Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2, \vec{a} \cdot \vec{b} = 3$ and $|\vec{a} \times \vec{b}|^2 = 75$. Then $|\vec{a}|^2$ is equal to

### Analyzing the Setup We are given two vectors, $\vec{a}$ and $\vec{b}$, originating from the same point. We are provided with three specific constraints: 1. $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$ 2. $\vec{a} \cdot \vec{b} = 3$ 3. $|\vec{a} \times \vec{b}|^2 = 75$ Our objective is to determine the value of $|\vec{a}|^2$. ### Phase 1: The Expansion We begin by utilizing the fundamental identity for the magnitude of a vector sum: $$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$$ Substituting this into our first given equation, we obtain: $$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + 2|\vec{b}|^2$$ By subtracting $|\vec{a}|^2$ from both sides, the expression simplifies significantly: $$|\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 2|\vec{b}|^2$$ Rearranging the terms to isolate the magnitude of $\vec{b}$, we find: $$2(\vec{a} \cdot \vec{b}) = |\vec{b}|^2$$ ### Phase 2: The Master Key Given that $\vec{a} \cdot \vec{b} = 3$, we can immediately calculate the squared magnitude of $\vec{b}$: $$|\vec{b}|^2 = 2(3) = 6$$ To bridge the gap between the dot product, the cross product, and the magnitudes, we employ **Lagrange's Identity**: $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$ Substituting our known values ($|\vec{a} \times \vec{b}|^2 = 75$, $\vec{a} \cdot \vec{b} = 3$, and $|\vec{b}|^2 = 6$) into this identity, we get: $$75 + (3)^2 = |\vec{a}|^2 (6)$$ ### Final Calculation Simplifying the arithmetic, we have: $$75 + 9 = 6|\vec{a}|^2$$ $$84 = 6|\vec{a}|^2$$ Dividing both sides by $6$, we arrive at the final result: $$|\vec{a}|^2 = \frac{84}{6} = 14$$ The squared magnitude of vector $\vec{a}$ is **14**.

Why can't we just use the angle $\theta$ between the vectors to solve this?

You absolutely could! You could find $\cos\theta$ from the dot product and $\sin\theta$ from the cross product, then use $\sin^2\theta + \cos^2\theta = 1$. However, Lagrange's Identity is essentially the algebraic form of that trigonometric identity. It is faster and less prone to calculation errors.

JEE Main 2022 (29 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two vectors such that $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}, a \cdot b = 3$ and $∣ a \times b ∣^{2} = 75$ . Then $∣ a ∣^{2}$ is equal to

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: When you see the magnitude of a sum squared, your first instinct should be to expand it using the dot product definition. Visualize the parallelogram formed by

a

and

b

—the diagonal length squared is not just the sum of the squares of the sides.

Hint 2

The Arsenal: Once you expand the first equation, you will notice that the term

∣ a ∣^{2}

appears on both sides. Cancel it out! This reveals a direct relationship between the dot product and the magnitude of

b

. You now have the value of

∣ b ∣^{2}

Hint 3

The Execution: Now that you have

∣ b ∣^{2}

and the dot product, you need to find

∣ a ∣^{2}

. The bridge connecting the dot product, cross product, and the magnitudes is Lagrange's Identity:

∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}

. Plug in your known values and solve for the unknown.

Visualized Solution

Visualizing the Vectors

Let $a$ and $b$ be two vectors.
Given: $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}$
Given: $a \cdot b = 3$
Given: $∣ a \times b ∣^{2} = 75$

Expanding $∣ a + b ∣^{2}$

Recall the vector identity for the magnitude of a sum:
$∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$

Equating the Expressions

Substitute the standard expansion into the given equation:
$∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b = ∣ a ∣^{2} + 2∣ b ∣^{2}$

Simplifying the Equation

Subtract $∣ a ∣^{2}$ from both sides:
$∣ b ∣^{2} + 2 a \cdot b = 2∣ b ∣^{2}$

Isolating $∣ b ∣^{2}$

Subtract $∣ b ∣^{2}$ from both sides to isolate it:
$2 a \cdot b = ∣ b ∣^{2}$

Finding $∣ b ∣^{2}$

We are given the dot product: $a \cdot b = 3$
Substitute this value into our simplified equation:
$∣ b ∣^{2} = 2 (3) = 6$

Introducing Lagrange's Identity

To connect dot product, cross product, and magnitudes, we use Lagrange's Identity:
$∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$

Substituting Known Values

Substitute the known values into the identity:
$∣ a \times b ∣^{2} = 75$
$a \cdot b = 3$
$∣ b ∣^{2} = 6$
$75 + (3)^{2} = ∣ a ∣^{2} (6)$

Simplifying the Relation

Calculate the square of the dot product:
$75 + 9 = 6∣ a ∣^{2}$
$84 = 6∣ a ∣^{2}$

Final Conclusion

Divide by $6$ to find the final magnitude squared:
$∣ a ∣^{2} = \frac{84}{6} = 14$
Final Answer: $14$

00:00 / 00:00

The Sigma Insight: Vector (Cross) Product

Solution Diagram

Analyzing the Setup

We are given two vectors,

a

and

b

, originating from the same point. We are provided with three specific constraints:

∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}

a \cdot b = 3

∣ a \times b ∣^{2} = 75

Our objective is to determine the value of

∣ a ∣^{2}

Phase 1

The Expansion

We begin by utilizing the fundamental identity for the magnitude of a vector sum:

∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 (a \cdot b)

Substituting this into our first given equation, we obtain:

∣ a ∣^{2} + ∣ b ∣^{2} + 2 (a \cdot b) = ∣ a ∣^{2} + 2∣ b ∣^{2}

By subtracting

∣ a ∣^{2}

from both sides, the expression simplifies significantly:

∣ b ∣^{2} + 2 (a \cdot b) = 2∣ b ∣^{2}

Rearranging the terms to isolate the magnitude of

b

, we find:

2 (a \cdot b) = ∣ b ∣^{2}

Phase 2

The Master Key

Given that

a \cdot b = 3

, we can immediately calculate the squared magnitude of

b

∣ b ∣^{2} = 2 (3) = 6

To bridge the gap between the dot product, the cross product, and the magnitudes, we employ Lagrange's Identity:

∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}

Substituting our known values (

∣ a \times b ∣^{2} = 75

a \cdot b = 3

, and

∣ b ∣^{2} = 6

) into this identity, we get:

75 + (3)^{2} = ∣ a ∣^{2} (6)

Final Calculation

Simplifying the arithmetic, we have:

75 + 9 = 6∣ a ∣^{2}

84 = 6∣ a ∣^{2}

Dividing both sides by

6

, we arrive at the final result:

∣ a ∣^{2} = \frac{84}{6} = 14

The squared magnitude of vector

a

is 14.

Common Traps & FAQs

Wait, why do we reject the negative root when finding the magnitude?

Remember, the magnitude of a vector is defined as the square root of the dot product of the vector with itself, which is always non-negative. We are dealing with

∣ a ∣^{2}

, which is inherently positive, so we don't need to worry about negative roots here.

Why can't we just use the angle

θ

between the vectors to solve this?

You absolutely could! You could find

cos θ

from the dot product and

sin θ

from the cross product, then use

sin^{2} θ + cos^{2} θ = 1

. However, Lagrange's Identity is essentially the algebraic form of that trigonometric identity. It is faster and less prone to calculation errors.

Is Lagrange's Identity always true for any two vectors?

Yes, it is a fundamental vector identity in 3D space. It is a powerful tool that often appears in JEE Advanced problems to link dot and cross products. Keep it in your toolkit!

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two vectors such that $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}, a \cdot b = 3$ and $∣ a \times b ∣^{2} = 75$ . Then $∣ a ∣^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors

Expanding $∣ a + b ∣^{2}$

Equating the Expressions

Simplifying the Equation

Isolating $∣ b ∣^{2}$

Finding $∣ b ∣^{2}$

Introducing Lagrange's Identity

Substituting Known Values

Simplifying the Relation

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Final Calculation

Similar Questions

Let $a = - \hat{i} + 2 \hat{j} + 2 \hat{k}, b = 8 \hat{i} + 7 \hat{j} - 3 \hat{k}$ and $c$ be vector such that $a \times c = b$ . If $c \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ , then $∣ a + c ∣^{2}$ is equal to :

Let $a$ and $b$ be two vector such that $∣ a ∣ = 14, ∣ b ∣ = 6$ and $∣ a \times b ∣ = 48$ . Then $(a \cdot b)^{2}$ is equal to

Let $a = 3 \hat{i} + \hat{j} - \hat{k}$ and $c = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ . If $b$ is a vector such that $a = b \times c$ and $∣ b ∣^{2} = 50$ , then $∣72 - ∣ b + c ∣^{2} ∣$ is equal to

Let $b = \hat{i} + \hat{j} + λ \hat{k}, λ \in R$ . If $a$ is a vector such that $a \times b = 13 \hat{i} - \hat{j} - 4 \hat{k}$ and $a \cdot b + 21 = 0$ , then $(b - a) \cdot (\hat{k} - \hat{j}) + (b + a) \cdot (\hat{i} - \hat{k})$ is equal to

Let $a$ and $b$ be two vectors such that $∣ b ∣ = 1$ and $∣ b \times a ∣ = 2$ . Then $∣ (b \times a) - b ∣^{2}$ is equal to

Let $a = \hat{i} + 2 \hat{j} + λ \hat{k}$ , $b = 3 \hat{i} - 5 \hat{j} - λ \hat{k}$ , $a \cdot c = 7$ , $2 b \cdot c + 43 = 0$ , $a \times c = b \times c$ . Then $∣ a \cdot b ∣$ is equal to

If $a$ and $b$ are vectors such that $∣ a + b ∣ = 29$ and $a \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times b$ , then a possible value of $(a + b) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k})$ is

Let $a = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $b = 2 \hat{i} - 3 \hat{j} + 5 \hat{k}$ . If $r \times a = b \times r$ , $r \cdot (α \hat{i} + 2 \hat{j} + \hat{k}) = 3$ and $r \cdot (2 \hat{i} + 5 \hat{j} - α \hat{k}) = - 1$ , $α \in R$ , then the value of $α + ∣ r ∣^{2}$ is equal to :

Let $a = \hat{i} + 4 \hat{j} + 2 \hat{k}$ , $b = 3 \hat{i} - \hat{j} + 7 \hat{k}$ and $c = 2 \hat{i} - \hat{j} + 4 \hat{k}$ . If a vector $d$ satisfies $d \times b = c \times b$ and $d \cdot a = 24$ , then $∣ d ∣^{2}$ 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 and b be two vectors such that ∣a+b∣2=∣a∣2+2∣b∣2,a⋅b=3 and ∣a×b∣2=75. Then ∣a∣2 is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a+b∣2

Equating the Expressions

Simplifying the Equation

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣b∣2

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣b∣2

Introducing Lagrange's Identity

Substituting Known Values

Simplifying the Relation

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=−i^+2j^​+2k^,b=8i^+7j^​−3k^ and c be vector such that a×c=b. If c⋅(i^+j^​+k^)=4, then ∣a+c∣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; } If a and b are vectors such that ∣a+b∣=29​ and a×(2i^+3j^​+4k^)=(2i^+3j^​+4k^)×b, then a possible value of (a+b)⋅(−7i^+2j^​+3k^) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+2j^​−3k^ and b=2i^−3j^​+5k^. If r×a=b×r, r⋅(αi^+2j^​+k^)=3 and r⋅(2i^+5j^​−αk^)=−1, α∈R, then the value of α+∣r∣2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+4j^​+2k^, b=3i^−j^​+7k^ and c=2i^−j^​+4k^. If a vector d satisfies d×b=c×b and d⋅a=24, then ∣d∣2 is equal to

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two vectors such that $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}, a \cdot b = 3$ and $∣ a \times b ∣^{2} = 75$ . Then $∣ a ∣^{2}$ is equal to

Expanding $∣ a + b ∣^{2}$

Isolating $∣ b ∣^{2}$

Finding $∣ b ∣^{2}$

Let $a = - \hat{i} + 2 \hat{j} + 2 \hat{k}, b = 8 \hat{i} + 7 \hat{j} - 3 \hat{k}$ and $c$ be vector such that $a \times c = b$ . If $c \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ , then $∣ a + c ∣^{2}$ is equal to :

Let $a$ and $b$ be two vector such that $∣ a ∣ = 14, ∣ b ∣ = 6$ and $∣ a \times b ∣ = 48$ . Then $(a \cdot b)^{2}$ is equal to

Let $a = 3 \hat{i} + \hat{j} - \hat{k}$ and $c = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ . If $b$ is a vector such that $a = b \times c$ and $∣ b ∣^{2} = 50$ , then $∣72 - ∣ b + c ∣^{2} ∣$ is equal to

Let $b = \hat{i} + \hat{j} + λ \hat{k}, λ \in R$ . If $a$ is a vector such that $a \times b = 13 \hat{i} - \hat{j} - 4 \hat{k}$ and $a \cdot b + 21 = 0$ , then $(b - a) \cdot (\hat{k} - \hat{j}) + (b + a) \cdot (\hat{i} - \hat{k})$ is equal to

Let $a$ and $b$ be two vectors such that $∣ b ∣ = 1$ and $∣ b \times a ∣ = 2$ . Then $∣ (b \times a) - b ∣^{2}$ is equal to

Let $a = \hat{i} + 2 \hat{j} + λ \hat{k}$ , $b = 3 \hat{i} - 5 \hat{j} - λ \hat{k}$ , $a \cdot c = 7$ , $2 b \cdot c + 43 = 0$ , $a \times c = b \times c$ . Then $∣ a \cdot b ∣$ is equal to

If $a$ and $b$ are vectors such that $∣ a + b ∣ = 29$ and $a \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times b$ , then a possible value of $(a + b) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k})$ is

Let $a = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $b = 2 \hat{i} - 3 \hat{j} + 5 \hat{k}$ . If $r \times a = b \times r$ , $r \cdot (α \hat{i} + 2 \hat{j} + \hat{k}) = 3$ and $r \cdot (2 \hat{i} + 5 \hat{j} - α \hat{k}) = - 1$ , $α \in R$ , then the value of $α + ∣ r ∣^{2}$ is equal to :

Let $a = \hat{i} + 4 \hat{j} + 2 \hat{k}$ , $b = 3 \hat{i} - \hat{j} + 7 \hat{k}$ and $c = 2 \hat{i} - \hat{j} + 4 \hat{k}$ . If a vector $d$ satisfies $d \times b = c \times b$ and $d \cdot a = 24$ , then $∣ d ∣^{2}$ is equal to