Question

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

Answer 1

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

$$\frac 1 2 (\overrightarrow{AB} \times \overrightarrow{AC})$$ $$\overrightarrow{AB}=(1-3)\overrightarrow{i}+(-1+1)\overrightarrow{j}+(2-2)\overrightarrow{k}=-2\overrightarrow{i}$$ $$\overrightarrow{AC}=(4-3)\overrightarrow{i}+(-2+1)\overrightarrow{j}+(1-2)\overrightarrow{k}=\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}$$ $$(\overrightarrow{AB} \times \overrightarrow{AC})=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ -2 & 0 & 0 \\ 1 & -1 & -1 \end{vmatrix} =0\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}=-2\overrightarrow{j}+2\overrightarrow{k}$$

Vector area of the triangle ABC:

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

Area of the triangle ABC:

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

(b)

Given that

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

Now , Differentiating it with respect to t:

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

Now, Unit tangent vector,

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

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

at t=1

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