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.