JEE Main 2023 (25 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: 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

Q: Wait, why can't I just assume $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve the system of equations?

You absolutely can! If you set $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$, you will get a system of linear equations. However, that approach is computationally tedious and prone to algebraic errors. The Vector Triple Product identity is a much more elegant and faster 'shortcut' that leverages the geometric structure of the vectors.

Q: Wait, why did we multiply the equation by 2 in the middle?

Great observation! We arrived at the equation $\vec{a} - 3\vec{b} = \hat{i} + \hat{j} + 2\hat{k}$. The question asks for $\vec{a} - 6\vec{b}$. By multiplying the entire equation by 2, we get $2\vec{a} - 6\vec{b} = 2\hat{i} + 2\hat{j} + 4\hat{k}$. This brings us one step closer to the target expression $\vec{a} - 6\vec{b}$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: When you see both a dot product (

a \cdot b = 1

) and a cross product (

a \times b = \hat{i} - \hat{j}

) involving the same unknown vector

b

, you are looking at a system of equations that constrains

b

in both its projection onto

a

and its perpendicular component. The key is to use the Vector Triple Product identity to combine these constraints.

Hint 2

Hint 2: Recall the Vector Triple Product identity:

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

. This identity is the 'skeleton key' for this problem, as it allows you to express the unknown vector

b

in terms of known quantities.

Hint 3

Hint 3: Start by taking the cross product of

a

with the given equation

a \times b = \hat{i} - \hat{j}

. This gives you

a \times (a \times b) = a \times (\hat{i} - \hat{j})

. Now, substitute the identity from Hint 2 and evaluate the right-hand side using a determinant.

Visualized Solution

Identify Given Information

Given: $a = - \hat{i} - \hat{j} + \hat{k}$
Given: $a \cdot b = 1$
Given: $a \times b = \hat{i} - \hat{j}$

The Strategy: Vector Triple Product

Strategy: Use the Vector Triple Product identity to isolate $b$ .
Identity: $a \times (a \times b) = (a \cdot b) a - (a \cdot a) b$

Applying the Identity

Taking cross product with $a$ on both sides:
$a \times (a \times b) = a \times (\hat{i} - \hat{j})$
Expanding LHS: $(a \cdot b) a - (a \cdot a) b = a \times (\hat{i} - \hat{j})$

Calculate $(a \cdot a)$

Calculate $a \cdot a$ :
$a \cdot a = (- 1)^{2} + (- 1)^{2} + (1)^{2}$
$a \cdot a = 1 + 1 + 1 = 3$

Setup Right-Hand Side Cross Product

Calculate $a \times (\hat{i} - \hat{j})$ :
$a \times (\hat{i} - \hat{j}) = \hat{i} - 1 1 \hat{j} - 1 - 1 \hat{k} 10$

Solve Determinant: $\hat{i}$ Component

Expanding for $\hat{i}$ :
$\hat{i} [(- 1) (0) - (1) (- 1)] = \hat{i} [0 + 1] = 1 \hat{i}$

Solve Determinant: $\hat{j}$ and $\hat{k}$ Components

Expanding for $\hat{j}$ and $\hat{k}$ :
$- \hat{j} [(- 1) (0) - (1) (1)] = - \hat{j} [0 - 1] = 1 \hat{j}$
$\hat{k} [(- 1) (- 1) - (- 1) (1)] = \hat{k} [1 + 1] = 2 \hat{k}$
Result: $a \times (\hat{i} - \hat{j}) = \hat{i} + \hat{j} + 2 \hat{k}$

Substitute Values into Identity

Substitute $a \cdot b = 1$ , $a \cdot a = 3$ , and the cross product result:
$(1) a - (3) b = \hat{i} + \hat{j} + 2 \hat{k}$
$a - 3 b = \hat{i} + \hat{j} + 2 \hat{k}$

Manipulate to find $a - 6 b$

We need to find $a - 6 b$ .
Multiply the equation by $2$ :
$2 (a - 3 b) = 2 (\hat{i} + \hat{j} + 2 \hat{k})$
$2 a - 6 b = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}$

Final Subtraction

Subtract $a$ from both sides to get $a - 6 b$ :
$a - 6 b = (2 a - 6 b) - a$
$a - 6 b = (2 \hat{i} + 2 \hat{j} + 4 \hat{k}) - (- \hat{i} - \hat{j} + \hat{k})$

Final Calculation and Result

Simplify the expression:
$a - 6 b = (2 - (- 1)) \hat{i} + (2 - (- 1)) \hat{j} + (4 - 1) \hat{k}$
$a - 6 b = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}$
$a - 6 b = 3 (\hat{i} + \hat{j} + \hat{k})$

Conclusion & Key Takeaway

Key Takeaway: The Vector Triple Product identity is a powerful tool to solve equations involving both dot and cross products of unknown vectors.
Final Answer: $3 (\hat{i} + \hat{j} + \hat{k})$

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The Sigma Insight: Vector Triple Product

Solution Diagram

Analyzing the Setup

We are given the vector

a = - \hat{i} - \hat{j} + \hat{k}

and an unknown vector

b

. The relationships provided are:

a \cdot b = 1

a \times b = \hat{i} - \hat{j}

These two equations define the projection of

b

onto

a

and the component of

b

perpendicular to

a

, respectively. Together, they uniquely determine

b

The Swiss Army Knife

Vector Triple Product

To solve for

b

, we utilize the Vector Triple Product identity:

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

Taking the cross product of

a

with the given equation

a \times b = \hat{i} - \hat{j}

, we obtain:

a \times (a \times b) = a \times (\hat{i} - \hat{j})

Evaluating the Components

First, we calculate the magnitude squared of

a

a \cdot a = (- 1)^{2} + (- 1)^{2} + (1)^{2} = 3

Next, we compute the cross product on the right-hand side using the determinant method:

a \times (\hat{i} - \hat{j}) = \hat{i} - 1 1 \hat{j} - 1 - 1 \hat{k} 10 = \hat{i} (0 - (- 1)) - \hat{j} (0 - 1) + \hat{k} (1 - (- 1)) = \hat{i} + \hat{j} + 2 \hat{k}

The Final Algebraic Dance

Substituting these values into the identity, we get:

(1) a - (3) b = \hat{i} + \hat{j} + 2 \hat{k}

To find the value of

a - 6 b

, we manipulate the equation above. First, multiply the entire equation by 2:

2 a - 6 b = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}

Subtracting

a

from both sides yields:

a - 6 b = (2 \hat{i} + 2 \hat{j} + 4 \hat{k}) - (- \hat{i} - \hat{j} + \hat{k})

Performing the vector subtraction, we arrive at the final result:

a - 6 b = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} = 3 (\hat{i} + \hat{j} + \hat{k})

Common Traps & FAQs

Wait, why can't I just assume

b = x \hat{i} + y \hat{j} + z \hat{k}

and solve the system of equations?

You absolutely can! If you set

b = x \hat{i} + y \hat{j} + z \hat{k}

, you will get a system of linear equations. However, that approach is computationally tedious and prone to algebraic errors. The Vector Triple Product identity is a much more elegant and faster 'shortcut' that leverages the geometric structure of the vectors.

How did you know to cross with

a

? Is there a rule?

Yes, there is a rule of thumb! Whenever you have a mixed product equation (involving both dot and cross products) and you need to isolate an unknown vector, crossing the equation with a known vector is the standard strategy. It transforms the cross product into a linear combination of vectors, which is much easier to handle.

Wait, why did we multiply the equation by 2 in the middle?

Great observation! We arrived at the equation

a - 3 b = \hat{i} + \hat{j} + 2 \hat{k}

. The question asks for

a - 6 b

. By multiplying the entire equation by 2, we get

2 a - 6 b = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}

. This brings us one step closer to the target expression

a - 6 b

Animated Solution for Mathematics - Vector Algebra: 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

Select Answer:

Visualized Solution

Identify Given Information

The Strategy: Vector Triple Product

Applying the Identity

Calculate $(a \cdot a)$

Setup Right-Hand Side Cross Product

Solve Determinant: $\hat{i}$ Component

Solve Determinant: $\hat{j}$ and $\hat{k}$ Components

Substitute Values into Identity

Manipulate to find $a - 6 b$

Final Subtraction

Final Calculation and Result

Conclusion & Key Takeaway

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Swiss Army Knife

Evaluating the Components

The Final Algebraic Dance

Similar Questions

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

Let $a = \hat{i} + \hat{j} + \hat{k}$ 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 $a = \hat{i} + \hat{j} + 2 \hat{k}$ 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 :

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

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 $a = \hat{i} - 2 \hat{j} + \hat{k}$ and $b = \hat{i} - \hat{j} + \hat{k}$ be two vectors. If $c$ is a vector such that $b \times c = b \times a$ and $c \cdot a = 0$ , then $c \cdot b$ is equal to

If $a \cdot b = 1, b \cdot c = 2$ and $c \cdot a = 3$ , then the value of $[a \times (b \times c), b \times (c \times a), c \times (b \times a)]$ is:

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

Let $a, b$ and $c$ be three non zero vectors such that $b \cdot c = 0$ and $a \times (b \times c) = \frac{b - c}{2}$ . If $d$ be a vector such that $b \cdot d = a \cdot b$ , then $(a \times b) \cdot (c \times d)$ 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:

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^​+k^,a⋅b=1 and a×b=i^−j^​. Then a−6b is equal to

Select Answer:

Visualized Solution

Identify Given Information

The Strategy: Vector Triple Product

Applying the Identity

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

Setup Right-Hand Side Cross Product

Solve Determinant: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i^ Component

Solve Determinant: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } j^​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k^ Components

Substitute Values into Identity

Manipulate to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a−6b

Final Subtraction

Final Calculation and Result

Conclusion & Key Takeaway

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Swiss Army Knife

Evaluating the Components

The Final Algebraic Dance

Similar Questions

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

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

.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

.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^−j^​−3k^,c=a×b and d=c×a. Then (a−b)⋅d is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^−2j^​+k^ and b=i^−j^​+k^ be two vectors. If c is a vector such that b×c=b×a and c⋅a=0, then c⋅b is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a⋅b=1,b⋅c=2 and c⋅a=3, then the value of [a×(b×c),b×(c×a),c×(b×a)] is:

.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

.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 non zero vectors such that b⋅c=0 and a×(b×c)=2b−c​. If d be a vector such that b⋅d=a⋅b, then (a×b)⋅(c×d) is equal to

Animated Solution for Mathematics - Vector Algebra: 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

Calculate $(a \cdot a)$

Solve Determinant: $\hat{i}$ Component

Solve Determinant: $\hat{j}$ and $\hat{k}$ Components

Manipulate to find $a - 6 b$

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

Let $a = \hat{i} + \hat{j} + \hat{k}$ 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 $a = \hat{i} + \hat{j} + 2 \hat{k}$ 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 :

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

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 $a = \hat{i} - 2 \hat{j} + \hat{k}$ and $b = \hat{i} - \hat{j} + \hat{k}$ be two vectors. If $c$ is a vector such that $b \times c = b \times a$ and $c \cdot a = 0$ , then $c \cdot b$ is equal to

If $a \cdot b = 1, b \cdot c = 2$ and $c \cdot a = 3$ , then the value of $[a \times (b \times c), b \times (c \times a), c \times (b \times a)]$ is:

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

Let $a, b$ and $c$ be three non zero vectors such that $b \cdot c = 0$ and $a \times (b \times c) = \frac{b - c}{2}$ . If $d$ be a vector such that $b \cdot d = a \cdot b$ , then $(a \times b) \cdot (c \times d)$ is equal to