Vertices help

**Unt0t** · August 19th 2007, 04:54 PM

a) A triangle ABC has vertices given by A(1,1,8), B(4,-3,-4) and C(-3,1,5)

(i) Determine the distance between B and C (in units).
(ii) Find the angle formed at the vertex A (2 decimal places)

Please help...

**Soroban** · August 19th 2007, 06:03 PM

Hello, Unt0t!

A triangle ABC has vertices given by A(1,1,8), B(4,-3,-4) and C(-3,1,5)

(a) Determine the distance between B and C (in units).

You don't know the Distance Formula?

$BC \:=\:\sqrt{(-3-4)^2 + (1-[-3])^2 + (5-[-4])^2} \:=\:\sqrt{49 + 16 + 81} \:=\:\sqrt{146}$

(b) Find the angle formed at the vertex A (2 decimal places)

The angle $\theta$ between two vectors, $\vec{u},\:\vec{v},$ is given by: . $\cos\theta\;=\;\frac{|\vec{u}\cdot\vec{v}|}{|\vec{ u}||\vec{v}|}$

Let: $\vec{u} \:=\:AB \:=\:\langle4-1,\:\text{-}3-1,\:\text{-}4-8\rangle \:=\:\langle3,\,\text{-}4,\,\text{-}12\rangle$

Let: $\vec{v} \:=\:AC\:=\:\langle\text{-}3-1,\:1-1,\:5-8\rangle \:=\:\langle\text{-}4,\:0,\:\text{-}3\rangle$

Then: . $\cos A \:=\:\frac{|\langle 3,\,\text{-}4,\,\text{-}12\rangle\cdot\langle\text{-}4,\,0,\,\text{-}3\rangle|}{\sqrt{3^2+(\text{-}4)^2+(\text{-}12)^2}\,\sqrt{(\text{-}4)^2+0^2 + (\text{-}3)^2}} \:=\:\frac{|(3)(\text{-}4) + (\text{-}4)(0) + (\text{-}12)(\text{-}3)|}{\sqrt{9 + 16 + 144}\,\sqrt{16 + 0 + 9}}$

. . . . . . . . $= \;\frac{-12 + 0 + 36}{\sqrt{169}\,\sqrt{25}} \;=\;\frac{24}{13\cdot5}$

Therefore: . $A \;=\;\cos^{-1}\left(\frac{24}{65}\right) \;\approx\;68.33^o$

**Unt0t** · August 19th 2007, 06:36 PM

Thanks, i've got another question now.

Let u = (1, -3, -1), v = (1, 1, 2), w = (-2, -2, 1), x = (-1, -1, 1).

a) By considering the dot product only, identify 2 vectors that will satisfy each of the following situations:

(i) the angle between them is less than (pie)/2;
(ii) the angle between them is greater than (pie)/2;

Justify your answers by considering the relationship between the dot product and the cosine of the angle.

b) Identify 2 vectors that will be perpendicular to each other. Justify your answer.

**red_dog** · August 19th 2007, 09:28 PM

If $\overrightarrow{u}=(x_1,y_1,z_1)$ and $\overrightarrow{v}=(x_2,y_2,z_2)$ then $\overrightarrow{u}\cdot\overrightarrow{v}=x_1x_2+y _1y_2+z_1z_2$

We have
$\overrightarrow{u}\cdot\overrightarrow{v}>0\Leftri ghtarrow 0\leq\widehat{\overrightarrow{u}\overrightarrow{v} }<\frac{\pi}{2}$
$\overrightarrow{u}\cdot\overrightarrow{v}<0\Leftri ghtarrow \frac{\pi}{2}<\widehat{\overrightarrow{u}\overrigh tarrow{v}}\leq\pi$
$\overrightarrow{u}\cdot\overrightarrow{v}=0\Leftri ghtarrow \widehat{\overrightarrow{u}\overrightarrow{v}}=\fr ac{\pi}{2}$ .
Now, using these, you can solve the problem.

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