I have the equation (a+kb)^2=c where a is a vector, b is a unit vector and k and c are both scalars. By ^2 I mean vector length (is this the correct notation?) I am trying to solve for k, but I am not sure where to start.

Audrey Arnold

Audrey Arnold

Answered question

2022-11-16

I have the equation ( a + k b ) 2 = c where a is a vector, b is a unit vector and k and c are both scalars. By 2 I mean vector length (is this the correct notation?)
I am trying to solve for k, but I am not sure where to start.
I have noticed when I plot this graph there is sometimes no solution for k (for example if k is less than a 2 and b is perpendicular to a), is it possible to solve the lower bound?

Answer & Explanation

mangoslush27fig

mangoslush27fig

Beginner2022-11-17Added 15 answers

You can use the scalar product and use | v | 2 = v v for any vector v. Then
| a + k b | 2 = ( a + k b ) ( a + k b ) = | a | 2 + 2 k a b + k 2 | b | 2 .
So your equation for k becomes the quadratic
| b | 2 k 2 + 2 ( a b ) k + | a | 2 c = 0.
You can now solve using the usual solution formula for quadratics (and you find out when there is no solution).

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