A=2i+3j-k, b=4i=2j-2k, find the vector x parallel to A but has magnitude of B
The magnitude of B is sqrt 24.
The magnitude of A is sqrt 14
Multiply vector A by sqrt(12/7)
You did not write vector B correctly. The I should be followed by + or -, not =.
thnx
To find the vector x parallel to vector A but with the magnitude of vector B, we will first find the unit vector in the direction of vector A. Then, we will multiply this unit vector by the magnitude of vector B to obtain the desired vector x.
1. Find the magnitude of vector A:
The magnitude of a vector A = √(a^2 + b^2 + c^2), where a, b, and c are the coefficients of the vector's components:
Magnitude of A = √(2^2 + 3^2 + (-1)^2)
= √(4 + 9 + 1)
= √14
2. Find the unit vector in the direction of vector A:
The unit vector in the direction of vector A can be found by dividing each component of vector A by its magnitude:
Unit vector u = A/|A|, where |A| represents the magnitude of A.
Unit vector u = (2/√14)i + (3/√14)j + (-1/√14)k
3. Find the magnitude of vector B:
The magnitude of vector B can be calculated similarly as we did for vector A:
Magnitude of B = √(4^2 + 2^2 + (-2)^2)
= √(16 + 4 + 4)
= √24
= 2√6
4. Obtain vector x:
Multiply the unit vector u by the magnitude of vector B:
Vector x = (2√6) * ((2/√14)i + (3/√14)j + (-1/√14)k)
Simplifying the expression, we have:
Vector x = (4√6/√14)i + (6√6/√14)j + (-2√6/√14)k
Therefore, vector x is (4√6/√14)i + (6√6/√14)j + (-2√6/√14)k.