Question #209796

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

Δ\Delta

b.) let Δ\Delta be the triangle with vertices the points P=(3, 1,-1), Q=(2, 0,3) and R=(1, 1,1). Determine whether Δ\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 θ\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>


1
Expert's answer
2021-06-25T05:45:57-0400

a)


AB=0(1),20,31=1,2,2\overrightarrow{AB}=\langle0-(-1),-2-0,3-1 \rangle=\langle1,-2,2 \rangle


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


AC=4(1),40,11=3,4,0\overrightarrow{AC}=\langle-4-(-1),4-0,1-1 \rangle=\langle-3,4,0 \rangle


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


BC=40,4(2),13=4,6,2\overrightarrow{BC}=\langle-4-0,4-(-2),1-3 \rangle=\langle-4,6,-2 \rangle


BC=(4)2+(6)2+(2)2=214BC=\sqrt{(-4)^2+(6)^2+(-2)^2}=2\sqrt{14}

b)


PQ=23,01,3(1)=1,1,4\overrightarrow{PQ}=\langle2-3,0-1,3-(-1) \rangle=\langle-1,-1,4 \rangle


PR=13,11,1(1)=2,0,2\overrightarrow{PR}=\langle1-3,1-1,1-(-1) \rangle=\langle-2,0,2 \rangle

QR=12,10,13=1,1,2\overrightarrow{QR}=\langle1-2,1-0,1-3 \rangle=\langle-1,1,-2 \rangle

PQPR=1(2)1(0)+4(2)=100\overrightarrow{PQ}\cdot \overrightarrow{PR}=-1(-2)-1(0)+4(2)=10\not=0

PQQR=1(1)1(1)+4(2)=80\overrightarrow{PQ}\cdot \overrightarrow{QR}=-1(-1)-1(1)+4(-2)=-8\not=0

QRPR=1(2)+1(0)2(2)=20\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)


uv=0,1,12,2,0=0+2+0=20\vec u\cdot\vec v=\langle 0,1,1 \rangle\langle 2,2,0 \rangle=0+2+0=2\not=0

uw=0,1,11,1,0=0+1+0=10\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1\not=0

wv=1,1,02,2,0=2+2+0=0\vec w\cdot\vec v=\langle -1,1,0 \rangle\langle 2,2,0 \rangle=-2+2+0=0

wv\vec w \perp \vec v

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



uw=0,1,11,1,0=0+1+0=1\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1

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

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

cosθ=uwuw=122=12\cos \theta=\dfrac{\vec u\cdot\vec w}{||\vec u||\cdot||\vec w||}=\dfrac{1}{\sqrt{2}\sqrt{2}}=\dfrac{1}{2}

θ=60°\theta=60\degree

d)


un=1,1,2x,y,z=x+y2z=0\vec u\cdot\vec n=\langle 1,1,-2 \rangle\langle x,y,z \rangle=x+y-2z=0

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

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


x=2yx=2y

z=x=2yz=x=2y

2y+y2y=02y+y-2y=0

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

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


e)


u×v=ijk011101\vec u\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 0 & -1 & -1 \\ 1 & 0 & -1 \\ \end{vmatrix}

=i1101j0111+k0110=\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}

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

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


e=33,33,33\vec e=\langle \dfrac{\sqrt{3}}{3}, -\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}\rangle


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