How would you find the unit vector along the line joining point (2, 4, 4) to point (-3, 2, 2)?

Question

Anna Allen · Accepted Answer

#(1/sqrt33) ((-5), (-2), (-2))# If point P is (2,4,4) and Q is (-3,2,2), the vector PQ would be (-5,-2,-2). To find the unit vector, divide vector PQ by its magnitude. #|| vec (PQ)||# would be #sqrt((-5)^2 +(-2)^2 + (-2)^2)) =sqrt33#. Hence unit vector would be #(1/sqrt33) ((-5), (-2), (-2))#

Chloe Alexander · Answer

Please see the explanation.

Given the points #(x_0, y_0, z_0) and (x_1, y_1, z_1)# You make a vector in the direction from one point to another by subtracting each starting coordinate from its respective ending coordinate: #barV = (x_1 - x_0)hati + (y_1 - y_0)hatj + (z_1 - z_0)hatk# To make it a unit vector, #hatV#, you divide each component of the vector by the magnitude, #|barV|#: #|barV| = sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)# Therefore the general form for a unit vector is: #hatV = (x_1 - x_0)/sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)hati + (y_1 - y_0)/sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)hatj + (z_1 - z_0)/sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)hatk# For the given points #(2, 4, 4)# and #(-3, 2, 2)# Compute the magnitude #|barV| =sqrt((x1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2) = # #sqrt((-3 - 2)^2 + (2 - 4)^2 + (2 - 4)^2) = # #sqrt(33)# We need to divide which it the same as multiplying by the reciprocal, #sqrt(33)/33# Substitute into the unit vector general form: #hatV = (-3 - 2)sqrt(33)/33hati + (2 - 4)sqrt(33)/33hatj + (2 - 4)sqrt(33)/33hatk# Simplify: #hatV = -(5sqrt(33))/33hati - (2sqrt(33))/33hatj - (2sqrt(33))/33hatk#

Xavier Alonso · Answer

undefined To find the unit vector along the line joining point (2, 4, 4) to point (-3, 2, 2), follow these steps:

1. Determine the direction vector by subtracting the coordinates of the two points: $ \vec{v} = (-3 - 2, 2 - 4, 2 - 4) = (-5, -2, -2) $.
2. Find the magnitude of the direction vector: $ |\vec{v}| = \sqrt{(-5)^2 + (-2)^2 + (-2)^2} = \sqrt{25 + 4 + 4} = \sqrt{33} $.
3. Divide the direction vector by its magnitude to obtain the unit vector: $ \hat{v} = \frac{\vec{v}}{|\vec{v}|} = \left( \frac{-5}{\sqrt{33}}, \frac{-2}{\sqrt{33}}, \frac{-2}{\sqrt{33}} \right) $.

Therefore, the unit vector along the line joining the two given points is $ \left( \frac{-5}{\sqrt{33}}, \frac{-2}{\sqrt{33}}, \frac{-2}{\sqrt{33}} \right) $.

Sophie Adams · Answer

undefined To find the unit vector along the line joining the points (2, 4, 4) and (-3, 2, 2), follow these steps:

1. Find the vector that represents the direction of the line by subtracting the coordinates of the second point from the coordinates of the first point.
2. Calculate the magnitude of the direction vector.
3. Divide each component of the direction vector by its magnitude to obtain the unit vector.

Let's denote the points as $ P_1(2, 4, 4) $ and $ P_2(-3, 2, 2) $.

1. The direction vector, $ \mathbf{v} $, is found by subtracting the coordinates of $ P_1 $ from the coordinates of $ P_2 $:
$$ \mathbf{v} = \langle -3 - 2, 2 - 4, 2 - 4 \rangle = \langle -5, -2, -2 \rangle $$

2. The magnitude of $ \mathbf{v} $, denoted as $ \| \mathbf{v} \| $, is calculated using the formula:
$$ \| \mathbf{v} \| = \sqrt{(-5)^2 + (-2)^2 + (-2)^2} = \sqrt{25 + 4 + 4} = \sqrt{33} $$

3. The unit vector, $ \mathbf{u} $, along the line is obtained by dividing each component of $ \mathbf{v} $ by its magnitude:
$$ \mathbf{u} = \left\langle \frac{-5}{\sqrt{33}}, \frac{-2}{\sqrt{33}}, \frac{-2}{\sqrt{33}} \right\rangle $$