How is the IBV3 Vector Applied in a Circle?

[karush]

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 $\frac{5}{2}\sqrt{11}$

 

[alyafey22]

So far , so good :)

 

[Plato2]

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 $$C:\binom{5}{\sqrt{11}}$$.

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

Thus what is the area of $$\Delta OAC~?$$

 

[karush]

Oops it triangle ABC. area=$6\sqrt{11}$