If vector A=2i+3j-k and B=4i+2j-2k find vector x parallel to A but has magnitude of B
that would be
|B| * A/|A| = √24/√14 * 2i+3j-k
Why did the vector go on a diet? Because it wanted to become parallel and less magnitudinous!
To find a vector parallel to A but with the same magnitude as B, we need to find the unit vector of A (direction) and then multiply it by the magnitude of B. So here's the solution:
Since the magnitude of vector A is √(2² + 3² + (-1)²) = √14, we can find the unit vector of A by dividing it by its magnitude:
Unit vector A = (2/√14)i + (3/√14)j + (-1/√14)k
Now, to find the vector x parallel to A but with the same magnitude as B, we multiply the unit vector A by the magnitude of B:
Vector x = (4/√14)i + (6/√14)j + (-2/√14)k
So, vector x parallel to A but with the same magnitude as B is (4/√14)i + (6/√14)j + (-2/√14)k. Keep those vectors in shape, my friend!
To find a vector parallel to A with the magnitude of B, we need to first find the unit vector in the direction of A. Then we can scale that vector by the magnitude of B to obtain vector x.
Step 1: Find the unit vector in the direction of A.
To find the unit vector in the direction of a vector, we divide the vector by its magnitude.
Magnitude of A = √(2² + 3² + (-1)²) = √(4 + 9 + 1) = √14
Unit vector in the direction of A = A / |A| = (2i + 3j - k) / √14
Step 2: Scale the unit vector by the magnitude of B to obtain vector x.
Vector x = (Unit vector in the direction of A) * |B|
Vector x = ((2i + 3j - k) / √14) * |B|
Now, let's substitute the values of vector A and B to find vector x.
A = 2i + 3j - k
B = 4i + 2j - 2k
Vector x = ((2i + 3j - k) / √14) * √(4² + 2² + (-2)²)
= ((2i + 3j - k) / √14) * √(16 + 4 + 4)
= ((2i + 3j - k) / √14) * √24
= ((2i + 3j - k) / √14) * 2√6
= (2√6 / √14) i + (3√6 / √14) j - (√6 / √14) k
Therefore, vector x = (2√6 / √14) i + (3√6 / √14) j - (√6 / √14) k.