To show: The value of d(u,projvu) is minimum among all the scalar multiples cv of the vector v. Given: The value of u is (6,2,4) and the value of v is (1,2,0).

Approach:
1. The orthogonal projection of u onto v is given as,
${\displaystyle pro{j}_{v}u=⟨u,v⟩⟨v,v⟩v\dots \left(1\right)}$
2. The value of ${\displaystyle \left|\left|v\right|\right|}$ is given as,
${\displaystyle \left|\left|v\right|\right|2=v\cdot v\dots \left(2\right)}$
3. The value of d(u,projvu) is given as,
${\displaystyle d(u,pro{j}_{v}u)=\left||u-pro{j}_{v}u|\right|\dots \left(3\right)}$
4. The value of d(u,cv) is given as,
${\displaystyle d(u,cv)=\left|\left|(u-cv)\right|\right|2=\left||(u-bv)+(b-c)v|\right|2\dots \left(4\right)}$
Calculation:
Consider,
${\displaystyle u=(6,2,4)}$ and ${\displaystyle v=(1,2,0)}$.
The value of ${\displaystyle u\cdot v}$ is given as,
${\displaystyle u\cdot v=(6,2,4)(1,2,0)=\left(6\right)\left(1\right)+\left(2\right)\left(2\right)+\left(4\right)\left(0\right)=6+4+0=10}$
Substitute (1,2,0) for v in equation (2).
${\displaystyle \left|\left|v\right|\right|2=(1,2,0)(1,2,0)=\left(1\right)\left(1\right)+\left(2\right)\left(2\right)+\left(0\right)\left(0\right)=1+4+0=5}$
Substitute (1,2,0) for v, 10 for ${\displaystyle u\cdot v}$, and 5 for ${\displaystyle v\cdot v}$ in equation (1).
${\displaystyle projvu=105(1,2,0)=2(1,2,0)=(2,4,0)}$
Substitute (2,4,0) for projvu and (6,2,4) for u in equation (3).
${\displaystyle d(u,pro{j}_{v}u)=\left||(6,2,4)-(2,4,0)|\right|=\left|\left|\begin{array}{ccc}6-2& 2-4& 4-0\end{array}\right|\right|=\left|\left|\begin{array}{ccc}4& -2& 4\end{array}\right|\right|}$
The value of ${\displaystyle \left|\left|\begin{array}{ccc}4& -2& 4\end{array}\right|\right|}$ is given as,
${\displaystyle \left|\left|\begin{array}{ccc}4& -2& 4\end{array}\right|\right|=\left(4\right)2+(-2)2+\left(4\right)2=16+4+16=36=6}$
So, the value of d(u,projvu) is 6.
Suppose, x is real number different from ${\displaystyle b=2=⟨u,v⟩⟨v,v⟩}$.
Here,