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in JEE by (40 points)
If \( \overrightarrow{ A }=2 \hat{ i }+\sqrt{7} \hat{ j } \) and \( \overrightarrow{ B }=5 \hat{ i }+\sqrt{7} \hat{ j }-3 \hat{ k } \), then the vector whose magnitude is equal to \( \overrightarrow{ A } \cdot \overrightarrow{ B } \) and parallel to \( \overrightarrow{ B }-\overrightarrow{ A } \) is : (A) \( \frac{17}{\sqrt{2}}(\hat{k}-\hat{j}) \) (B) \( \frac{17}{\sqrt{2}}(\hat{i}-\hat{k}) \) (C) \( 3 \hat{i}-3 \hat{k} \) (D) \( 3 \hat{k}-3 \hat{i} \)

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by (120 points)

\(\vec{B} - \vec{A} = 3\hat{i} - 3\hat{k}\) (direction)

\(\vec{A}.\vec{B} = (2.5 + \sqrt{7}\sqrt{7}) = 17\) (magnitude)

\(\frac{17}{\sqrt{2}} (\hat{i} - \hat{k})\)  Ans Option B

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