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#### karush

##### Well-known member

- Jan 31, 2012

- 2,716

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

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

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