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 \(\displaystyle C:\binom{5}{\sqrt{11}}\).

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]

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

 

[CellCoree]

I can provide a mathematical explanation for the given content. The content mentions an IBV3 vector in a circle, which can be represented by the coordinates of points A, B, and C. Point A is located at (6,0) on the x-axis, while point B is located at (-6,0) on the x-axis. Point C is located at (5, √11) which implies that the distance from point O (the origin) to point C is 6, as √(5^2 + 11) = 6. This can be seen as the radius of the circle.

In part (b), it is mentioned that vector AC can be from the origin, which means that the vector OC is equal to vector OA plus vector AC. This can be written as OC = OA + AC, which simplifies to AC = (OC - OA). In this case, vector AC is equal to (-1, √11).

In part (c), the dot product of vectors OA and OC is calculated and divided by the product of their magnitudes. The dot product is the sum of the products of the corresponding components of two vectors, which in this case is (-6)(5) + (0)(√11) = -30. The magnitude of vector OA is 6 and the magnitude of vector OC is √(5^2 + 11) = 6. Therefore, the result of the equation is 30/36 = 5/6.

Finally, in part (d), the area of the circle is calculated by using the formula A = πr^2, where r is the radius of the circle. In this case, the radius is 6, so the area is (5/2)√11. This is the mathematical explanation for the given content.