Let $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j} + 3\hat{k}$. Then the vector product $(\vec{a} + \vec{b}) \times ((\vec{a} \times ((\vec{a} - \vec{b}) \times \vec{b})) \times \vec{b})$ is equal to : Options: (A) $5(34\hat{i} - 5\hat{j} + 3\hat{k})$ (B) $7(34\hat{i} - 5\hat{j} + 3\hat{k})$ (C) $7(30\hat{i} - 5\hat{j} + 7\hat{k})$ (D) $5(30\hat{i} - 5\hat{j} + 7\hat{k})$

### Analyzing the Innermost Term Our journey begins with the innermost term: $(\vec{a} - \vec{b}) \times \vec{b}$. By applying the distributive property of the cross product, this expression expands to: $$(\vec{a} \times \vec{b}) - (\vec{b} \times \vec{b})$$ Since the cross product of any vector with itself is the zero vector, the second term vanishes. We are left simply with $\vec{a} \times \vec{b}$. ### Applying the Vector Triple Product The expression now simplifies to $\vec{a} \times (\vec{a} \times \vec{b})$. This is a classic **Vector Triple Product**, which we resolve using the identity: $$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$ Applying this to our specific case, we transform the expression into: $$(\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$$ ### Substituting Known Values We are given the dot products: $\vec{a} \cdot \vec{b} = 7$ and $\vec{a} \cdot \vec{a} = 6$. Substituting these values into our derived equation, the 'monster' expression collapses into: $$7\vec{a} - 6\vec{b}$$ ### Final Calculation The final phase requires evaluating $(\vec{a} + \vec{b}) \times (7(\vec{a} \times \vec{b}))$. First, pull the scalar $7$ out of the cross product. Then, compute the cross product of the sum $(\vec{a} + \vec{b})$ with the result of $(\vec{a} \times \vec{b})$ using the determinant method. By staying calm and trusting your vector identities, the complexity dissolves, leading you directly to the **final result**.

JEE Main 2021 (27 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + \hat{j} + 2 \hat{k}$ and $b = - \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Then the vector product $(a + b) \times ((a \times ((a - b) \times b)) \times b)$ is equal to :

Select Answer:

Progressive Hints

Hint 1

The secret to conquering this 'monster' expression is to resist the urge to calculate everything at once. Start from the very innermost parenthesis: (\vec{a} - \vec{b}) \times \vec{b}. Remember the distributive property of the cross product, and recall that the cross product of any vector with itself is the zero vector. This will collapse the innermost term instantly.

Hint 2

Once you have simplified the inner part, you will be left with a Vector Triple Product of the form \vec{a} \times (\vec{a} \times \vec{b}). This is a classic JEE Advanced pattern. Use the identity \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}. This will transform a complex cross product into a simple linear combination of vectors \vec{a} and \vec{b}.

Hint 3

After applying the identity, the entire expression will reduce to (\vec{a} + \vec{b}) \times (7(\vec{a} \times \vec{b})). Now, simply calculate the vector \vec{a} + \vec{b} and the vector \vec{a} \times \vec{b} separately using the determinant method. Finally, take their cross product and multiply by the scalar 7 to arrive at the answer.

Visualized Solution

Define Vectors $a$ and $b$

Given vectors:
$a = \hat{i} + \hat{j} + 2 \hat{k}$
$b = - \hat{i} + 2 \hat{j} + 3 \hat{k}$

Simplify Innermost Term

Focus on the innermost bracket: $(a - b) \times b$
Using distributive property:
$= a \times b - b \times b$

Cross Product of a Vector with Itself

We know that the cross product of any vector with itself is zero.
$b \times b = 0$
So, the term simplifies to: $a \times b$

Apply Vector Triple Product

The expression now becomes: $a \times (a \times b)$
Using the Vector Triple Product (VTP) formula:
$a \times (b \times c) = (a \cdot c) b - (a \cdot b) c$
Here: $a \times (a \times b) = (a \cdot b) a - (a \cdot a) b$

Calculate Dot Products

Let's find $a \cdot b$ :
$(1) (- 1) + (1) (2) + (2) (3) = - 1 + 2 + 6 = 7$
Let's find $a \cdot a$ :
$(1)^{2} + (1)^{2} + (2)^{2} = 1 + 1 + 4 = 6$

Substitute into VTP

Substituting the dot products back:
$a \times (a \times b) = 7 a - 6 b$

Simplify the Next Outer Bracket

The next part of our main expression is: $(7 a - 6 b) \times b$
Distributing the cross product:
$= 7 (a \times b) - 6 (b \times b)$
Since $b \times b = 0$ , this leaves us with $7 (a \times b)$

Final Expression Structure

The full monster expression has now collapsed to:
$(a + b) \times 7 (a \times b)$
Pulling the scalar $7$ out:
$= 7 [(a + b) \times (a \times b)]$

Calculate $a + b$

Add the components of $a$ and $b$ :
$a + b = (1 - 1) \hat{i} + (1 + 2) \hat{j} + (2 + 3) \hat{k}$
$= 3 \hat{j} + 5 \hat{k}$

Calculate $a \times b$

Using the determinant method for cross product:
$a \times b = \hat{i} 1 - 1 \hat{j} 12 \hat{k} 23$
$= \hat{i} (3 - 4) - \hat{j} (3 - (- 2)) + \hat{k} (2 - (- 1))$
$= - \hat{i} - 5 \hat{j} + 3 \hat{k}$

Compute Final Cross Product

Now, compute $(a + b) \times (a \times b)$ :
$= (3 \hat{j} + 5 \hat{k}) \times (- \hat{i} - 5 \hat{j} + 3 \hat{k})$
$= \hat{i} 0 - 1 \hat{j} 3 - 5 \hat{k} 53$

Evaluate the Determinant

Expanding along the first row:
$= \hat{i} (9 - (- 25)) - \hat{j} (0 - (- 5)) + \hat{k} (0 - (- 3))$
$= \hat{i} (9 + 25) - \hat{j} (5) + \hat{k} (3)$
$= 34 \hat{i} - 5 \hat{j} + 3 \hat{k}$

Final Result

Multiply by the scalar $7$ we pulled out earlier:
Result $= 7 (34 \hat{i} - 5 \hat{j} + 3 \hat{k})$
This perfectly matches Option (2).

00:00 / 00:00

The Sigma Insight: Vector Triple Product

Solution Diagram

Analyzing the Innermost Term

Our journey begins with the innermost term:

(a - b) \times b

By applying the distributive property of the cross product, this expression expands to:

(a \times b) - (b \times b)

Since the cross product of any vector with itself is the zero vector, the second term vanishes. We are left simply with

a \times b

Applying the Vector Triple Product

The expression now simplifies to

a \times (a \times b)

. This is a classic Vector Triple Product, which we resolve using the identity:

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

Applying this to our specific case, we transform the expression into:

(a \cdot b) a - (a \cdot a) b

Substituting Known Values

We are given the dot products:

a \cdot b = 7

and

a \cdot a = 6

Substituting these values into our derived equation, the 'monster' expression collapses into:

7 a - 6 b

Final Calculation

The final phase requires evaluating

(a + b) \times (7 (a \times b))

First, pull the scalar

7

out of the cross product. Then, compute the cross product of the sum

(a + b)

with the result of

(a \times b)

using the determinant method.

By staying calm and trusting your vector identities, the complexity dissolves, leading you directly to the final result.

Common Traps & FAQs

Wait, why does (\vec{a} - \vec{b}) \times \vec{b} simplify to \vec{a} \times \vec{b}? Shouldn't there be a negative sign?

Great question! When you distribute the cross product, you get (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{b}). Since the angle between \vec{b} and itself is 0 degrees, the magnitude of \vec{b} \times \vec{b} is |\vec{b}||\vec{b}|\sin(0) = 0. Thus, the term vanishes, leaving you with just \vec{a} \times \vec{b}. It is a beautiful simplification that saves you from unnecessary work!

Is it mandatory to use the Vector Triple Product formula? Can't I just expand the cross products directly?

Technically, you could, but it would be a nightmare of algebra! The Vector Triple Product formula is a powerful tool designed specifically to save you time and prevent calculation errors. In JEE Advanced, efficiency is just as important as accuracy. Embrace the formula; it is your best friend in vector geometry.

Why do we pull the scalar 7 out of the cross product?

The cross product is linear with respect to scalar multiplication. This means for any scalar k and vectors \vec{u} and \vec{v}, k(\vec{u} \times \vec{v}) = (k\vec{u}) \times \vec{v} = \vec{u} \times (k\vec{v}). Pulling the scalar out keeps your numbers smaller and your calculations cleaner, which significantly reduces the chance of a silly arithmetic mistake.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + \hat{j} + 2 \hat{k}$ and $b = - \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Then the vector product $(a + b) \times ((a \times ((a - b) \times b)) \times b)$ is equal to :

Select Answer:

Visualized Solution

Define Vectors $a$ and $b$

Simplify Innermost Term

Cross Product of a Vector with Itself

Apply Vector Triple Product

Calculate Dot Products

Substitute into VTP

Simplify the Next Outer Bracket

Final Expression Structure

Calculate $a + b$

Calculate $a \times b$

Compute Final Cross Product

Evaluate the Determinant

Final Result

The Sigma Insight: Vector Triple Product

Analyzing the Innermost Term

Applying the Vector Triple Product

Substituting Known Values

Final Calculation

Similar Questions

If $a = \frac{1}{10} (3 \hat{i} + \hat{k})$ and $b = \frac{1}{7} (2 \hat{i} + 3 \hat{j} - 6 \hat{k})$ , then the value of $(2 a - b) \cdot [(a \times b) \times (a + 2 b)]$ is

If $a = (\hat{i} + \hat{j} + \hat{k})$ , $a \cdot b = 1$ and $a \times b = \hat{j} - \hat{k}$ , then $b$ is

Let $a = - \hat{i} - \hat{j} + \hat{k}, a \cdot b = 1$ and $a \times b = \hat{i} - \hat{j}$ . Then $a - 6 b$ is equal to

Let $a = - \hat{i} + \hat{j} + 2 \hat{k}$ and $b = \hat{i} - \hat{j} - 3 \hat{k}, c = a \times b$ and $d = c \times a$ . Then $(a - b) \cdot d$ is equal to :

Let $a = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $b = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $c = (((a \times b) \times \hat{i}) \times \hat{i}) \times \hat{i}$ . Then $c \cdot (- \hat{i} + \hat{j} + \hat{k})$ is equal to

If $a$ and $b$ are vectors in space given by $a = \frac{i ^ - 2 j ^}{5}$ and $b = \frac{2 i ^ + j ^ + 3 k ^}{14}$ , then find the value of $(2 a + b) \cdot [(a \times b) \times (a - 2 b)]$ .

Let $λ \in R, a = λ \hat{i} + 2 \hat{j} - 3 \hat{k}, b = \hat{i} - λ \hat{j} + 2 \hat{k}$ . If $((a + b) \times (a \times b)) \times (a - b) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$ , then $∣ λ (a + b) \times (a - b) ∣^{2}$ is equal to

Let $a = \hat{i} + \hat{j} + \hat{k}$ and $b = \hat{j} - \hat{k}$ . If $c$ is a vector such that $a \times c = b$ and $a \cdot c = 3$ , then $a \cdot (b \times c)$ is equal to :

Let $a = 2 \hat{i} - \hat{j} + 5 \hat{k}$ and $b = α \hat{i} + β \hat{j} + 2 \hat{k}$ . If $((a \times b) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}$ , then $∣ b \times 2 \hat{j} ∣$ 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:

Animated Solution for Mathematics - Vector Algebra: .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=−i^+2j^​+3k^. Then the vector product (a+b)×((a×((a−b)×b))×b) is equal to :

Select Answer:

Visualized Solution

Define Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Simplify Innermost Term

Cross Product of a Vector with Itself

Apply Vector Triple Product

Calculate Dot Products

Substitute into VTP

Simplify the Next Outer Bracket

Final Expression Structure

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a×b

Compute Final Cross Product

Evaluate the Determinant

Final Result

The Sigma Insight: Vector Triple Product

Analyzing the Innermost Term

Applying the Vector Triple Product

Substituting Known Values

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=10​1​(3i^+k^) and b=71​(2i^+3j^​−6k^), then the value of (2a−b)⋅[(a×b)×(a+2b)] is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=−5i^+j^​−3k^, b=i^+2j^​−4k^ and c=(((a×b)×i^)×i^)×i^. Then c⋅(−i^+j^​+k^) is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let λ∈R,a=λi^+2j^​−3k^,b=i^−λj^​+2k^. If ((a+b)×(a×b))×(a−b)=8i^−40j^​−24k^, then ∣λ(a+b)×(a−b)∣2 is equal to

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.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:

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + \hat{j} + 2 \hat{k}$ and $b = - \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Then the vector product $(a + b) \times ((a \times ((a - b) \times b)) \times b)$ is equal to :

Define Vectors $a$ and $b$

Calculate $a + b$

Calculate $a \times b$

If $a = \frac{1}{10} (3 \hat{i} + \hat{k})$ and $b = \frac{1}{7} (2 \hat{i} + 3 \hat{j} - 6 \hat{k})$ , then the value of $(2 a - b) \cdot [(a \times b) \times (a + 2 b)]$ is

If $a = (\hat{i} + \hat{j} + \hat{k})$ , $a \cdot b = 1$ and $a \times b = \hat{j} - \hat{k}$ , then $b$ is

Let $a = - \hat{i} - \hat{j} + \hat{k}, a \cdot b = 1$ and $a \times b = \hat{i} - \hat{j}$ . Then $a - 6 b$ is equal to

Let $a = - \hat{i} + \hat{j} + 2 \hat{k}$ and $b = \hat{i} - \hat{j} - 3 \hat{k}, c = a \times b$ and $d = c \times a$ . Then $(a - b) \cdot d$ is equal to :

Let $a = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $b = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $c = (((a \times b) \times \hat{i}) \times \hat{i}) \times \hat{i}$ . Then $c \cdot (- \hat{i} + \hat{j} + \hat{k})$ is equal to

If $a$ and $b$ are vectors in space given by $a = \frac{i ^ - 2 j ^}{5}$ and $b = \frac{2 i ^ + j ^ + 3 k ^}{14}$ , then find the value of $(2 a + b) \cdot [(a \times b) \times (a - 2 b)]$ .

Let $λ \in R, a = λ \hat{i} + 2 \hat{j} - 3 \hat{k}, b = \hat{i} - λ \hat{j} + 2 \hat{k}$ . If $((a + b) \times (a \times b)) \times (a - b) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$ , then $∣ λ (a + b) \times (a - b) ∣^{2}$ is equal to

Let $a = \hat{i} + \hat{j} + \hat{k}$ and $b = \hat{j} - \hat{k}$ . If $c$ is a vector such that $a \times c = b$ and $a \cdot c = 3$ , then $a \cdot (b \times c)$ is equal to :

Let $a = 2 \hat{i} - \hat{j} + 5 \hat{k}$ and $b = α \hat{i} + β \hat{j} + 2 \hat{k}$ . If $((a \times b) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}$ , then $∣ b \times 2 \hat{j} ∣$ 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: