find the direction cosines of the vector joining the two points (4,2,2,) & (7,6,14)
(this is new to me, used to think it was just one point not two.)
change in x = 7 - 4 = 3
change in y = 6 - 2 = 4
change i z = 14 - 2 = 12
so V = 3 i + 4 j + 12 k is in the direction
magnitude:
sqrt (9+16+144) = 13
dot product with unit x vector for example
V dot 1 i = 13 * 1 * cos angle
so
cos angle = 3/13
then y axis
cos angle = 4/13
and x
cos angle = 12/13
in the text book, the answer is -2i -7j -18k . could it be a mistake has being made? i mean your result or the one from the text
I thinks the one on the textbook is correct but why are you doing change in x,y&z
Two points can indeed define a vector! To find the direction cosines of the vector joining the points (4, 2, 2) and (7, 6, 14), we can first find the components of the vector by subtracting the corresponding coordinates:
∆x = 7 - 4 = 3
∆y = 6 - 2 = 4
∆z = 14 - 2 = 12
Now, to find the direction cosines, we divide each component by the magnitude of the vector. The magnitude of the vector is given by:
|m| = √(∆x² + ∆y² + ∆z²)
|m| = √(3² + 4² + 12²)
= √(9 + 16 + 144)
= √169
= 13
Now we can find the direction cosines by dividing each component by the magnitude:
L₁ = ∆x / |m| = 3 / 13
L₂ = ∆y / |m| = 4 / 13
L₃ = ∆z / |m| = 12 / 13
So, the direction cosines of the vector are approximately:
L₁ ≈ 0.23
L₂ ≈ 0.31
L₃ ≈ 0.92
Now, the vector will take you to the direction of terrible puns and silly jokes! Enjoy your journey!
To find the direction cosines of a vector joining two points, we need to first find the components of the vector.
The components of the vector are obtained by subtracting the corresponding coordinates of the two points.
Given points: A(4, 2, 2) and B(7, 6, 14)
The components of the vector AB can be found as follows:
AB = (x2 - x1, y2 - y1, z2 - z1)
= (7 - 4, 6 - 2, 14 - 2)
= (3, 4, 12)
Next, we need to find the magnitudes of the components of the vector AB.
Magnitude of a 3D vector can be found using the formula:
|AB| = sqrt(x^2 + y^2 + z^2)
= sqrt(3^2 + 4^2 + 12^2)
= sqrt(9 + 16 + 144)
= sqrt(169)
= 13
Finally, the direction cosines of the vector AB are calculated by dividing each component of the vector by its magnitude.
Direction cosines: (l, m, n)
l = x/|AB| = 3/13
m = y/|AB| = 4/13
n = z/|AB| = 12/13
So, the direction cosines of the vector joining the points A(4, 2, 2) and B(7, 6, 14) are (3/13, 4/13, 12/13).