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[SOLVED] IBV3 vector in a circle

karush

Well-known member
Jan 31, 2012
2,774
View attachment 1143

(a)

if $r=6$ and $\displaystyle \pmatrix { 6 \\ 0 } $ then $A$ is $6$ from $0,0$ on the $x$ axis
and if $\displaystyle \pmatrix { -6 \\ 0 }$ then $B$ is $-6$ from $0,0$ on the $x$ axis

and if $\displaystyle \pmatrix { 5 \\ \sqrt{11} }$ implies $\sqrt{5^2 + 11}=6 = OC$

(b) I presume $\vec{AC}$ can be from origin so
$\displaystyle \vec{OC}-{OA} = \vec{AC} = \pmatrix{-1 \\ \sqrt{11}}$



(c) $\displaystyle\frac{\vec{OA}\cdot\vec{OC}}{||OA||\ ||OC||}
=\frac{30}{36}=\frac{5}{6}
$

(d) area is just $\frac{5}{2}\sqrt{11}$
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
So far , so good :)
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
View attachment 1143
(a) if $r=6$ and $\displaystyle \pmatrix { 6 \\ 0 } $ then $A$ is $6$ from $0,0$ on the $x$ axis
and if $\displaystyle \pmatrix { -6 \\ 0 }$ then $B$ is $-6$ from $0,0$ on the $x$ axis

and if $\displaystyle \pmatrix { 5 \\ \sqrt{11} }$ implies $\sqrt{5^2 + 11 }=6 = OC$

(b) I presume $\vec{AC}$ can be from origin so
$\displaystyle \vec{OC}-{OA} = \vec{AC} = \pmatrix{-1 \\ \sqrt{11}}$

(c) $\displaystyle\frac{\vec{OA}\cdot\vec{OC}}{||OA||\ ||OC||}
=\frac{30}{36}=\frac{5}{6}$

(d) area is just $\frac{5}{2}\sqrt{11}$
I cannot tell how much I dislike the question. I am sure that whoever wrote it is so proud of her/himself.
Look. you know that the coordinates of \(\displaystyle C:\binom{5}{\sqrt{11}}\).

So in the triangle \(\displaystyle \Delta OAC\) the altitude from \(\displaystyle C\) has length \(\displaystyle \sqrt{11}\).

Thus what is the area of \(\displaystyle \Delta OAC~?\)
 

karush

Well-known member
Jan 31, 2012
2,774
Oops it triangle ABC. area=$6\sqrt{11}$