vector A =icap+jcap +2kcap. find a vector B which is perpendicular to vector A and whose magnitude is 5.
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A = <1,1,2>
You want B = <a,b,c> such that
A•B = a+b+2c = 0
Should not be hard to figure out some values, such as
<1,1,-2> or <3,5,-4>
To find a vector B that is perpendicular to vector A and has a magnitude of 5, we can use the cross product between vector A and vector B.
1. Given vector A = î + ĵ + 2k̂.
2. Let vector B = xî + yĵ + zk̂.
3. Since vector B is perpendicular to vector A, the dot product of A and B will be zero.
(A ⋅ B = 0)
4. Taking the dot product of A and B:
(î + ĵ + 2k̂) ⋅ (xî + yĵ + zk̂) = 0
(x + y + 2z) = 0
5. Solving for z in terms of x and y:
z = -(x + y) / 2
6. We can choose any values for x and y, but to ensure that vector B has a magnitude of 5, we need to consider the Pythagorean theorem.
Magnitude of B = √(x² + y² + z²) = 5
(x² + y² + z²) = 5²
(x² + y² + (-(x + y) / 2)²) = 25
7. Simplifying the equation and solving for y:
x² + y² + (x² + 2xy + y²) / 4 = 25
4x² + 4y² + x² + 2xy + y² = 100
5x² + 3y² + 2xy = 100
3y² + (2x)y + (5x² - 100) = 0
8. This is a quadratic equation in terms of y. We can solve it using the quadratic formula:
y = [-2x ± √((2x)² - 4(3)(5x² - 100))] / (2(3))
9. Simplifying further:
y = [-2x ± √(4x² - 60x² + 1200)] / 6
y = [-2x ± √(-56x² + 1200)] / 6
10. B = xî + yĵ + zk̂, which gives two possible vectors B.
B₁ = x₁î + y₁ĵ + (-((x₁ + y₁) / 2))k̂
B₂ = x₂î + y₂ĵ + (-((x₂ + y₂) / 2))k̂
Where:
B₁ = [x₁, -2x₁ ± √(-56x₁² + 1200), -((x₁ ± √(-56x₁² + 1200)) / 2)]
B₂ = [x₂, -2x₂ ± √(-56x₂² + 1200), -((x₂ ± √(-56x₂² + 1200)) / 2)]
The values of x₁, y₁, and x₂, y₂ can be chosen arbitrarily as long as they satisfy the equations above.