write MN as the sum of unit vectors for M(-3/4, 5, 2/3) and N(6, -9, 3/5).
Someone help me please?
vector MN = [6+3/4 , -9-5 , 3/5-2/3]
= [ 27/4 , - 14 , -1/15]
magnitude of MN = √(729/16+ 196 + 1/225)
= √(870354/3600)
= appr. 15.549
so a unit vector in the direction of MN
= 1/15.549 [27/4, -14, -1/15]
= [ .434 , -.900 , -.004]
check: .434^2 + (-.9)^2 + (.004)^2 = .99918 (very close to 1, that's good)
so [ .434 , -.900 , -.004]
= .434[1,0,0] - .9[0,1,0] - .004[0,0,1]
Thank you so much for providing a clear explanation!! :)
To write MN as the sum of unit vectors, we can first find the vector MN, normalize it to obtain a unit vector, and then express it as a sum of unit vectors.
1. Find the vector MN:
The vector MN is obtained by subtracting the coordinates of point M from those of point N.
MN = N - M = (6, -9, 3/5) - (-3/4, 5, 2/3)
To subtract the coordinates, we subtract corresponding components:
MN = (6 - (-3/4), -9 - 5, 3/5 - 2/3)
MN = (6 + 3/4, -9 - 5, 3/5 - 2/3)
MN = (27/4, -14, 9/15 - 10/15)
MN = (27/4, -14, -1/15)
2. Normalize the vector MN:
To normalize the vector MN, divide each component by the magnitude of MN.
Magnitude of MN (|MN|) is calculated as:
|MN| = √((27/4)^2 + (-14)^2 + (-1/15)^2)
|MN| = √((729/16) + 196 + 1/225)
|MN| = √((729*225 + 16*196 + 16) / (16*225))
|MN| = √((729*225 + 3136 + 16) / (16*225))
|MN| = √((164025 + 3152) / (3600))
|MN| = √(167177/3600)
|MN| ≈ 5.679
Normalize MN by dividing each component by |MN|
MN = (27/4, -14, -1/15) / 5.679
MN ≈ (27/4*5.679, -14/5.679, -1/15*5.679)
MN ≈ (0.744, -2.468, -0.379)
3. Express MN as the sum of unit vectors:
MN can be expressed as a sum of unit vectors by dividing each component by its magnitude.
MN = 0.744 * (1, 0, 0) + (-2.468) * (0, 1, 0) + (-0.379) * (0, 0, 1)
MN ≈ (0.744, 0, 0) + (0, -2.468, 0) + (0, 0, -0.379)
Therefore, MN can be written as the sum of unit vectors as (0.744, -2.468, -0.379).
To write MN as the sum of unit vectors, we first need to find the direction vector of MN and then express it as a unit vector.
Step 1: Finding the direction vector of MN:
The direction vector of MN is obtained by subtracting the position vector of M from the position vector of N.
MN = N - M
MN = (6, -9, 3/5) - (-3/4, 5, 2/3)
= (6 + 3/4, -9 - 5, 3/5 - 2/3)
= (24/4 + 3/4, -9 - 5, 3/5 - 2/3)
= (27/4, -14, 9/15 - 10/15)
= (27/4, -14, -1/15)
So, the direction vector of MN is (27/4, -14, -1/15).
Step 2: Expressing the direction vector as a unit vector:
To find the unit vector of MN, we need to divide the direction vector by its magnitude.
Magnitude of MN = √((27/4)^2 + (-14)^2 + (-1/15)^2)
= √(729/16 + 196 + 1/225)
= √(131769/3600)
= 11.49 (approximately)
Unit vector of MN = (27/4, -14, -1/15) / 11.49
= (27/4 * 1/11.49, -14/11.49, -1/15 * 1/11.49)
= (27/183.53, -14/11.49, -1/172.35)
Therefore, MN can be expressed as the sum of unit vectors as:
MN ≈ (27/183.53)i - (14/11.49)j - (1/172.35)k