Question

If $|\overline{x}|=7,|\overline{y}|=\sqrt{2},\overline{x}\times \overline{y}=(6,2,3)$ then $|\overline{x}.\overline{y}{|}^2=$ __________

• A. $49$
• B. $7$
• C. $147$
• D. $98$

Answer 1

The correct option is A $49$

$x=|\overrightarrow{x}|=7$ and $y=|\overrightarrow{y}|=\sqrt{2}$

$|\overrightarrow{x}\times \overrightarrow{y}|=xy\sin \theta$

$⟹\sqrt{6^2+2^2+3^2}=7\sqrt{2}\sin \theta =\sqrt{49}$

$⟹\sin \theta =\frac{1}{\sqrt{2}}$

$⟹\cos \theta =\sqrt{1-{\sin }^2\theta }=\frac{1}{\sqrt{2}}$

$\overrightarrow{x}.\overrightarrow{y}=xy\cos \theta =7\sqrt{2}\frac{1}{\sqrt{2}}=7$

$⟹|\overrightarrow{x}.\overrightarrow{y}{|}^2=49$