Let’s discuss a beautiful problem useful for Physics Olympiad based on Vector Analysis.

Vector Analysis Problem:

Let (\vec{a}=6\vec{i}-3\vec{j}-6\vec{k}) and (\vec{d}=\vec{i}+\vec{j}+\vec{k}). Suppose that (\vec{a}=\vec{b}+\vec{c}) where (\vec{b}) is parallel to (\vec{d}) and (\vec{c}) is perpendicular to (\vec{d}). Then (\vec{c}) is
(A)(5\vec{i}-4\vec{j}-\vec{k})
(B)   ( 7\vec{i}-2\vec{j}-5\vec{k})
(C)    (4\vec{i}-5\vec{j}+\vec{k})
(D)    (3\vec{i}+6\vec{j}-9\vec{k})

Discussion:

In the given problem, (\vec{a})=(6\vec{i}-3\vec{j}-6\vec{k})
$$\vec{d}=\vec{i}+\vec{j}+\vec{k}$$ and $$\vec{a}=\vec{b}+\vec{c}…(i)$$
Now, let us consider (\vec{b}=\lambda\vec{d}) and (\vec{c}).(\vec{d})=)0.
Therefore, (b=\lambda\vec{i}+\lambda\vec{j}+\lambda\vec{j})
From (1),
$$6\vec{i}-3\vec{j}-6\vec{k}=\lambda\vec{i}+\lambda\vec{j}+\lambda\vec{k}+c$$
$$\Rightarrow\vec{c}=(6-\lambda)\vec{i}-(3+\lambda)\vec{j}-(6+\lambda)\vec{k}$$
Now,
$$\vec{c}.\vec{d}=0$$
$$\Rightarrow\lambda=-1$$
Hence,
$$\vec{c}=7\vec{i}-2\vec{j}-5\vec{k}$$