Determine whether the lines given by the vector equations r1=2i+3j+k+s(4i−k) and r2=2i+3j+k+t(2i+2j+k) intersect. If they intersect, give the coordinates of their point of intersection.

They do. the initial point is same in both lines.

formally,

$\vec{r_2}=(2\hat i +3 \hat j +\hat k)+t(2\hat i+2\hat j +\hat k)$

if they intersect then $\begin{vmatrix} 4 & 0 & -1 \\ 2 & 2 & 1 \\ (2-2) & (3-3) & (1-1) \end{vmatrix}=0$

(which obviously is zero)

the answer is b i just did that

step-by-step explanation:

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680