JEE Mains · Maths · STD 12 - 10. vector algebra
In a triangle \(\mathrm{ABC}\), if \(|\overrightarrow{\mathrm{BC}}|=3,|\overrightarrow{\mathrm{C}}|=5\) and \(|\overrightarrow{\mathrm{BA}}|=7\), then the projection of the vector \(\overline{\mathrm{BA}}\) on \(\overline{\mathrm{BC}}\) is equal to:
- A \(\frac{11}{2}\)
- B \(\frac{13}{2}\)
- C \(\frac{19}{2}\)
- D \(\frac{15}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{11}{2}\)
Step-by-step Solution
Detailed explanation
Projection of \(\overline{B A}\) on \(\overrightarrow{\mathrm{BC}}\) is equal to \(=|\vec{BA}| \cos \, A B C\) \(\cos \theta=\frac{49+9-25}{2.7 .3}=\frac{11}{14}\) \(\therefore B A \cos \theta=\frac{11}{2}\)
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