Question

If two vectors are Ā= î+ j hat + k hat and B = - î - j hat - k hat , then find the angle between (Ā-B) and Ā.
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Answer 1

Given:

$\sf{\vec{A} = \hat{\imath} + \hat{\jmath}+ \hat{k}}$

$\sf{\vec{B} = \hat{\imath} - \hat{\jmath} + 2 \hat{k}} {\mathfrak{\underline{Solution:}}}$

Dot product formula:

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

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

➣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:

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

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

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

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

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

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