Find a vector of length 3 that is perpendicular to [2,1,-2]
First we need a perpendicular, any one will do as long as the dot product is zero. How about
(2,1,-2)dot(1,4 a) = 0
2 + 4 - 2a = 0
-2a = -6
a = 3
so (1,4,3) is a normal
now we have to stretch to a length of 3
length of my choice = (1/√26)
so a unit vector is (1/√26)[1,4,3)
and one of length 3 is
(3/√26([1,4,3]
of course this answer is not unique, there would be an infinite number of answers.
Why did the mathematician get a job at the circus? Because he was outstanding in his field!
To find a vector that is perpendicular to [2,1,-2], we can take the cross product of [2,1,-2] with any other vector. Let's take the cross product with [1,0,1]:
[2,1,-2] x [1,0,1] = (1*-2 - (-2)*0) i + (2*1 - (-2)*1) j + (2*0 - 1*1) k
= -2i + 4j - 1k
So, a vector that is perpendicular to [2,1,-2] is [-2, 4, -1]. It has a length of 3.
To find a vector that is perpendicular to [2, 1, -2], we can use the cross product of two vectors.
Let's choose [1, 0, 0] as one of the vectors. The cross product of [2, 1, -2] and [1, 0, 0] will give us a vector that is perpendicular to both.
Using the formula for cross product:
[i, j, k]
[2, 1, -2]
[1, 0, 0]
= ((1 * -2) - (0 * -2))i - ((2 * 0) - (1 * 0))j + ((2 * 0) - (1 * 1))k
= (-2)i - 0j - 1k
= -2i - k
So, the vector [-2, 0, -1] is perpendicular to [2, 1, -2].
To find a vector that is perpendicular to the given vector [2, 1, -2], we need to find a vector that satisfies the condition of their dot product being zero.
Let's consider the vector [x, y, z] as the vector we are looking for.
The dot product of two vectors can be calculated by multiplying their corresponding components and summing the results. So, the dot product of [2, 1, -2] and [x, y, z] can be calculated as follows:
2x + 1y + (-2z) = 0
Now, we need to find the values of x, y, and z that satisfy this equation.
Let's assume x = 1 to start with. Substituting x = 1 into the equation, we get:
2(1) + 1y - 2z = 0
2 + y - 2z = 0
To keep things simple, let's assume z = 0. Substituting z = 0 into the equation, we get:
2 + y - 2(0) = 0
2 + y = 0
Now, solve for y:
y = -2
So, we have found one possible vector that is perpendicular to [2, 1, -2], which is [1, -2, 0].
But note that there are actually infinitely many vectors that are perpendicular to [2, 1, -2]. This is because we can choose different values for x, y, and z as long as they satisfy the equation 2x + y - 2z = 0.
So, the vector of length 3 that is perpendicular to [2, 1, -2] can be represented as [x, y, z], where x, y, and z are any values that satisfy the equation 2x + y - 2z = 0.