Question

Find the angle between vec A=hat i+2hat j-hat k and vec B=-hat i+hat j-2hat k

Answer 1

Answer:

k k liye ek shahyri the intended recipient the reader of length of the year again when ever you want to do the needful at earliest you can win in

Explanation:

no problem I will discuss with the water of length is density in site content of 300 kg and breadth of knowledge and breadth of experience in the intended recipient or the intended recipient or the other side of length of time to kya hai kya kar rahe ho gaye the aur

Answer 2

Answer:

$\boxed{\mathfrak{Angle \ between \ \overrightarrow{A} \ and \ \overrightarrow{B} \ (\theta) = 60\degree}}$

Given:

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

Explanation:

$\rm \overrightarrow{A}. \overrightarrow{B} = (\hat{i} + 2\hat{j} - \hat{k}). (-\hat{i} + \hat{j} -2 \hat{k}) \\ \\ \rm \overrightarrow{A}. \overrightarrow{B} = - 1 + 2 + 2 \\ \\ \rm \overrightarrow{A}. \overrightarrow{B} = 3$

$\rm | \overrightarrow{A} | = \sqrt{ {1}^{2} + {2}^{2} + {( - 1)}^{2} } \\ \\ \rm | \overrightarrow{A} | = \sqrt{1 + 4 + 1} \\ \\ \rm | \overrightarrow{A} | = \sqrt{6} \\ \\ \\ \rm | \overrightarrow{B} | = \sqrt{ { ( - 1)}^{2} + {1}^{2} + {( - 2)}^{2} } \\ \\ \rm | \overrightarrow{B} | = \sqrt{1 + 1 + 4} \\ \\ \rm | \overrightarrow{B} | = \sqrt{6}$

Let $\rm \theta$ be the angle between $\rm \overrightarrow{A}$ & $\rm \overrightarrow{B}$

$\rm \implies \overrightarrow{A}. \overrightarrow{B} = | \overrightarrow{A} | | \overrightarrow{B} | cos \theta \\ \\ \rm \implies cos \theta = \dfrac{\overrightarrow{A}. \overrightarrow{B} }{| \overrightarrow{A} | | \overrightarrow{B} |} \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{\overrightarrow{A}. \overrightarrow{B} }{| \overrightarrow{A} | | \overrightarrow{B} |} ) \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{3 }{ \sqrt{6} \times \sqrt{6} } ) \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{3 }{ 6} ) \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{1 }{ 2} ) \\ \\ \rm \implies \theta = {60}^{ \circ}$