Let ABCD be a parallelogram such that AB = q, AD = p and ∠BAD be an acute angle. If r is the vector

Q: Wait, why is $\vec{BA} = -\vec{q}$? Shouldn't it be $\vec{q}$?

That is a fantastic question! Remember, vectors are defined by both magnitude and direction. We are given $\vec{AB} = \vec{q}$. When we move from $B$ to $A$, we are traversing the exact same path but in the opposite direction. Therefore, $\vec{BA}$ must be the negative of $\vec{AB}$, which is $-\vec{q}$. Always keep an eye on the arrowheads!

Q: Why do we use the dot product formula for projection instead of just using trigonometry?

You absolutely could use trigonometry, but in the world of JEE Advanced, we want to be efficient and general. The dot product formula $\text{proj}_{\vec{p}} \vec{q} = \frac{(\vec{q} \cdot \vec{p})}{|\vec{p}|^2} \vec{p}$ is a powerful tool because it works regardless of whether the angle is acute or obtuse, and it keeps everything in terms of the vectors $\vec{p}$ and $\vec{q}$ without needing to calculate the angle $\theta$ explicitly.

Q: Is the altitude vector $\vec{r}$ always pointing in the same direction as $\vec{p}$?

Not quite! The altitude vector $\vec{r}$ is perpendicular to $\vec{p}$. However, the component $\vec{AX}$ (the projection) is in the direction of $\vec{p}$. The vector $\vec{r}$ is the 'vertical' component, while $\vec{AX}$ is the 'horizontal' component relative to the base $\vec{p}$. Distinguishing between these two is the key to solving this problem.

Question

Let $ABCD$ be a parallelogram such that $\vec{AB} = \vec{q}$, $\vec{AD} = \vec{p}$ and $\angle BAD$ be an acute angle. If $\vec{r}$ is the vector that coincide with the altitude directed from the vertex $B$ to the side $AD$, then $\vec{r}$ is given by:

Options:

(A) $\vec{r} = 3\vec{q} - \frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})}\vec{p}$

(B) $\vec{r} = -\vec{q} + \frac{(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})}\vec{p}$

(C) $\vec{r} = \vec{q} - \frac{(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})}\vec{p}$

(D) $\vec{r} = -3\vec{q} - \frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})}\vec{p}$

Sigma Percentile · Accepted Answer

### The Geometry of Shadows: Unlocking the Altitude Vector

Welcome, future engineer. Today, we are not just solving a vector problem; we are learning to see the hidden architecture of space. When you look at a parallelogram, most students see four lines. I want you to see a playground of vectors.

We are given a parallelogram $ABCD$ with side vectors $\vec{AD} = \vec{p}$ and $\vec{AB} = \vec{q}$. Our mission is to find the altitude vector $\vec{r}$ dropped from vertex $B$ to the side $AD$. This isn't just about algebra; it is about decomposing motion.

### Phase 1: The Triangle Law

Imagine standing at vertex $A$. You have two paths to get to $B$: you can go directly along $\vec{q}$, or you can traverse the path of the altitude. Let's drop a perpendicular from $B$ to the line $AD$, hitting at point $X$.

Now, look at the triangle $	riangle ABX$. It is a right-angled triangle. This is our anchor.

By the triangle law of vector addition, we know that the vector $\vec{BX}$ (which is our altitude vector $\vec{r}$) can be expressed as the sum of two vectors: $\vec{BX} = \vec{BA} + \vec{AX}$.

Pause here. This is where many students stumble. We know $\vec{AB} = \vec{q}$. But our path in the triangle starts at $B$ and goes to $A$.

Since we are reversing the direction of $\vec{q}$, we must write $\vec{BA} = -\vec{q}$. Our equation now stands as $\vec{r} = -\vec{q} + \vec{AX}$. We have successfully reduced the problem to finding the vector $\vec{AX}$.

### Phase 2: The Power of Projection

Now, what is $\vec{AX}$? Geometrically, if you were to shine a light from directly above $B$ perpendicular to the base $AD$, the segment $AX$ is the shadow cast by the side $AB$.

In the language of linear algebra, $\vec{AX}$ is the **vector projection** of $\vec{q}$ onto $\vec{p}$.

This is a beautiful concept. We are taking a vector $\vec{q}$ that is tilted at an angle and finding how much of it 'lies' along the direction of $\vec{p}$. The formula for the projection of a vector $\vec{a}$ onto $\vec{b}$ is a standard tool in your JEE arsenal:

$$	ext{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} ight) \vec{b}$$

Applying this to our specific vectors, where $\vec{a} = \vec{q}$ and $\vec{b} = \vec{p}$, we get:

$$\vec{AX} = \left( \frac{\vec{q} \cdot \vec{p}}{|\vec{p}|^2} ight) \vec{p}$$

### Phase 3: The Final Synthesis

We are almost there. The problem asks for the answer in terms of dot products, not magnitudes. Recall that the square of the magnitude of a vector is simply the dot product of the vector with itself: $|\vec{p}|^2 = \vec{p} \cdot \vec{p}$.

Furthermore, the dot product is commutative, so $\vec{q} \cdot \vec{p} = \vec{p} \cdot \vec{q}$. Substituting these into our expression for $\vec{AX}$, we get:

$$\vec{AX} = \frac{(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$$

Now, we bring it all home. Substitute this back into our original equation for the altitude vector $\vec{r}$:

$$\vec{r} = -\vec{q} + \vec{AX}$$

$$\vec{r} = -\vec{q} + \frac{(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$$

Look at that expression. It is elegant, precise, and perfectly matches our target. You have just decomposed a complex geometric altitude into a simple sum of vector components.

This is the essence of JEE Advanced physics and mathematics—taking a complex, intimidating shape and breaking it down into the fundamental language of vectors. Keep this intuition, and no geometry problem will ever be able to stop you.

Animated Solution for Mathematics - Vector Algebra: Let $A B C D$ be a parallelogram such that $A B = q$ , $A D = p$ and $∠ B A D$ be an acute angle. If $r$ is the vector that coincide with the altitude directed from the vertex $B$ to the side $A D$ , then $r$ is given by:

Select Answer:

Visualized Solution

Defining the Parallelogram and Vectors

Introducing the Altitude Vector $r$

Applying the Triangle Law of Vector Addition

Setting up the Vector Equation

Substituting Known Vectors

Geometric Interpretation of $A X$

Formula for Vector Projection

Calculating the Projection $A X$

Simplifying the Denominator

Final Substitution for $r$

Conclusion and Matching Options

The Sigma Insight: Scalar (Dot) Product

The Geometry of Shadows

Phase 1

Phase 2

Phase 3

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