JEE Main 2004

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Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be non-zero vectors such that $(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a$ . If $θ$ is the acute angle between the vectors $b$ and $c$ , then $sin θ$ equals

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Visualized Solution

The Vector Triple Product Equation

Given: $(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a$
Goal: Find $sin θ$ , where $θ$ is the angle between $b$ and $c$ .

Vector Triple Product Expansion

Formula: $(a \times b) \times c = (a \cdot c) b - (b \cdot c) a$

Substituting the Expansion

Substitute into given equation:
$(a \cdot c) b - (b \cdot c) a = \frac{1}{3} ∣ b ∣∣ c ∣ a$

Grouping Like Vectors

Rearrange terms:
$(a \cdot c) b - (b \cdot c + \frac{1}{3} ∣ b ∣∣ c ∣) a = 0$

Comparing Coefficients: Vector $b$

Coefficient of $b$ must be zero:
$a \cdot c = 0$
This implies $a ⊥ c$ .

Comparing Coefficients: Vector $a$

Coefficient of $a$ must be zero:
$- (b \cdot c) = \frac{1}{3} ∣ b ∣∣ c ∣$

Expanding the Dot Product

Recall: $b \cdot c = ∣ b ∣∣ c ∣ cos θ$
Substitute: $- ∣ b ∣∣ c ∣ cos θ = \frac{1}{3} ∣ b ∣∣ c ∣$

Solving for $cos θ$

Cancel $∣ b ∣∣ c ∣$ from both sides (since vectors are non-zero).
$cos θ = - \frac{1}{3}$

Calculating $sin θ$

Use trigonometric identity: $sin^{2} θ + cos^{2} θ = 1$
$sin θ = 1 - cos^{2} θ$
$sin θ = 1 - (- \frac{1}{3})^{2}$

Final Simplification

$sin θ = 1 - \frac{1}{9}$
$sin θ = \frac{8}{9}$
$sin θ = \frac{2 2}{3}$

Conclusion

Final Answer: $sin θ = \frac{2 2}{3}$
Key Takeaway: The Vector Triple Product expansion is a powerful tool to convert complex cross products into solvable linear combinations.

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The Sigma Insight: Vector Triple Product

Solution Diagram

Analyzing the Setup

We are given the vector equation:

(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a

This expression involves a vector triple product. Recall that the vector triple product expansion is defined as:

(a \times b) \times c = (a \cdot c) b - (b \cdot c) a

The Master Equation

By substituting the expansion into our original equation, we obtain:

(a \cdot c) b - (b \cdot c) a = \frac{1}{3} ∣ b ∣∣ c ∣ a

Rearranging the terms to group the vectors

a

and

b

, we get:

(a \cdot c) b - ((b \cdot c) + \frac{1}{3} ∣ b ∣∣ c ∣) a = 0

Applying Linear Independence

Assuming

a

and

b

are non-collinear, the coefficients of the linear combination must vanish independently. First, for the coefficient of

b

to be zero:

a \cdot c = 0

This implies that

a

and

c

are orthogonal. Next, we set the coefficient of

a

to zero:

- (b \cdot c) = \frac{1}{3} ∣ b ∣∣ c ∣

Final Calculation

Using the definition of the dot product,

b \cdot c = ∣ b ∣∣ c ∣ cos θ

, where

θ

is the angle between

b

and

c

- ∣ b ∣∣ c ∣ cos θ = \frac{1}{3} ∣ b ∣∣ c ∣

Dividing by the magnitudes (assuming non-zero vectors), we find:

cos θ = - \frac{1}{3}

To find

sin θ

, we use the identity

sin^{2} θ + cos^{2} θ = 1

sin θ = 1 - (- \frac{1}{3})^{2} = 1 - \frac{1}{9} = \frac{8}{9}

The final result is:

sin θ = \frac{2 2}{3}

Common Traps & FAQs

Why can we equate the coefficients of

a

and

b

to zero? Is that always allowed?

That is a fantastic question! We can equate coefficients because

a

and

b

are linearly independent. If we have an equation of the form

x a + y b = 0

, and

a

and

b

are not parallel, the only way for this to be true is if

x = 0

and

y = 0

. In our problem, if

a

and

b

were collinear, their cross product would be zero, which contradicts the problem statement. Thus, they must be independent, and comparing coefficients is mathematically rigorous.

I got

cos θ = - 1/3

, but the question asks for the angle between vectors. Does the negative sign matter?

The negative sign is absolutely critical! It tells us that the angle

θ

is obtuse (between 90 and 180 degrees). However, when we calculate

sin θ

, we use the identity

sin^{2} θ + cos^{2} θ = 1

. Since

sin θ

is always positive for an angle between 0 and 180 degrees, the square root will always yield a positive value. You are on the right track; don't let the negative cosine shake your confidence!

Is there a geometric way to solve this without the expansion formula?

While you could attempt to use the geometric properties of cross products (like the fact that

(a \times b) \times c

lies in the plane of

a

and

b

), the algebraic expansion is the most efficient and robust path for JEE Advanced. Geometry is great for intuition, but algebra is your safety net. Mastering the expansion formula is a hallmark of a top-tier student.

Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be non-zero vectors such that $(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a$ . If $θ$ is the acute angle between the vectors $b$ and $c$ , then $sin θ$ equals

Select Answer:

Visualized Solution

The Vector Triple Product Equation

Vector Triple Product Expansion

Substituting the Expansion

Grouping Like Vectors

Comparing Coefficients: Vector $b$

Comparing Coefficients: Vector $a$

Expanding the Dot Product

Solving for $cos θ$

Calculating $sin θ$

Final Simplification

Conclusion

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Master Equation

Applying Linear Independence

Final Calculation

Similar Questions

Let $a, b$ and $c$ be three non-zero vectors such that no two of them are collinear and $(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a$ . If $θ$ is the angle between vectors $b$ and $c$ , then a value of $sin θ$ is :

Let $a, b$ and $c$ be three unit vectors such that $a \times (b \times c) = \frac{3}{2} (b + c)$ . If $b$ is not parallel to $c$ , then the angle between $a$ and $b$ is:

If $(a \times b) \times c = a \times (b \times c)$ where $a, b$ and $c$ are any three vectors such that $a \cdot b \neq = 0, b \cdot c \neq = 0$ then $a$ and $c$ are

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣$ , $∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $\frac{π}{6}$ , then the value of $∣ (a \times b) \times c ∣$ is :

Let $x, y$ and $z$ be three vectors each of magnitude $2$ and the angle between each pair of them is $\frac{π}{3}$ . If $a$ is a non-zero vector perpendicular to $x$ and $y \times z$ and $b$ is a non-zero vector perpendicular to $y$ and $z \times x$ , then

If $a, b, c$ are non coplanar unit vectors such that $a \times (b \times c) = \frac{( b + c )}{2}$ , then the angle between $a$ and $b$ is

Let $a$ , $b$ and $c$ be three vectors such that $∣ a ∣ = 3$ , $∣ b ∣ = 5$ , $b \cdot c = 10$ and angle between $b$ and $c$ is $\frac{π}{3}$ . If $a$ is perpendicular to the vector $b \times c$ , then $∣ a \times (b \times c) ∣$ is equal to

Let $a = 6 \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $∣ c ∣ \geq 6, a \cdot c = 6∣ c ∣, ∣ c - a ∣ = 22$ and the angle between $a \times b$ and $c$ is $6 0^{\circ}$ , then $∣ (a \times b) \times c ∣$ is equal to:

Let $a, b$ and $c$ be three vectors such that $a = b \times (b \times c)$ . If magnitudes of the vectors $a, b$ and $c$ are $2, 1$ and 2 respectively and the angle between $b$ and $c$ is $θ (0 < θ < \frac{π}{2})$ , then the value of $1 + tan θ$ is equal to:

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . Let $c$ be a vector such that $∣ c - a ∣ = 3$ , $∣ (a \times b) \times c ∣ = 3$ and the angle between $c$ and $a \times b$ be $3 0^{\circ}$ . Then $a \cdot c$ is equal to:

Select Answer:

Visualized Solution

The Vector Triple Product Equation

Vector Triple Product Expansion

Substituting the Expansion

Grouping Like Vectors

Comparing Coefficients: Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Comparing Coefficients: Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Expanding the Dot Product

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

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

Final Simplification

Conclusion

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Master Equation

Applying Linear Independence

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b and c be three unit vectors such that a×(b×c)=23​​(b+c). If b is not parallel to c, then the angle between a and b is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (a×b)×c=a×(b×c) where a,b and c are any three vectors such that a⋅b=0,b⋅c=0 then a and c are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are non coplanar unit vectors such that a×(b×c)=2​(b+c)​, then the angle between a and b is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+j^​−2k^ and b=i^+j^​. Let c be a vector such that ∣c−a∣=3, ∣(a×b)×c∣=3 and the angle between c and a×b be 30∘. Then a⋅c is equal to:

Comparing Coefficients: Vector $b$

Comparing Coefficients: Vector $a$

Solving for $cos θ$

Calculating $sin θ$

Let $a, b$ and $c$ be three unit vectors such that $a \times (b \times c) = \frac{3}{2} (b + c)$ . If $b$ is not parallel to $c$ , then the angle between $a$ and $b$ is:

If $(a \times b) \times c = a \times (b \times c)$ where $a, b$ and $c$ are any three vectors such that $a \cdot b \neq = 0, b \cdot c \neq = 0$ then $a$ and $c$ are

If $a, b, c$ are non coplanar unit vectors such that $a \times (b \times c) = \frac{( b + c )}{2}$ , then the angle between $a$ and $b$ is

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . Let $c$ be a vector such that $∣ c - a ∣ = 3$ , $∣ (a \times b) \times c ∣ = 3$ and the angle between $c$ and $a \times b$ be $3 0^{\circ}$ . Then $a \cdot c$ is equal to: