# If the vertices of a triangle are $A(1,4,2),\:B(-2,1,2)\:C(2,-3,4)$ then the angle $B$ = ?

$(a)\:cos^{-1}1/\sqrt 3\:\:\:\qquad\:(b)\:90^{\circ}\:\:\:\qquad\:(c)\:cos^{-1}\sqrt 3\:\:\:\qquad\:(d)\:cos^{-1}\sqrt 6/3$

## 1 Answer

Angle $B$ is angle between $\overrightarrow {BA}$ and $\overrightarrow {BC}$
Given: $A(1,4,2),\:B(-2,1,2)\:C(2,-3,4)$
$\overrightarrow {BA}=(3,3,0)\:\:and\:\:\overrightarrow {BC}=(4,-4,2)$
Angle $B$ is given by $cos^{-1}\bigg(\large\frac{\overrightarrow {BA}.\overrightarrow {BC}}{|\overrightarrow {BA}||\overrightarrow {BC}|}\bigg)$
$=cos^{-1}\bigg(\large\frac{(3,3,0).(4,-4,2)}{\sqrt {18}.6}\bigg)$
$=cos^{-1}0=90^{\circ}$
answered Jan 11, 2014

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