Question

Find the angles between the vectors Ā - î+j^+k^&B=i^- j^ + 2 k^.​

Answer 1

Answer:

Here is your answer :-)

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Answer 2

Answer:

Given:

• $\sf{\vec{A} = \hat{\imath} + \hat{\jmath}+ \hat{k}}$
• $\sf{\vec{B} = \hat{\imath} - \hat{\jmath} + 2 \hat{k}}$

${\mathfrak{\underline{Solution:}}}$

In order to find the angle , we can use the formula for dot product of vectors. Then the angle can be found by solving dot product in component form

⚕Dot product formula:

${\boxed{\sf{\vec{A} \ . \ \vec{B} = AB \ cos \theta}}}$

➣Using this formula:

➫$\sf{cos \theta = \dfrac{\vec{A} \ . \ \vec{B}}{AB}}$

➫$\sf{cos \theta = \dfrac{\vec{A} \ . \ \vec{B}}{|\vec{A}| \vec{B}|}}$

✦Now finding dot product in component form :

$\boxed{\sf{\vec{A} \ . \ \vec{B} = A_xB_x + A_yB_y + A_zB_z}}$

→$\sf{\vec{A} \ . \ \vec{B} = (1)(1) + 1(-1) + 1(2)}$

→$\sf{\vec{A} \ . \ \vec{B} = 1 - 1 + 2}$

→$\sf{\vec{A} \ . \ \vec{B} = 2}$

✦Now finding magnitude in component form :

$\boxed{\sf{|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}}}$

For $\sf{\vec{A}}$ :

→$\sf{|\vec{A}| = \sqrt{1^2 + 1^2 + 1^2}}$

→$\sf{| \vec{A}| = \sqrt{3}}$

For $\sf{\vec{B}}$:

→$\sf{|\vec{B}| = \sqrt{1^2 + (-1)^2 + 2^2}}$

→$\sf{|\vec{B}| = \sqrt{1 + 1 + 4}}$

→$\sf{|\vec{B}| = \sqrt{6}}$

Now putting all these values in the dot product formula:

➫$\sf{cos \theta = \dfrac{\vec{A} \ . \ \vec{B}}{|\vec{A}| |\vec{B}|}}$

➫$\sf{cos \theta = \dfrac{2}{\sqrt3 \times \sqrt{6}}}$

➫$\sf{cos \theta = \dfrac{2}{3\sqrt2}}$

➫$\sf{\theta = cos^{-1} \dfrac{\sqrt2}{3}}$

•°•The angle is $\sf{\bf{cos^{-1} \dfrac{\sqrt2}{3}}}$.

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