Does the order of the vectors matter? Like, is the projection of $\vec{BA}$ on $\vec{BC}$ the same as $\vec{BC}$ on $\vec{BA}$?

Absolutely not! The projection of $\vec{u}$ on $\vec{v}$ is $|\vec{u}| \cos \theta$, while the projection of $\vec{v}$ on $\vec{u}$ is $|\vec{v}| \cos \theta$. They are different unless the magnitudes are equal. Always identify which vector is the "base" (the one being projected onto).

JEE Main 2021 (20 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: In a triangle ABC, if $∣ B C ∣ = 3, ∣ C A ∣ = 5$ and $∣ B A ∣ = 7$ , then the projection of the vector $B A$ on $B C$ is equal to

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

Given side lengths:
$∣ B C ∣ = 3$
$∣ C A ∣ = 5$
$∣ B A ∣ = 7$

The Concept of Projection

Projection of $B A$ on $B C$ is given by:
$Projection = ∣ B A ∣ cos B$
Where $B$ is the angle between $B A$ and $B C$ .

Applying the Cosine Rule

Using the Cosine Rule in $△ A B C$ :
$cos B = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c}$

Substitution of Values

Substitute $a = 3, b = 5, c = 7$ :
$cos B = \frac{3 ^{2} + 7 ^{2} - 5 ^{2}}{2 ( 3 ) ( 7 )}$

Calculating the Numerator

Numerator calculation:
$3^{2} + 7^{2} - 5^{2} = 9 + 49 - 25 = 33$

Calculating the Denominator

Denominator calculation:
$2 (3) (7) = 42$

Simplifying $cos B$

$cos B = \frac{33}{42}$
Simplifying by dividing by $3$ :
$cos B = \frac{11}{14}$

Final Projection Calculation

Substitute $∣ B A ∣ = 7$ and $cos B = \frac{11}{14}$ into the projection formula:
$Projection = 7 \cdot \frac{11}{14}$

The Final Result

$Projection = \frac{11}{2}$
The projection of $B A$ on $B C$ is $\frac{11}{2}$ units.

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The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

Imagine you are standing on a vast, flat plane, and before you lies a triangle

A B C

. You are given the lengths of its sides:

∣ B C ∣ = 3

∣ C A ∣ = 5

, and

∣ B A ∣ = 7

Your mission is to find the projection of vector

B A

onto vector

B C

. When we talk about the projection of one vector onto another, we are essentially asking for the length of the shadow cast by

B A

onto the line containing

B C

Mathematically, this scalar projection is defined as

∣ B A ∣ cos B

, where

B

is the angle between the two vectors.

The Bridge

The Cosine Rule

We have the side lengths, but we do not have the angle

B

. To bridge the gap between side lengths and angles, we use the Cosine Rule.

The Cosine Rule states that for any triangle with sides

a, b, c

and angle

B

opposite to side

b

, the relationship is:

cos B = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c}

In our specific case, we assign the values

a = ∣ B C ∣ = 3

b = ∣ C A ∣ = 5

, and

c = ∣ B A ∣ = 7

. This formula acts as our golden key to transform static lengths into the trigonometric value we require.

The Execution

Precision and Elegance

Let us perform the substitution into the Cosine Rule:

cos B = \frac{3 ^{2} + 7 ^{2} - 5 ^{2}}{2 ( 3 ) ( 7 )}

The numerator simplifies to

9 + 49 - 25 = 33

. The denominator is

2 \times 3 \times 7 = 42

Thus, we find:

cos B = \frac{33}{42} = \frac{11}{14}

Now, we return to our projection formula:

Projection = ∣ B A ∣ cos B

. Substituting our known values, we get:

Projection = 7 \cdot (\frac{11}{14})

The

7

in the numerator and the

14

in the denominator simplify, leaving us with the final result:

Projection

= \frac{11}{2}

Common Traps & FAQs

Wait, why is the projection formula just

∣ B A ∣ cos B

? Shouldn't it be a vector?

Great catch! The term "projection" can refer to either the scalar projection or the vector projection. In this context, since the options are scalars, we are looking for the scalar projection, which is the signed magnitude of the shadow. The formula

∣ B A ∣ cos B

gives us exactly that scalar value.

Does the order of the vectors matter? Like, is the projection of

B A

B C

the same as

B C

B A

Absolutely not! The projection of

u

v

∣ u ∣ cos θ

, while the projection of

v

u

∣ v ∣ cos θ

. They are different unless the magnitudes are equal. Always identify which vector is the "base" (the one being projected onto).

Why did we use the Cosine Rule instead of just drawing a perpendicular and using Pythagoras?

You could use Pythagoras by dropping an altitude, but that often involves solving for the height and the segments of the base, which is more tedious. The Cosine Rule is a direct, elegant shortcut that gets you the cosine value immediately. Efficiency is key in JEE!

Animated Solution for Mathematics - Vector Algebra: In a triangle ABC, if $∣ B C ∣ = 3, ∣ C A ∣ = 5$ and $∣ B A ∣ = 7$ , then the projection of the vector $B A$ on $B C$ is equal to

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

The Concept of Projection

Applying the Cosine Rule

Substitution of Values

Calculating the Numerator

Calculating the Denominator

Simplifying $cos B$

Final Projection Calculation

The Final Result

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Bridge

The Execution

Similar Questions

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:

Consider a $Δ A B C$ where $A (1, 2, 3)$ , $B (- 2, 8, 0)$ and $C (3, 6, 7)$ . If the angle bisector of $∠ B A C$ meets the line $B C$ at $D$ , then the length of the projection of the vector $A D$ on the vector $A C$ is:

Let $a = \hat{i} - \hat{j} + 2 \hat{k}$ and $b$ be a vector such that $a \times b = 2 \hat{i} - \hat{k}$ and $a \cdot b = 3$ . Then the projection of $b$ on the vector $a - b$ is :-

Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ and $c = \hat{i} + \hat{j} - 2 \hat{k}$ be three vectors. A vector in the plane of $b$ and $c$ , whose projection on $a$ is of magnitude $2/3$ , is:

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is

Let $u, v, w$ be such that $∣ u ∣ = 1, ∣ v ∣ = 2, ∣ w ∣ = 3$ . If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v, w$ are perpendicular to each other then $∣ u - v + w ∣$ equals

Let $a = 2 \hat{i} - \hat{j} - \hat{k}, b = \hat{i} + 3 \hat{j} - \hat{k}$ and $c = 2 \hat{i} + \hat{j} + 3 \hat{k}$ . Let $v$ be the vector in the plane of the vectors $a$ and $b$ , such that the length of its projection on the vector $c$ is $\frac{1}{14}$ . Then $∣ v ∣$ is equal to

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∣=3,∣CA∣=5 and ∣BA∣=7, then the projection of the vector BA on BC is equal to

Select Answer:

Visualized Solution

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

The Concept of Projection

Applying the Cosine Rule

Substitution of Values

Calculating the Numerator

Calculating the Denominator

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosB

Final Projection Calculation

The Final Result

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Bridge

The Execution

Similar Questions

.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:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^−j^​+2k^ and b be a vector such that a×b=2i^−k^ and a⋅b=3. Then the projection of b on the vector a−b is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^−j^​+k^, b=i^+2j^​−k^ and c=i^+j^​−2k^ be three vectors. A vector in the plane of b and c, whose projection on a is of magnitude 2/3​, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+j^​+k^, b=i^−j^​+k^ and c=i^−j^​−k^ be three vectors. A vector v in the plane of a and b, whose projection on c is 3​1​, is given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+2j^​+k^, b=i^−j^​+k^ and c=i^+j^​−k^. A vector in the plane of a and b whose projection on c is 3​1​, is

Animated Solution for Mathematics - Vector Algebra: In a triangle ABC, if $∣ B C ∣ = 3, ∣ C A ∣ = 5$ and $∣ B A ∣ = 7$ , then the projection of the vector $B A$ on $B C$ is equal to

Visualizing Triangle $A B C$

Simplifying $cos B$

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:

Let $a = \hat{i} - \hat{j} + 2 \hat{k}$ and $b$ be a vector such that $a \times b = 2 \hat{i} - \hat{k}$ and $a \cdot b = 3$ . Then the projection of $b$ on the vector $a - b$ is :-

Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ and $c = \hat{i} + \hat{j} - 2 \hat{k}$ be three vectors. A vector in the plane of $b$ and $c$ , whose projection on $a$ is of magnitude $2/3$ , is:

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is