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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,
$$\vec{c}=7\vec{i}-2\vec{j}-5\vec{k}$$