Question

Three vectors a, b and c are given. Find the equation of a vector that lies in the plane of vector a and vector b and whose projection on vector c is 1/√3.

Answer 1

To find the equation of a vector that lies in the plane of vectors a and b and has a projection of 1/√3 on vector c, we can use vector projection and vector addition.

Let's denote the vector we want to find as v.

The projection of v on vector c is given by:

proj_c(v) = (v . c) / ||c||

Since we want the projection to be 1/√3, we can write the equation:

(v . c) / ||c|| = 1/√3

Next, we know that the vector v lies in the plane of vectors a and b. Therefore, it can be expressed as a linear combination of vectors a and b:

v = λa + μb

where λ and μ are scalar coefficients.

Now, we can substitute v into the projection equation:

((λa + μb) . c) / ||c|| = 1/√3

Expanding the dot product:

(λ(a . c) + μ(b . c)) / ||c|| = 1/√3

Since v lies in the plane of a and b, we can also write:

v = (a x b) x a

where x denotes the cross product. This means that v is perpendicular to a and lies in the plane of a and b. Therefore, we can take the cross product of a and v to obtain b:

v x a = a x b

Solving for v:

v = (a x b) x a / ||a||^2

Substituting this expression for v into the projection equation:

(((a x b) x a) . c) / (||a||^2 ||c||) = 1/√3

Expanding the double cross product:

((b . a) a - (a . a) b) . c / (||a||^2 ||c||) = 1/√3

Simplifying:

((b . a) / ||a||^2 - 1) (a . c) / ||c|| = 1/√3

This gives us a relationship between the scalar coefficients λ and μ:

(μ(b . a) / ||a||^2 - λ) (a . c) / ||c|| = 1/√3

Therefore, we can choose any value for μ that satisfies this equation and then solve for the corresponding value of λ. The vector v will then be given by v = λa + μb.