Question

Do 19,21 from above picture Please ​

Answer 1

(19)

Let,

$\longrightarrow\sf{\vec A=6\hat i+6\hat j-3\hat k}\\\\\\\longrightarrow\sf{\vec B=7\hat i+4\hat j+4\hat k}$

Then,

$\longrightarrow\sf{A=\sqrt{6^2+6^2+(-3)^2}}\\\\\\\longrightarrow\sf{A=9}$

And,

$\longrightarrow\sf{B=\sqrt{7^2+4^2+4^2}}\\\\\\\longrightarrow\sf{B=9}$

Taking the scalar product,

$\longrightarrow\sf{\vec A\cdot\vec B=(6\hat i+6\hat j-3\hat k)\cdot(7\hat i+4\hat j+4\hat k)}\\\\\\\longrightarrow\sf{AB\cos\theta=6\cdot7+6\cdot4+(-3)\cdot4}\\\\\\\longrightarrow\sf{81\cos\theta=54}\\\\\\\longrightarrow\sf{\cos\theta=\dfrac{2}{3}}\\\\\\\Longrightarrow\sf{\sin\theta=\sqrt{1-\left (\dfrac{2}{3}\right)^2}\quad\quad\left [\sin^2\theta+\cos^2\theta=1\right]}\\\\\\\longrightarrow\sf{\sin\theta=\dfrac{\sqrt5}{3}}\\\\\\\longrightarrow\underline {\underline {\sf{\theta=\sin^{-1}\left (\dfrac{\sqrt5}{3}\right)}}}$

(21)

$\longrightarrow\sf{W=\vec F\cdot\vec s}\\\\\\\longrightarrow\sf{W=(4\hat i+5\hat j+0\hat k)\cdot(3\hat i+0\hat j+6\hat k)}\\\\\\\longrightarrow\sf{W=(4\cdot3+5\cdot0+0\cdot6)}\\\\\\\longrightarrow\sf{\underline {\underline {W=4\times 3}}}$

Answer 2

Answer:

My snapchat id aaradhya00762 this my I'd for sex