How to scale a fat matrix to ensure its rows are orthonormal?

Question

I have the following matrix

$${\bf B} = \begin{bmatrix} 0&-4.2423&4.2423&1.4871\\ 1.6532&-1.2735& -1.2735&0.0024\\ 0 & -0.2805 & 0.2805 & -0.8823 \end{bmatrix}$$

I need to find a coordinate change necessary to scale $\bf B$ so that ${\bf B} {\bf B}^\top = {\bf I}_3$ and therefore its Euclidian norm is $\|{\bf B}\| = 1$. In other words i need to Transform B such that the TRANSFORMED B would satisfy ${\bf B} {\bf B}^\top = {\bf I}_3$ and therefore $\|{\bf B}\| = 1$.

$${\bf \mathbf{BB}^\top } = \begin{bmatrix} 38.2057 & 0.0036 & 1.0679 \\ 0.0036 & 5.9767 & -0.0021 \\ 1.0679 & -0.0021 & 0.9358 \end{bmatrix}$$

and i need to find a coordinate change necessary to scale $\bf B$ so that ${\bf B} {\bf B}^\top = {\bf I}_3$

I can't figure out a way to perform this operation.

for additional information about what im actually trying to do here I'm adding additional information.

Actually the B matrix in question is (a part of) the input distribution matrix (the Actual B matrix is divided into two parts B1 and B2 but we are concerned here with only B2 and therefore im referring to it as B instead of B2) of an ADMIRE Aircraft model and I am trying to design a fault tolerant control system for this system. I am following this research paper in which the author have mentioned

It will be assumed without loss of generality that the states of the system have been transformed so that ${\bf B} {\bf B}^\top = {\bf I}_3$ and therefore $\|{\bf B}\| = 1$. . This is always possible since rank(B) = l.

Here l is 3.

Im trying to follow this instruction of the author. i need to Transform B such that the TRANSFORMED B would satisfy ${\bf B} {\bf B}^\top = {\bf I}_3$ and therefore $\|{\bf B}\| = 1$.

i think that the matlab code is not much relevant but still i'm posting the code. There is a simulink model that uses the variables etc from the matlab workspace.

A=[-0.5432 0.0137 0 0.9778 0;
        0 -0.1179 0.2215 0 -0.9661;
        0 -10.5128 -0.9967 0 0.6176;
        2.6221 -0.0030 0 -0.5057 0;
        0 0.7075 -0.0939 0 -0.2127];
    % This B is the overall input distribution matrix of the system. This B is divided into B1 and B2 below. 
    B = [0.0069 -0.0866 -0.0866 0.0004; 
        0 0.0119 -0.0119 0.0287;
        0 -4.2423 4.2423 1.4871;
        1.6532 -1.2735 -1.2735 0.0024;
        0 -0.2805 0.2805 -0.8823];
    
    C= [0 0 1 0 0;
        0 0 0 0 1;
        0 0 0 1 0];
    B1 = B(1:2,:);  % here B is broken into B1 and B2
    B2 = B(3:5,:); 
    Cc= [eye(3) zeros(3,2)];
    Aa = [zeros(3,3) -Cc;
        zeros(5,3) A];
    Ba = [zeros(3,4) ; B];
    Br = [eye(3) ; zeros(5,3)];
    
    Q = diag([20 20 20 7 10 10 1 1]);
    Sa = [-0.0002 -0.0844 -1.4117 0.0001 0.0701 1 0 0;
        -4.4721 0.0074 0.0001 3.5011 0.0058 0 1 0;
        0.0074 4.4642 -0.2668 -0.0058 -4.2606 0 0 1];
    Lxa = [0.0015 0.7592 12.7053 -0.0012 9.8061 -9.4305 -0.0001 -0.5499;
        40.2492 -0.0665 -0.0006 -36.7020 -0.0895 -0.0012 -11.9176 0.0056;
        -0.0664 -40.1775 2.4009 0.0562 41.5999 0.7709 0.0056 -12.9035];
    Lr = [0.0002 0.0844 1.4117;
        4.4721 -0.0074 -0.0001;
        -0.0074 -4.4642 0.2668];
    Tau = -20 * eye(3);
    BResult= (B2*B2');
    norm(B21)
    norm(B1)