If possible, find a, b, and c so that
v =
[a
b
c]
is orthogonal to both
w =
[1
6
1]
and x =
[1
−9
1]
A vector that is orthogonal to two given vectors is the cross product, namely
|i j k|
|x1 y1 z1|
|x2 y2 z2|
which in this case is
|i j k|
|1 6 1|
|1 -9 1|
Expanding the determinant, we get the following vector:
<6*1+1*9, 1*1-1*1, 1*-9-6*1>
<15, 0, -15>
which can be reduced to
<1,0,-1>
Check:
<1,6,1>. <1,0,-1>=6-0-6=0
<1,-9,1>. <1,0,-1>=1+0-1=0
Thus <1,0,-1> is orthogonal to both given vectors.
<15,0
how did you expand the determiannt
thank you
To find values for a, b, and c such that vector v is orthogonal to both vectors w and x, we need to find the dot product of v with both w and x, and set the dot products equal to zero.
The dot product of two vectors is given by the sum of the products of their corresponding components. In this case, the dot product of v with w would be:
v · w = a(1) + b(6) + c(1)
Similarly, the dot product of v with x would be:
v · x = a(1) + b(-9) + c(1)
Since we want both dot products to equal zero, we can set up the following system of equations:
a + 6b + c = 0 (Equation 1)
a - 9b + c = 0 (Equation 2)
To solve this system, we can use elimination or substitution. Here, we will use elimination.
First, let's subtract Equation 2 from Equation 1:
(a + 6b + c) - (a - 9b + c) = 0 - 0
Simplifying, we get:
15b = 0
Dividing both sides by 15, we find:
b = 0
Substituting b = 0 back into Equation 1, we get:
a + c = 0
Now, we have two variables, a and c, with one equation. We can choose any value for a or c and solve for the other variable.
Let's set a = 1. Substituting this into the equation a + c = 0, we find:
1 + c = 0
Therefore, c = -1.
So, with a = 1, b = 0, and c = -1, we have the vector v:
v = [1
0
-1]