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Step-by-Step Solution
Step 1: Understand what is being asked
We are given a triangle ABC, with lengths:
• |BC| = 3
• |CA| = 5
• |BA| = 7
We want to find the projection of the vector $ \overrightarrow{BA} $ on $ \overrightarrow{BC} $.
Step 2: Recall the formula for projection of one vector onto another
The projection of a vector $ \mathbf{u} $ on another vector $ \mathbf{v} $ is given by:
$$
\text{Proj}_{\mathbf{v}}(\mathbf{u}) \;=\; \frac{\mathbf{u}\cdot \mathbf{v}}{\lvert \mathbf{v} \rvert}.
$$
So we need $ \overrightarrow{BA} \cdot \overrightarrow{BC} $ and the magnitude $ \lvert \overrightarrow{BC} \rvert $ to compute this projection.
Step 3: Express the dot product in terms of magnitudes and the angle
The dot product can also be written as:
$$
\overrightarrow{BA} \cdot \overrightarrow{BC}
\;=\;
\lvert \overrightarrow{BA} \rvert
\lvert \overrightarrow{BC} \rvert
\cos(\angle ABC).
$$
We know $ \lvert \overrightarrow{BA} \rvert = 7 $ and $ \lvert \overrightarrow{BC} \rvert = 3 $, so we need to find $ \cos(\angle ABC). $
Step 4: Apply the Law of Cosines to find $ \cos(\angle ABC) $
In a triangle, the Law of Cosines for side $ AC $ is:
$$
|AC|^2
\;=\;
|AB|^2
+ |BC|^2
- 2\,(|AB|)\,(|BC|)\,\cos(\angle ABC).
$$
Plugging in the given lengths, we get:
$$
5^2
\;=\;
7^2 + 3^2 - 2 \times 7 \times 3 \times \cos(\angle ABC).
$$
$$
25 \;=\; 49 + 9 - 42\,\cos(\angle ABC).
$$
$$
25 \;=\; 58 - 42\,\cos(\angle ABC).
$$
$$
42\,\cos(\angle ABC) = 58 - 25 = 33.
$$
$$
\cos(\angle ABC) = \frac{33}{42} = \frac{11}{14}.
$$
Step 5: Compute $ \overrightarrow{BA} \cdot \overrightarrow{BC} $
Using $ \lvert \overrightarrow{BA} \rvert = 7 $, $ \lvert \overrightarrow{BC} \rvert = 3 $, and $ \cos(\angle ABC) = \tfrac{11}{14} $:
$$
\overrightarrow{BA} \cdot \overrightarrow{BC}
\;=\;
7 \times 3 \times \frac{11}{14}
\;=\;
21 \times \frac{11}{14}
\;=\;
\frac{21 \times 11}{14}
\;=\;
\frac{231}{14}
\;=\;
\frac{33}{2}.
$$
Step 6: Calculate the projection of $ \overrightarrow{BA} $ on $ \overrightarrow{BC} $
Finally, the projection is
$$
\frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{\lvert \overrightarrow{BC} \rvert}
\;=\;
\frac{\frac{33}{2}}{3}
\;=\;
\frac{33}{2} \times \frac{1}{3}
\;=\;
\frac{33}{6}
\;=\;
\frac{11}{2}.
$$
Step 7: Conclude the answer
The projection of the vector $ \overrightarrow{BA} $ on $ \overrightarrow{BC} $ is $ \tfrac{11}{2} $.
Reference Image
