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Vector Analysis

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

July 3, 2017

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