Consider the plane P in the $x-y-z$ space determined by the equation $x + y + z = 0$. Given the vector $\vec{V} =(2,3,4) $, in default position. Find vectors $\vec{A}$ and $\vec{B}$, such that vector $\vec{A}$ lies on the plane $P$, vector $\vec{B}$ is in the same or opposite direction as normal vector to P, and $\vec{V} = \vec{A} + \vec{B}$
Im not sure how to do this question. Can someone help me with this?
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Hint:
what you are searching is the orthogonal decomposition of the vector $\vec V$ in the spaces $W= span (1,1,1)$ and its orthogonal complement $W^\bot$ ( that is the given plane).
Do you see this?
If the vector lies on the plane then must be orthogonal to a normal vector of the plane $\langle 1,1,1 \rangle$.
So we have
$$\langle 1,1,1 \rangle \cdot \langle a_1,a_2,a_3 \rangle=0$$
We have that $\vec B$ is parallel (or equal to) a normal vector of the plane,
$$c \langle 1,1,1 \rangle= \langle b_1,b_2,b_3 \rangle$$
We have $\vec A+\vec B=\vec V$,
$$\langle a_1+b_1,a_2+b_2,a_3+b_3 \rangle= \langle 2,3,4 \rangle$$
Combine this information to solve for $\vec A$ and $\vec B$.