If vector u = [1,-3,-2] and vector v = [4,-5,3] find:
a) |vector u + vector v - 5j |
b) Find Q, where P (-7,5,-1] and vector PQ=2(vector u)
c)find vector w which is orthogonal to both vector u and vector v
if you are using the standard notiation to label a vector with [i, j, k]
then 5j would be [0, 5, 0]
a) |vector u + vector v - 5j |
= |[1,-3,-2] + [4,-5,3] - [0,5,0] |
= | [5, -13, 1]
= √(25+169+1)
= √195
b) let Q be (x,y,z)
then PQ = [x+7,y-5,z+1]
but PQ = 2u = [2,-6,-4]
then:
x+7=2 --->x = -5
y-5=-6 --->y = -1
z+1 = -4 --> z = -5
Q is (-5,-1,-5)
c) you want u x u, the cross-product
I don't know what method you have learned, but you must have one.
See if you can come up with [19,11,-7]
thanks! Does, vector u x vector v = vector w?
a) To find the magnitude of vector u + vector v - 5j, you can follow these steps:
Step 1: Add vector u and vector v:
u + v = [1, -3, -2] + [4, -5, 3]
= [1 + 4, -3 + (-5), -2 + 3]
= [5, -8, 1]
Step 2: Subtract 5j from the resulting vector:
u + v - 5j = [5, -8, 1] - [0, 5, 0]
= [5, -8 - 5, 1 - 0]
= [5, -13, 1]
Step 3: Find the magnitude of the resulting vector:
|u + v - 5j| = sqrt(5^2 + (-13)^2 + 1^2)
= sqrt(25 + 169 + 1)
= sqrt(195)
Therefore, |u + v - 5j| = sqrt(195).
b) To find point Q, you can use the formula Q = P + vector PQ. Given that P = (-7, 5, -1) and vector PQ = 2(vector u), you can proceed with the following steps:
Step 1: Multiply vector u by 2:
2(vector u) = 2[1, -3, -2]
= [2, -6, -4]
Step 2: Add vector PQ to point P:
Q = [-7, 5, -1] + [2, -6, -4]
= [-7 + 2, 5 + (-6), -1 + (-4)]
= [-5, -1, -5]
Therefore, Q = (-5, -1, -5).
c) To find a vector that is orthogonal to both u and v, you can take their cross product. Here's how:
Step 1: Take the cross product of u and v:
u x v = [1, -3, -2] x [4, -5, 3]
Step 2: Compute the determinant of the resulting matrix:
|i j k |
|1 -3 -2 |
|4 -5 3 |
= i(-3*3 - (-2*-5)) - j(1*3 - (-2*4)) + k(1*(-5) - (-3*4))
= i(-9 + 10) - j(3 + 8) + k(-5 + 12)
= i(1) - j(11) + k(7)
= [1, -11, 7]
Therefore, vector w = [1, -11, 7] is orthogonal to both u and v.