Answer to Question #189994 in Mechanics | Relativity for Akash

Question #189994

a) Calculate the area of a triangle whose vertices are given by (3, −1, 2), (1, −1, 2) 

and (4, −2, 1). (5)

b) Determine the unit tangent vector to the following curve at 

t = 1

r i j 5 k


1
Expert's answer
2021-05-07T08:46:23-0400

(a) For A(3,-1,2), B(1,-1,2) and C(4,-2,1) vector area of triangle is:



12(AB×AC)\frac 1 2 (\overrightarrow{AB} \times \overrightarrow{AC})AB=(13)i+(1+1)j+(22)k=2i\overrightarrow{AB}=(1-3)\overrightarrow{i}+(-1+1)\overrightarrow{j}+(2-2)\overrightarrow{k}=-2\overrightarrow{i}AC=(43)i+(2+1)j+(12)k=ijk\overrightarrow{AC}=(4-3)\overrightarrow{i}+(-2+1)\overrightarrow{j}+(1-2)\overrightarrow{k}=\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}(AB×AC)=ijk200111=0i2j+2k=2j+2k(\overrightarrow{AB} \times \overrightarrow{AC})=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ -2 & 0 & 0 \\ 1 & -1 & -1 \end{vmatrix} =0\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}=-2\overrightarrow{j}+2\overrightarrow{k}

Vector area of the triangle ABC:



12(AB×AC)=j+k\frac 1 2 (\overrightarrow{AB} \times \overrightarrow{AC})=-\overrightarrow{j}+\overrightarrow{k}

Area of the triangle ABC:



Area=12(AB×AC)=(1)2+12=2Area=|\frac 1 2 (\overrightarrow{AB} \times \overrightarrow{AC})|=\sqrt{(-1)^2+1^2}=\sqrt 2



(b)

Given that

 r=(1+t2)i^+(4t3)j^+(2t26t)k^r =(1+t^2)\hat{i} + (4t-3)\hat{j} + (2t^2-6t)\hat{k}

Now , Differentiating it with respect to t:

r=2ti^+4j^+(4t6)k^r' = 2t\hat{i} + 4\hat{j} + (4t-6)\hat{k}


Now, Unit tangent vector,


r^=rr2ti^+4j^+(4t6)k^4t2+16+(4t6)2\hat{r'} = \dfrac{\vec{r'}}{|r'|} \\\Rightarrow \dfrac{2t\hat{i} + 4\hat{j} + (4t-6)\hat{k}}{\sqrt{4t^2 + 16 + (4t-6)^2}} 


2ti^+4j^+(4t6)k^(20t212t+52)ti^+2j^+(2t3)k^(5t23t+13)\Rightarrow \dfrac{2t\hat{i} + 4\hat{j} + (4t-6)\hat{k}}{\sqrt{(20t^2-12t+52)}} \\\Rightarrow \dfrac{t\hat{i} + 2\hat{j} + (2t-3)\hat{k}}{\sqrt{(5t^2-3t+13)}}


at t=1

the unit tangent vector = i^+2j^+(1)k^15=115i^+215j^115k^\dfrac{\hat i+2\hat j+(-1)\hat k}{\sqrt{15}}=\frac{1}{\sqrt{15}}\hat i+\frac{2}{\sqrt {15}}\hat j - \frac{1}{\sqrt{15}}\hat k


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