FIND A VECTOR EQUATION OF THE LINE THAT CONTAINS(-1,3,0)& IS PARALLEL TO 2i-3j-k.
To find the vector equation of a line that is parallel to a given vector and passes through a given point, you can express the line in parametric form using the formula:
r = l + tn
where r represents the position vector of any point on the line, l is the position vector of the known point on the line, t is a scalar parameter, and n is the direction vector (the vector parallel to the line).
Given:
Point on the line: (-1, 3, 0)
Direction vector: 2i - 3j - k
Let's find the vector equation of the line:
Step 1: Determine the position vector l
The position vector l represents the known point on the line. In this case, the known point is (-1, 3, 0). Therefore, we have:
l = (-1)i + (3)j + (0)k
= -i + 3j + 0k
= -i + 3j
Step 2: Determine the direction vector n
The direction vector n represents the vector parallel to the line. In this case, the direction vector is 2i - 3j - k.
n = 2i - 3j - k
Step 3: Write the vector equation of the line
Using the formula r = l + tn, we can substitute the position vector l and the direction vector n into the equation to get the vector equation of the line:
r = (-i + 3j) + t(2i - 3j - k)
Simplifying further, we have:
r = -i + 3j + 2ti - 3tj - tk
Combining like terms:
r = (-i + 2ti) + (3j - 3tj) - tk
Finally, we can separate the components:
x = -1 + 2t
y = 3 - 3t
z = -t
Therefore, the vector equation of the line that contains the point (-1, 3, 0) and is parallel to 2i - 3j - k is:
r = (-1 + 2t) i + (3 - 3t) j - tk