Question

Let vector a= i+j+k, b=i-j+k and c=i-j-k be three vectors. what is vector v which is in the plane of a and b, whose projection on c is 1/3?

Answer 1

Given: Three vectors a= i+j+k, b=i-j+k and c=i-j-k

To find: What is vector v which is in the plane of a and b, whose projection on c is 1/3?

Solution:

• Now the vector in the plane a and b will be:

                r = a + λb = i + j + k + λ( i - j + k )

                r = (1 + λ)i + (1 - λ)j + (1 + λ)k

• Projection this r on the vector c will be:

                r.c / |c|

                = (1 + λ)i + (1 - λ)j + (1 + λ)k . i - j - k / √1 + 1 + 1

                = 1 +  λ + 1 - λ + 1 + λ / √3

                = 3 + λ / √3

• Now this will be equal to the magnitude 1/3, so:

                3 + λ / √3 = ± 1/3

                3 + λ =  ± √3/3

                3 + λ =  ±1/√3

                λ = (1/√3) - 3 or (-1/√3) - 3

• Hence

        r = a + λb

        r = i + j + k + {(1/√3) - 3( i - j + k )} or r = i + j + k + {-(1/√3) - 3( i - j + k )}

Answer:

• So the vector is

             r = i + j + k + {(1/√3) - 3( i - j + k )}

             r = i + j + k + {-(1/√3) - 3( i - j + k )}