Suppose vector A=3i-2j, B= -i-4j and vector C are three vectors in xy plane with the property that vectors A+B-C. What is the unit vector along vector C?
Don't you mean that A + B = C ?
A + B - C is not a property. It is a vector that depends upon what C is.
A "unit vector along C" will depend upon how C is defined. It will be a combination of i and j. It is hard to believe they are really asking for that.
I bet Dr WLS is right and that - sign should be an = sign
making that assumption:
A + B = C = 2 i - 6 j
so
we need a unit vector in the C direction
Find magnitude of i - 3 j which is in C direction
sqrt (1^2 + 3^2) = sqrt (10)
so
(1/sqrt 10) i - (3/sqrt 10) j
To find the unit vector along vector C, we first need to find vector C itself. We know that vector A + vector B - vector C = 0, so we can rearrange this equation to solve for vector C.
Given vector A = 3i - 2j and vector B = -i - 4j, we can substitute these values into the equation:
(3i - 2j) + (-i - 4j) - C = 0
Combine like terms:
(3i - i) + (-2j - 4j) - C = 0
2i - 6j - C = 0
Now, isolate vector C by moving it to the other side:
C = 2i - 6j
Now that we have the vector C, we can find its magnitude to normalize it and obtain the unit vector.
Magnitude of vector C, denoted as ||C||, is given by the formula:
||C|| = √(C_x² + C_y²)
Substituting C_x = 2 and C_y = -6:
||C|| = √((2)² + (-6)²)
||C|| = √(4 + 36)
||C|| = √40
||C|| = 2√10
To find the unit vector along vector C, divide vector C by its magnitude:
Unit vector C = C / ||C||
Unit vector C = (2i - 6j) / (2√10)
Simplifying by dividing each component:
Unit vector C = i / √10 - 3j / √10
Therefore, the unit vector along vector C is (1/√10)i - (3/√10)j.