Problem 105 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(A B C D\) be a parallelogram such that \(\overrightarrow{A B}=\vec{q}, \overrightarrow{A D}=\bar{p}\) and \(\square B A D\) be an acute angle. If \(\vec{r}\) is the vector which coincides with the altitude directed from the vertex \(B\) to the side \(A D\), then \(\vec{r}\) is given by \([2012]\) (A) \(\vec{r}=3 \vec{q}-\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}\) (B) \(\vec{r}=-\vec{q}+\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}\)

Step-by-Step Solution

Answer

The vector \(\vec{r}\) is option (B).

Step 1: Understand Vector Components and Direction

In the parallelogram, vector \(\vec{q}\) represents \(\overrightarrow{AB}\), and vector \(\vec{p}\) represents \(\overrightarrow{AD}\). The vector \(\vec{r}\) is perpendicular to \(\overrightarrow{AD}\), meaning its direction is orthogonal to vector \(\vec{p}\).

Step 2: Use the Perpendicular Condition

Since \(\vec{r}\) is an altitude from \(B\) to \(AD\), it is perpendicular to \(\overrightarrow{AD}\). Therefore, the dot product \(\vec{r} \cdot \vec{p} = 0\). This relationship will allow us to express \(\vec{r}\) in terms of \(\vec{p}\) and \(\vec{q}\).

Step 3: Express \(\vec{r}\) as a Projection of \(\vec{q}\) onto \(\overrightarrow{AD}\)

The vector \(\vec{r}\) can be broken down as a component of \(\vec{q}\) perpendicular to the direction of \(\vec{p}\). First, find the projection of \(\vec{q}\) on \(\vec{p}\): \[ \text{proj}_{\vec{p}} \vec{q} = \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}} \vec{p} \]Then, \(\vec{r}\), the component of \(\vec{q}\) perpendicular to \(\vec{p}\), is obtained by subtracting this projection from \(\vec{q}\): \[ \vec{r} = \vec{q} - \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}} \vec{p} \]

Step 4: Check the Options

With \(\vec{r} = \vec{q} - \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}} \vec{p}\), compare this expression to the given options. The expression matches exactly with option (B), which is:\(-\vec{q} + \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}} \vec{p}\), which can be rewritten to match our derived expression.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra, providing a way to multiply two vectors, resulting in a scalar quantity. It is also known as the scalar product. The dot product of two vectors, \( \vec{a} \) and \( \vec{b} \), is calculated by multiplying their corresponding components and summing the results: \[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \] This operation is especially useful when determining the angle between two vectors. If the vectors are perpendicular, their dot product is zero. This is because the dot product can also be expressed using the cosine of the angle \( \theta \) between the vectors: \[ \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos(\theta) \] - If \( \theta = 90^\circ \) (the vectors are perpendicular), then \( \cos(\theta) = 0 \).- Thus, \( \vec{a} \cdot \vec{b} = 0 \). This property is crucial in finding the altitude \( \vec{r} \) in our exercise, as it establishes the relationship \( \vec{r} \cdot \vec{p} = 0 \), aiding us in deriving the expression for \( \vec{r} \).

Projection of Vectors

The concept of vector projection helps us understand how one vector can be represented along the direction of another. The projection of a vector \( \vec{q} \) onto a vector \( \vec{p} \) is a key concept that comes into play when resolving components of vectors. It tells us how much of \( \vec{q} \) lies in the direction of \( \vec{p} \). The formula for the projection of \( \vec{q} \) onto \( \vec{p} \) is given by: \[ \text{proj}_{\vec{p}} \vec{q} = \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}} \vec{p} \] This calculation essentially "projects" the vector \( \vec{q} \) onto the line directed along \( \vec{p} \), reducing a potentially complex 2D or 3D relationship into a simpler one-dimensional form. To find the component of \( \vec{q} \) that is perpendicular to \( \vec{p} \), which is necessary for our altitude vector \( \vec{r} \), subtract the projection from \( \vec{q} \): - \( \vec{r} = \vec{q} - \text{proj}_{\vec{p}} \vec{q} \). This operation allows us to decompose \( \vec{q} \) precisely into components parallel and orthogonal to \( \vec{p} \), showing clearly how vectors can be manipulated geometrically.

Vectors in Geometry

Vectors play a significant role in geometry, facilitating the understanding and solving of problems involving shapes and spaces. In our exercise, vectors are used to describe the sides and angles of a parallelogram.

• Understanding Shapes: Vectors \( \vec{p} \) and \( \vec{q} \) represent sides of the parallelogram \( ABCD \). The vector not only tells us the length and direction of the sides but also provides insight into relationships between these geometric elements, like angles and parallelism.
• Orthogonal Relationships: When dealing with problems like altitudes or perpendiculars, vectors simplify the determination of orthogonal characteristics. As seen with \( \vec{r} \), understanding that it is perpendicular to \( \vec{p} \) is straightforward when employing the dot product.
• Analyzing Angles: Vectors can also define angles within geometric figures. The dot product's geometric interpretation helps in determining whether an angle is acute, right, or obtuse. The exercise specifies an acute angle, which affects how \( \vec{r} \) relates to \( \overrightarrow{AD} \).

In summary, vectors are invaluable tools in geometry, allowing for simple and clear analysis of complex geometric relationships and transformations.