Answer to Question #209796 in Analytic Geometry for Tshego

Question

a.) Consider the point A=(-1,0,1), B=(0, - 2,3) and C = (-4,4,1) to be vertices of a triangle $\Delta$ . Evaluate all side lengths of

$\Delta$

b.) let $\Delta$ be the triangle with vertices the points P=(3, 1,-1), Q=(2, 0,3) and R=(1, 1,1). Determine whether $\Delta$ is a right triangle. If it is not, explain with reason, why?

c.) let u=<0,1,1>, v=<2,2,0> and w=<-1,1,0> be three vectors in standard form. (i) Determine which two vectors form a right angle triangle? (ii) find $\theta$ =uw, the angel between the given two vectors.

d.) let x<0.find the vector n=<x, Y, z> that is orthogonal to all three vectors u=<1,1,-2>,v=<-1,2,0> and w=<-1,0,1>.

e.)find a unit vector that is orthogonal to both u =<0,-1,-1> and v=<1,0,-1>

Answer 1

a)

$$\overrightarrow{AB}=\langle0-(-1),-2-0,3-1 \rangle=\langle1,-2,2 \rangle$$

$$AB=\sqrt{(1)^2+(-2)^2+(2)^2}=3$$

$$\overrightarrow{AC}=\langle-4-(-1),4-0,1-1 \rangle=\langle-3,4,0 \rangle$$

$$AC=\sqrt{(-3)^2+(-4)^2+(0)^2}=5$$

$$\overrightarrow{BC}=\langle-4-0,4-(-2),1-3 \rangle=\langle-4,6,-2 \rangle$$

$$BC=\sqrt{(-4)^2+(6)^2+(-2)^2}=2\sqrt{14}$$

b)

$$\overrightarrow{PQ}=\langle2-3,0-1,3-(-1) \rangle=\langle-1,-1,4 \rangle$$

$$\overrightarrow{PR}=\langle1-3,1-1,1-(-1) \rangle=\langle-2,0,2 \rangle$$

$$\overrightarrow{QR}=\langle1-2,1-0,1-3 \rangle=\langle-1,1,-2 \rangle$$

$$\overrightarrow{PQ}\cdot \overrightarrow{PR}=-1(-2)-1(0)+4(2)=10\not=0$$

$$\overrightarrow{PQ}\cdot \overrightarrow{QR}=-1(-1)-1(1)+4(-2)=-8\not=0$$

$$\overrightarrow{QR}\cdot \overrightarrow{PR}=-1(-2)+1(0)-2(2)=-2\not=0$$

There is no a pair of orthogonal vectors, hence $\vartriangle$ is not right.

c)

$$\vec u\cdot\vec v=\langle 0,1,1 \rangle\langle 2,2,0 \rangle=0+2+0=2\not=0$$

$$\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1\not=0$$

$$\vec w\cdot\vec v=\langle -1,1,0 \rangle\langle 2,2,0 \rangle=-2+2+0=0$$

$$\vec w \perp \vec v$$

Vectors $\vec w$ and $\vec v$ form a right angle triangle.

$$\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1$$

$$||\vec u||=\sqrt{(0)^2+(1)^2+(1)^2}=\sqrt{2}$$

$$||\vec w||=\sqrt{(-1)^2+(1)^2+(0)^2}=\sqrt{2}$$

$$\cos \theta=\dfrac{\vec u\cdot\vec w}{||\vec u||\cdot||\vec w||}=\dfrac{1}{\sqrt{2}\sqrt{2}}=\dfrac{1}{2}$$

$$\theta=60\degree$$

d)

$$\vec u\cdot\vec n=\langle 1,1,-2 \rangle\langle x,y,z \rangle=x+y-2z=0$$

$$\vec v\cdot\vec n=\langle -1,2,0 \rangle\langle x,y,z\rangle=-x+2y=0$$

$$\vec w\cdot\vec n=\langle -1,0,1 \rangle\langle x,y,z\rangle=-x+z=0$$

$$x=2y$$

$$z=x=2y$$

$$2y+y-2y=0$$

$$x=0, y=0, z=0$$

Since $x<0,$ then such vector $\vec n$ does not exist.

e)

$$\vec u\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 0 & -1 & -1 \\ 1 & 0 & -1 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} -1 & -1 \\ 0 & -1 \end{vmatrix}-\vec j\begin{vmatrix} 0 & -1 \\ 1 & -1 \end{vmatrix}+\vec k\begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix}$$

$$=\vec i-\vec j+\vec k$$

$$\sqrt{(1)^2+(-1)^2+(1)^2}=\sqrt{3}$$

$$\vec e=\langle \dfrac{\sqrt{3}}{3}, -\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}\rangle$$