Question

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

and (4,-2, 1).

b) Determine the unit tangent vector to the following curve at t = 1

-2i+(-4))+ (5t-Pk

Answer 1

1.(a) Given vertices are: $(3,-1,2),(1,-1,2) \text{and} (4,-2,1)$

Area of triangle =$\begin{vmatrix}x_1&x_2&x_3\\y_1&y_2&y_3\\z_1&z_2&z_3\end{vmatrix}$

=$\begin{vmatrix}3&-1&2\\1&-1&2\\4&-2&1\end{vmatrix}$

=$3(-1+4)+1(1-8)+2(-2+4)=3(3)+(-7)+2(2)$

=$9-7+4=6 \text{sq. units}$

1.(b) Given

$t=(1-2)i-4j+5t-pk\Rightarrow 5t-t=i+4j+pk$

$t=\dfrac{i}{4}+j+\dfrac{p}{4}k$

Tangent unit vector is given by,-

$\hat{t}=\dfrac{1}{4}(\dfrac{i+4j+pk}{\sqrt{1^2+4^2+p^2}})$

=$\dfrac{1}{4}(\dfrac{i+4j+pk}{\sqrt{p^2+17}})$