Key Concepts and Formulas

• Vector Projection: The projection of vector $\vec{a}$ onto vector $\vec{b}$ is given by $\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = |\vec{a}| \cos \theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
• Cosine Rule: In a triangle with sides $a, b, c$ and angles $A, B, C$ opposite to them respectively, the cosine of an angle is given by:
  • $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$

Step-by-Step Solution

Step 1: Identify the vectors and the required projection. We are asked to find the projection of vector $\overrightarrow{BA}$ on vector $\overrightarrow{BC}$. Let $\vec{a} = \overrightarrow{BA}$ and $\vec{b} = \overrightarrow{BC}$. The magnitudes are given as: $|\overrightarrow{BC}| = 3$ $|\overrightarrow{CA}| = 5$ $|\overrightarrow{BA}| = 7$

Using standard triangle notation where $a, b, c$ are the lengths of the sides opposite to vertices $A, B, C$ respectively: $a = |\overrightarrow{BC}| = 3$ $b = |\overrightarrow{CA}| = 5$ $c = |\overrightarrow{BA}| = 7$

The projection of $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is given by the formula: $\text{Proj}_{\overrightarrow{BC}} \overrightarrow{BA} = |\overrightarrow{BA}| \cos(\angle ABC)$ We need to find the value of $\cos(\angle ABC)$.

Step 2: Apply the Cosine Rule to find $\cos(\angle ABC)$. The Cosine Rule relates the lengths of the sides of a triangle to the cosine of one of its angles. For angle $B$ (which is $\angle ABC$), the rule is: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$ Substituting the given side lengths: $a = |\overrightarrow{BC}| = 3$ $b = |\overrightarrow{CA}| = 5$ $c = |\overrightarrow{BA}| = 7$

$\cos(\angle ABC) = \frac{3^2 + 7^2 - 5^2}{2 \times 3 \times 7}$ $\cos(\angle ABC) = \frac{9 + 49 - 25}{42}$ $\cos(\angle ABC) = \frac{58 - 25}{42}$ $\cos(\angle ABC) = \frac{33}{42}$ Simplifying the fraction by dividing the numerator and denominator by 3: $\cos(\angle ABC) = \frac{11}{14}$

Step 3: Calculate the projection of $\overrightarrow{BA}$ on $\overrightarrow{BC}$. Now we use the projection formula with the values we have: $\text{Proj}_{\overrightarrow{BC}} \overrightarrow{BA} = |\overrightarrow{BA}| \cos(\angle ABC)$ We know $|\overrightarrow{BA}| = 7$ and $\cos(\angle ABC) = \frac{11}{14}$.

$\text{Proj}_{\overrightarrow{BC}} \overrightarrow{BA} = 7 \times \frac{11}{14}$ $\text{Proj}_{\overrightarrow{BC}} \overrightarrow{BA} = \frac{7 \times 11}{14}$ $\text{Proj}_{\overrightarrow{BC}} \overrightarrow{BA} = \frac{77}{14}$ Simplifying the fraction by dividing the numerator and denominator by 7: $\text{Proj}_{\overrightarrow{BC}} \overrightarrow{BA} = \frac{11}{2}$

Common Mistakes & Tips

• Correct Angle Identification: Ensure the angle used in the projection formula is the actual angle between the two vectors when placed tail-to-tail. In this case, $\overrightarrow{BA}$ and $\overrightarrow{BC}$ both originate from $B$, so $\angle ABC$ is the correct angle.
• Cosine Rule Application: Carefully match the sides with the angles in the Cosine Rule. For $\cos B$, the side opposite to $B$ (which is $b$ or $|\overrightarrow{CA}|$) is subtracted, and the adjacent sides ($a$ and $c$, or $|\overrightarrow{BC}|$ and $|\overrightarrow{BA}|$) are used in the positive terms and the denominator.
• Scalar Projection vs. Vector Projection: The question asks for the projection, which is a scalar quantity (a length). The formula $|\vec{a}| \cos \theta$ gives this scalar projection. Do not confuse this with the vector projection, which would be $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$.

Summary

The problem requires calculating the scalar projection of one vector onto another in a triangle. This is achieved by first identifying the magnitudes of the sides and the angle between the vectors. The Cosine Rule is then applied to find the cosine of the angle $\angle ABC$ using the given side lengths. Finally, the scalar projection is computed by multiplying the magnitude of the vector being projected ($\overrightarrow{BA}$) by the cosine of the angle between the two vectors.

The final answer is $\boxed{{{11} \over 2}}$.