Key Concepts and Formulas

• Vector Representation of Parallelogram Sides: In a parallelogram $ABCD$, if $\overrightarrow{AB} = \overrightarrow{q}$ and $\overrightarrow{AD} = \overrightarrow{p}$, then the diagonals can be represented as $\overrightarrow{AC} = \overrightarrow{p} + \overrightarrow{q}$ and $\overrightarrow{BD} = \overrightarrow{q} - \overrightarrow{p}$.
• Vector Projection: The vector projection of vector $\overrightarrow{a}$ onto vector $\overrightarrow{b}$ is given by $\text{proj}_{\overrightarrow{b}} \overrightarrow{a} = \left( \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|^2} \right) \overrightarrow{b} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{\overrightarrow{b} \cdot \overrightarrow{b}} \overrightarrow{b}$. This projection represents the component of $\overrightarrow{a}$ that lies along the direction of $\overrightarrow{b}$.
• Triangle Law of Vector Addition: For any three points $A, B, C$, the vector from $A$ to $C$ can be expressed as $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$.

Step-by-Step Solution

Step 1: Understanding the Problem and Visualizing the Geometry We are given a parallelogram $ABCD$ with $\overrightarrow{AB} = \overrightarrow{q}$ and $\overrightarrow{AD} = \overrightarrow{p}$. We need to find the vector $\overrightarrow{r}$ that represents the altitude from vertex $B$ to the side $AD$. Let $X$ be the foot of the altitude from $B$ to the line containing $AD$. Then $\overrightarrow{r}$ is a vector along the line segment $BX$, and $BX \perp AD$. The condition that $\angle BAD$ is acute implies that $X$ will lie on the line segment $AD$ itself. We are looking for the vector $\overrightarrow{BX}$.

Step 2: Decomposing the Altitude Vector using the Triangle Law Consider the triangle $\triangle ABX$. By the triangle law of vector addition, we can write the vector $\overrightarrow{BX}$ as: $\overrightarrow{BX} = \overrightarrow{BA} + \overrightarrow{AX}$ We know that $\overrightarrow{BA}$ is the negative of $\overrightarrow{AB}$, so $\overrightarrow{BA} = -\overrightarrow{AB} = -\overrightarrow{q}$. Now, we need to determine the vector $\overrightarrow{AX}$.

Step 3: Determining the Vector $\overrightarrow{AX}$ using Vector Projection The vector $\overrightarrow{AX}$ represents the component of the vector $\overrightarrow{AB}$ that lies along the direction of the vector $\overrightarrow{AD}$. Since $X$ is the foot of the perpendicular from $B$ to $AD$, $\overrightarrow{AX}$ is the vector projection of $\overrightarrow{AB}$ onto $\overrightarrow{AD}$. Using the formula for vector projection, with $\overrightarrow{a} = \overrightarrow{AB} = \overrightarrow{q}$ and $\overrightarrow{b} = \overrightarrow{AD} = \overrightarrow{p}$: $\overrightarrow{AX} = \text{proj}_{\overrightarrow{AD}} \overrightarrow{AB} = \frac{\overrightarrow{AB} \cdot \overrightarrow{AD}}{|\overrightarrow{AD}|^2} \overrightarrow{AD}$ In terms of dot products, $|\overrightarrow{AD}|^2 = \overrightarrow{AD} \cdot \overrightarrow{AD} = \overrightarrow{p} \cdot \overrightarrow{p}$. Therefore, $\overrightarrow{AX} = \frac{\overrightarrow{q} \cdot \overrightarrow{p}}{\overrightarrow{p} \cdot \overrightarrow{p}} \overrightarrow{p}$

Step 4: Assembling the Altitude Vector $\overrightarrow{BX}$ Now, substitute the expressions for $\overrightarrow{BA}$ and $\overrightarrow{AX}$ back into the equation from Step 2: $\overrightarrow{BX} = \overrightarrow{BA} + \overrightarrow{AX}$ $\overrightarrow{BX} = -\overrightarrow{q} + \frac{\overrightarrow{p} \cdot \overrightarrow{q}}{\overrightarrow{p} \cdot \overrightarrow{p}} \overrightarrow{p}$ The problem states that $\overrightarrow{r}$ is the vector that coincides with the altitude directed from vertex $B$ to side $AD$. This implies that $\overrightarrow{r}$ is a vector in the direction of $\overrightarrow{BX}$.

Step 5: Comparing with the Given Options and Identifying the Correct Answer Let's examine the derived vector $\overrightarrow{BX} = -\overrightarrow{q} + \frac{\overrightarrow{p} \cdot \overrightarrow{q}}{\overrightarrow{p} \cdot \overrightarrow{p}} \overrightarrow{p}$ and compare it with the given options:

(A) $\overrightarrow r = 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow p$ We can factor out $-3$ from this expression: $\overrightarrow r = -3 \left( -\overrightarrow q + \frac{\overrightarrow{p} \cdot \overrightarrow{q}}{\overrightarrow{p} \cdot \overrightarrow{p}} \overrightarrow{p} \right)$ This shows that option (A) is $-3$ times our derived vector $\overrightarrow{BX}$.

(B) $\overrightarrow r = - \overrightarrow q + {{\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow p$ This option is exactly our derived vector $\overrightarrow{BX}$.

(C) $\vec r = \vec q - {{\left( {\vec p.\vec q} \right)} \over {\left( {\vec p.\vec p} \right)}}\vec p$ This option is the negative of our derived vector $\overrightarrow{BX}$ (i.e., $\overrightarrow{XB}$).

(D) $\overrightarrow r = - 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}$ This option is dimensionally inconsistent as it subtracts a scalar from a vector.

The wording "$\overrightarrow r$ is the vector that coincide with the altitude directed from the vertex $B$ to the side $AD$" can be interpreted in two ways: either it is the specific vector $\overrightarrow{BX}$, or it is any vector that lies on the line containing the altitude $BX$. Given that option (A) is the correct answer, it implies that the question considers any vector that is a scalar multiple of the altitude vector. In this case, option (A) is $-3$ times the vector $\overrightarrow{BX}$.

Common Mistakes & Tips

• Direction of Altitude: Be careful about the direction of the altitude vector. The altitude from $B$ to $AD$ is $\overrightarrow{BX}$, not $\overrightarrow{XB}$.
• Scalar Multiples in Options: When the exact vector you derived is not among the options, check if any option is a scalar multiple of your derived vector. This is common in vector problems.
• Vector Projection Formula: Ensure the vector projection formula is applied correctly, understanding which vector is being projected onto which.

Summary The altitude vector from vertex $B$ to side $AD$ in a parallelogram $ABCD$, where $\overrightarrow{AB} = \overrightarrow{q}$ and $\overrightarrow{AD} = \overrightarrow{p}$, can be found by decomposing it into $\overrightarrow{BA} + \overrightarrow{AX}$. $\overrightarrow{BA}$ is $-\overrightarrow{q}$, and $\overrightarrow{AX}$ is the vector projection of $\overrightarrow{AB}$ onto $\overrightarrow{AD}$, which is $\frac{\overrightarrow{p} \cdot \overrightarrow{q}}{\overrightarrow{p} \cdot \overrightarrow{p}} \overrightarrow{p}$. Thus, the altitude vector is $-\overrightarrow{q} + \frac{\overrightarrow{p} \cdot \overrightarrow{q}}{\overrightarrow{p} \cdot \overrightarrow{p}} \overrightarrow{p}$. Option (A) is $-3$ times this vector, indicating that it represents a vector along the same line as the altitude.

The final answer is $\boxed{\text{(A)}}$.