Find the unit vector parallel to the vector : C=12i+24j-8k.
A.9/7i+5/7j+1/7k
B.3/7i+6/7j-2/7k.
I think B is correct.
Please tell me correct option and method of finding.
first make it easier by dividing by 4
3 i + 6 j - 2 k is parallel but easier
now divide all coefs by
sqrt(9+36+4) = 7 :)
3/7 i + 6/7 j - 2/7 k
The correct unit vector parallel to the vector C = 12i + 24j - 8k can be found by dividing the vector by its magnitude.
Magnitude of vector C = √(12² + 24² + (-8)²) = √(144 + 576 + 64) = √784 = 28
Now, dividing each component of vector C by its magnitude, we get:
Ĉ = (12/28)i + (24/28)j + (-8/28)k
Simplifying further, we have:
Ĉ = (3/7)i + (6/7)j + (-2/7)k
Therefore, the correct option for the unit vector parallel to the vector C is B: (3/7)i + (6/7)j - (2/7)k.
To find the unit vector parallel to a given vector, you need to normalize the vector by dividing each component by its magnitude.
First, find the magnitude of the vector C:
|C| = sqrt((12)^2 + (24)^2 + (-8)^2)
|C| = sqrt(144 + 576 + 64)
|C| = sqrt(784)
|C| = 28
Next, divide each component of the vector C by its magnitude:
C-unit = (12/|C|)i + (24/|C|)j + (-8/|C|)k
C-unit = (12/28)i + (24/28)j + (-8/28)k
C-unit = 3/7i + 6/7j - 2/7k
Comparing the options:
Option A: 9/7i + 5/7j + 1/7k
Option B: 3/7i + 6/7j - 2/7k
Since the correct unit vector is C-unit = 3/7i + 6/7j - 2/7k, your choice B is correct.