NIMCET 2023 — Mathematics PYQ

$A\times B=<br>\left|\begin{array}{ccccccc}<br>\hat{i}& amp;\hat{j}& amp;\hat{k}\text{<}br>2& amp;1& amp;-2\text{<}br>1& amp;1& amp;0<br>\end{array}\right|$

$=\hat{i}(2)-\hat{j}(2)+\hat{k}(1)$

$A\times B=2\hat{i}-2\hat{j}+\hat{k}$

$|C-A|=3$

$|C-A{|}^2=9$

$|C{|}^2+|A{|}^2-2C\cdot A=9;\Rightarrow ;|C{|}^2-2C\cdot A=0\text{.....(i)}$

Now

$|(A\times B)\times C|=|A\times B|,|C|=\sqrt{4+4+1},|C|\times \sin 30^{\circ }=3$

Given $\theta =30^{\circ }$

$\Rightarrow |C|=2$

Put this value in equation (i)

We get $;C\cdot A=2$