The vector equation of the line joining the 2 points $i-2j+k$ and $-2j+3k$
A.$r=t\left(i+j+k\right)$
B.$r=t_1\left(i-2j+k\right)+t_2\left(3k-2j\right)$
C.$r=\left(i-2j+k\right)+t\left(2k-i\right)$
D.$r=t\left(2k-i\right)$

Hint: In this question, the position vectors of 2 points are given. We need to find the vector equation of the line joining the given 2 points.



Formula Used:To find the equation of the line joining the given 2 points is
$r=a+t\left(b-a\right).$
Where $a\ is the position vector of a point say A and b\ $is the position vector of a point say B.



Complete step by step solution:The equation of the line joining the given 2 points is
r=a+t(b-a).
Now according to the question, the position vector of point a is i-2j+k and position vector of point b is -2j+3k. Assign the values of a and b to the formula of r. Therefore,
$r=\left(i-2j+k\right)+t\left(\left(-2j+3k\right)-\left(i-2j+k\right)\right)$
Take the minus sign inside.
$r=\left(i-2j+k\right)+t\left(-2j+3k-i+2j-k\right)$
Now carefully do the addition and subtraction.
$r=\left(i-2j+k\right)+t(2k-i)$
Therefore, the vector equation of line joining the points is$ r=\left(i-2j+k\right)+t(2k-i)$



Option ‘C’ is correct



Note: Since the vector equations of the two points are already given, the vector equation of the line joining them can be easily found by using the formula $r=a+t\left(b-a\right).$