Question:
In a triangle ABC, if \(\left|\overrightarrow{BC}\right|=3\), \(\left|\overrightarrow{CA}\right|=5\) and \(\left|\overrightarrow{BA}\right|=7\), then the projection of the vector \(\overrightarrow{BA}\) on \(\overrightarrow{BC}\) is equal to :
In a triangle ABC, if \(\left|\overrightarrow{BC}\right|=3\), \(\left|\overrightarrow{CA}\right|=5\) and \(\left|\overrightarrow{BA}\right|=7\), then the projection of the vector \(\overrightarrow{BA}\) on \(\overrightarrow{BC}\) is equal to :
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Projection \(=\frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BC}|}\); get the dot product from the cosine rule.
Updated On: Nov 4, 2026
- \(\frac{19}{2}\)
- \(\frac{13}{2}\)
- \(\frac{11}{2}\)
- \(\frac{15}{2}\)
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The Correct Option is C
Solution and Explanation
\(\overrightarrow{CA}=\overrightarrow{BA}-\overrightarrow{BC}\), so \(|CA|^2=|BA|^2+|BC|^2-2\,\overrightarrow{BA}\cdot\overrightarrow{BC}\).
\(25=49+9-2\,\overrightarrow{BA}\cdot\overrightarrow{BC}\Rightarrow\overrightarrow{BA}\cdot\overrightarrow{BC}=\frac{33}{2}\).
Projection \(=\frac{33/2}{3}=\frac{11}{2}\).
Answer: (3) \(\frac{11}{2}\)
\(25=49+9-2\,\overrightarrow{BA}\cdot\overrightarrow{BC}\Rightarrow\overrightarrow{BA}\cdot\overrightarrow{BC}=\frac{33}{2}\).
Projection \(=\frac{33/2}{3}=\frac{11}{2}\).
Answer: (3) \(\frac{11}{2}\)
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