q are complex numbers such that |p|=7√2 and |p+q|=12√3 .

Answer 1

Answer:

Option B is correct

Step-by-step explanation:

Given: $\left | p \right |=7\sqrt{2}\,,\,\left | p+q \right |=12\sqrt{3}$

To find: interval on which $\left | q \right |$ must fall

Solution:

$\left | p \right |=7\sqrt{2}\\-7\sqrt{2}\leq p\leq 7\sqrt{2}\,\,(i)$

$\left | p+q \right |=12\sqrt{3}\\-12\sqrt{3}\leq p+q\leq 12\sqrt{3}\,\,(ii)$

Subtract (ii) from (i)

$-12\sqrt{3}+7\sqrt{2}\leq p+q-p\leq 12\sqrt{3}-7\sqrt{2}\\-12\sqrt{3}+7\sqrt{2}\leq q\leq 12\sqrt{3}-7\sqrt{2}\\\left | q \right |=12\sqrt{3}-7\sqrt{2}$

So, $\left | q \right |$ must fall in interval $[12\sqrt{3}-7\sqrt{2},\infty)$

Therefore, option B is correct.