If A = 3i + 6j -2k, then unit vector parallel to A will be
answer is 3/7i + 6/7j - 2/7k
solve it please, because i don't know how he got that answe thanks
Unit vector = V/|V|
if V = ai + bj + ck
|V| = sqrt(a^2 + b^2 + c^2) = length of the vector
@GanonTEK thanks bro , it helped me alot in studying my MCAT but still i don't understand that if there is -ve sign then why we have added all three? and why divided each term by 7? thanks
ahh ok.
well,
|A| = sqrt[ (3)^2 + (6)^2 + (-2)^2 ]
-ve signs don't affect the length of the vector. That only tells you about the direction of it.
|A| = sqrt[ 9 + 36 + 4 ] = sqrt [49] = 7
The length of the vector is 7. Unit vectors have length 1. So we need to divide A by 7 to make it a length of 1.
@GanonTEK , Thanks alot, I got how it works bro, btw your physics is nice and your concepts are good in physics , THanks sir
To find the unit vector parallel to vector A, you need to divide vector A by its magnitude.
First, let's find the magnitude of vector A using the formula:
|A| = √(A_x² + A_y² + A_z²)
Given A = 3i + 6j - 2k, we can substitute the values into the formula:
|A| = √((3)² + (6)² + (-2)²)
= √(9 + 36 + 4)
= √49
= 7
So, the magnitude of vector A is 7.
Next, to find the unit vector parallel to A, divide vector A by its magnitude:
Unit vector = A / |A| = (3i + 6j - 2k) / 7
Now, divide each component of vector A by 7:
(3/7)i + (6/7)j + (-2/7)k
Therefore, the unit vector parallel to vector A is 3/7i + 6/7j - 2/7k.