In a triangle ABC, if |BC| = 8, |CA| = 7, |AB| = 10, then the projection of the vector AB on AC is e

Q: Wait, why is the projection formula $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$ and not just the dot product?

Great question! The dot product $\vec{u} \cdot \vec{v}$ gives you a scalar value that includes the magnitudes of both vectors and the cosine of the angle between them. By dividing by $|\vec{v}|$, you are essentially 'normalizing' the vector $\vec{v}$, leaving you with the magnitude of the component of $\vec{u}$ along the direction of $\vec{v}$. Think of it as stripping away the length of the base vector so only the 'shadow' length remains.

Q: Why did we use $\vec{AC}$ instead of $\vec{CA}$ in the formula?

The problem specifically asks for the projection of $\vec{AB}$ on $\vec{AC}$. Since $\vec{AC} = -\vec{CA}$, the direction is reversed. If you use $\vec{CA}$ by mistake, you will get the correct magnitude but the wrong sign. Always check the vector orientation carefully!

Q: Can I use the Law of Cosines instead of vector algebra?

Absolutely! The Law of Cosines is essentially the geometric cousin of the dot product. You could find $\cos(A)$ using the side lengths and then use the definition of projection as $|\vec{AB}| \cos(A)$. However, mastering the vector approach is a superpower for JEE Advanced because it works even in 3D space where 'angles' are harder to visualize.

Q: Why did we get a negative dot product? Does that mean the length is negative?

A negative dot product simply tells us that the angle between the two vectors is obtuse (greater than 90 degrees). The projection itself is a signed scalar—it tells you whether the shadow falls in the direction of the vector or opposite to it. In this specific geometry, the projection value is positive because we are projecting onto $\vec{AC}$ and the dot product calculation accounts for the directionality.

Q: Is there a shortcut to remember the sign in the projection formula?

Think of it physically: if the angle is acute, the shadow falls in the same direction as the vector, resulting in a positive projection. If the angle is obtuse, the shadow falls 'behind' the starting point of the vector, resulting in a negative projection. Always trust the algebra—the dot product sign will handle the geometry for you!

Question

In a triangle ABC, if $|\vec{BC}| = 8$, $|\vec{CA}| = 7$, $|\vec{AB}| = 10$, then the projection of the vector $\vec{AB}$ on $\vec{AC}$ is equal to:

Options:

(A) $\frac{25}{4}$

(B) $\frac{85}{14}$

(C) $\frac{127}{20}$

(D) $\frac{115}{16}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are not just solving a problem; we are uncovering the hidden architecture of a triangle.

When you look at a triangle $ABC$ with side lengths $|\vec{BC}| = 8$, $|\vec{CA}| = 7$, and $|\vec{AB}| = 10$, you might see a simple shape. To a physicist, this is a system of vectors waiting to be decoded.

We are tasked with finding the projection of $\vec{AB}$ onto $\vec{AC}$. Imagine standing at vertex $A$, shining a light perpendicular to the line $AC$; the 'shadow' cast by the segment $AB$ onto the line $AC$ is exactly what we need to calculate.

### The Mathematical Toolkit

To find the projection of a vector $\vec{u}$ onto $\vec{v}$, we use the fundamental definition:

$$Projection = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$$

In our case, we want the projection of $\vec{AB}$ onto $\vec{AC}$. Let us define our vectors clearly: let $\vec{c} = \vec{AB}$, $\vec{b} = \vec{CA}$, and $\vec{a} = \vec{BC}$.

Since $\vec{AC} = -\vec{CA} = -\vec{b}$, our formula becomes:

$$Projection = \frac{\vec{c} \cdot (-\vec{b})}{|-\vec{b}|} = \frac{-\vec{c} \cdot \vec{b}}{7}$$

This is where the magic happens. We do not know the angle between these vectors, but we know the lengths of all three sides, which is a classic invitation to use the Triangle Law of Vector Addition.

### The Power of the Triangle Law

In any closed triangle, the sum of the vectors forming the sides must be zero. Thus, $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$.

Substituting our variables, we get $\vec{c} + \vec{a} + \vec{b} = \vec{0}$. We need to find the dot product $\vec{b} \cdot \vec{c}$.

Let us isolate $\vec{a}$ and square the equation:

$$\vec{a} = -(\vec{b} + \vec{c})$$

$$|\vec{a}|^2 = |-(\vec{b} + \vec{c})|^2$$

$$|\vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c})$$

This step is the bridge between geometry and algebra. We are essentially using the Law of Cosines in disguise, successfully extracting the dot product term by squaring the vector sum.

### The Final Calculation

Now, we plug in our known magnitudes: $|\vec{a}| = 8$, $|\vec{b}| = 7$, and $|\vec{c}| = 10$. The equation transforms into:

$$8^2 = 7^2 + 10^2 + 2(\vec{b} \cdot \vec{c})$$

$$64 = 49 + 100 + 2(\vec{b} \cdot \vec{c})$$

$$64 = 149 + 2(\vec{b} \cdot \vec{c})$$

Solving for the dot product, we find $2(\vec{b} \cdot \vec{c}) = 64 - 149 = -85$, which means $\vec{b} \cdot \vec{c} = -\frac{85}{2}$.

Returning to our projection formula, we substitute this value back in:

$$Projection = \frac{-(-\frac{85}{2})}{7} = \frac{85}{14}$$

The negative signs cancel out, leaving us with a clean, positive result. You have successfully navigated the vector space to arrive at the **final answer of $\frac{85}{14}$**.

Animated Solution for Mathematics - Vector Algebra: In a triangle ABC, if $∣ B C ∣ = 8$ , $∣ C A ∣ = 7$ , $∣ A B ∣ = 10$ , then the projection of the vector $A B$ on $A C$ is equal to:

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

Defining the Goal

The Projection Formula

Vector Representation

Adjusting the Formula

The Triangle Law

Isolating the Unknowns

Squaring the Equation

Substituting Magnitudes

Evaluating the Squares

Solving for the Dot Product

Final Substitution

The Final Answer

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Mathematical Toolkit

The Power of the Triangle Law

The Final Calculation

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle ABC, if ∣BC∣=8, ∣CA∣=7, ∣AB∣=10, then the projection of the vector AB on AC is equal to: