The angle between two vectors given by 6hat i | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) $\cos 	heta = \frac{{\vec A\vec B}}{{AB}} = \frac{{42 + 24 - 12}}{{\sqrt {36 + 36 + 9} \sqrt {49 + 16 + 16} }}$$ = \frac{{56}}{{9\sqrt {71} }}$

Vedclass · Accepted Answer

${\sin ^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} \right)$

Vedclass · Answer

(d) $\cos 	heta = \frac{{\vec A\vec B}}{{AB}} = \frac{{42 + 24 - 12}}{{\sqrt {36 + 36 + 9} \sqrt {49 + 16 + 16} }}$$ = \frac{{56}}{{9\sqrt {71} }}$

$\cos 	heta = \frac{{56}}{{9\sqrt {71} }}$ 

$	herefore \sin 	heta = \frac{{\sqrt 5 }}{3}$ or $	heta = {\sin ^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} ight)$

The angle between two vectors given by $6\hat i + 6\hat j - 3\hat k$ and $7\hat i + 4\hat j + 4\hat k$ is

Similar Questions

Find angle between $\vec A = 3\hat i - \hat j + 4\hat k$ and $Z-$ axis

The area of the parallelogram represented by the vectors $\overrightarrow A = 2\hat i + 3\hat j$ and $\overrightarrow B = \hat i + 4\hat j$ is.......$units$

The two vectors $\vec A = -2\widehat i + \widehat j + 3\widehat k$ and $\vec B = 7\widehat i + 5\widehat j + 3\widehat k$ are :-

Explain right hand screw law.

Two vector $A$ and $B$ have equal magnitudes. Then the vector $\mathop A\limits^ \to + \mathop B\limits^ \to $ is perpendicular to