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Geometry 3D Line of Intersection Coordinates

Hi attached question and mark scheme answer . I understand everything up to part (d)

I don't follow how using [ λ ,1/2-1/2 λ , 3-λ] dot product [2,-1,-2]=0 is used . Are these two vectors perpendicular .

I understand it evaluates to λ=13/9 and gives the coordinates .

I am also struggling with the alternative answer of differentiation for OP^2.

Thank you

Re: Geometry 3D Line of Intersection Coordinates

Quote:

Hi attached question and mark scheme answer . I understand everything up to part (d)

I don't follow how using [ λ ,1/2-1/2 λ , 3-λ] dot product [2,-1,-2]=0 is used . Are these two vectors perpendicular .

I understand it evaluates to λ=13/9 and gives the coordinates .

I am also struggling with the alternative answer of differentiation for OP^2.

Thank you

Let P be the point on the line, l, closest to the origin, O. Then the vector, OP, is perpendicular to l, which has direction [2,-1,-2]. Assuming P has coordinates (λ , 1/2-1/2 λ, 3-λ), we have OP = [λ , 1/2-1/2 λ, 3-λ]. But OP is perpendicular to [2,-1,-2], or put another way OP.[2,-1,-2] = 0.

Alternatively, minimizing the magnitude of OP, or the square of it for simplicity, will also yield the same result.

|OP|^{2} = λ^{2} + (1/2 - 1/2 λ)^{2} + (3 - λ)^{2}

= λ^{2} + 1/4 - 1/2 λ + 1/4 λ^{2} + 9 - 6λ + λ^{2}

= 9/4 λ^{2} - 13/2 λ + 37/4

= 1/4(9λ^{2} - 26 λ + 37)

We set the derivative of this to 0 to find the critical value of λ. The derivative is

1/4(18 λ - 26).

Setting it to 0 gives the result λ = 13/9.

Hope this helped.