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Find the distance from a vector $v$ which is $(1,2,3,4)$ onto a subspace spanned by $v_1=(1,-1,1,0)^T$ and $v_2=(1,2,1,1)^T$

What I tried was finding the distance $||v-w||$ where $w$ is the projection of $v$ onto the subspace $P_E(v)$.
I came across a different method in which we first need to find the basis of the orthogonal complement of the two vectors that span $E$, say, $x$ and then find the projection of $v$ onto $x$ to find the distance $||P_x(w)||$
I tried the two methods and some questions, the answer actually matched while for a few it did not. I'm not sure what I'm missing here. How do we calculate the distance between a vector and a subspace? Do the two methods yield same answers? If not, why? Please try to explain in the context of the question given above. Thanks.

Hint.

Given $vec v_1, vec v_2,vec v_3$ with the usual scalar product, $vec ucdot vec v = sum u_i v_i$ we have

$$
d^2 = ||vec v -alphavec v_1-beta vec v_2||^2 = f(alpha,beta)
$$

now calculating

$$
min_{alpha,beta} f(alpha,beta)
$$

we will solve this problem.

NOTE

$||vec u|| = sqrt{vec ucdotvec u}$

In the finite-dimensional case just use the Gram-determinant $G$, see https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant.

$sqrt{G(v,v_1,v_2)}$ ist the $3$-volume of the parallelepiped spanned by $v$, $v_1$ and $v_2$, whereas $sqrt{G(v_1,v_2)}$ ist the $2$-volume of the parallelepiped spanned by $v_1$ and $v_2$, its base. Now use that a parallelepiped's volume is its base area times its height, so the desired distance is $sqrt{G(v,v_1,v_2)}/sqrt{G(v_1,v_2)}$.