# [SOLVED]IBV3 vector in a circle

#### karush

##### Well-known member
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:

MHB Math Helper
So far , so good

#### Plato

##### Well-known member
MHB Math Helper
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
Oops it triangle ABC. area=$6\sqrt{11}$