Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{c} = x\hat{i} + 2\hat{j} + 3\hat{k}, x \in \mathbb{R}$. If $\vec{d}$ is the unit vector in the direction of $\vec{b} + \vec{c}$ such that $\vec{a} \cdot \vec{d} = 1$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to Options: (A) 11 (B) 3 (C) 9 (D) 6

### Analyzing the Setup We are given three vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}$, and $\vec{c} = x\hat{i} + 2\hat{j} + 3\hat{k}$. These vectors represent the edges of a parallelepiped in three-dimensional space. ### The Bridge of the Unit Vector We are introduced to a unit vector $\vec{d}$, defined as the direction of the sum $\vec{b} + \vec{c}$. First, we calculate the sum: $$\vec{b} + \vec{c} = (x + 2)\hat{i} + 6\hat{j} - 2\hat{k}$$ The problem provides the condition $\vec{a} \cdot \vec{d} = 1$. Since $\vec{d}$ is the unit vector along $\vec{b} + \vec{c}$, we define it as: $$\vec{d} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|}$$ Substituting this into the condition, we obtain: $$\vec{a} \cdot \left( \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} \right) = 1$$ By multiplying both sides by the magnitude, we arrive at the fundamental relation: $$\vec{a} \cdot (\vec{b} + \vec{c}) = |\vec{b} + \vec{c}|$$ ### The Algebraic Unfolding Next, we compute the dot product $\vec{a} \cdot (\vec{b} + \vec{c})$ using the components: $$\vec{a} \cdot (\vec{b} + \vec{c}) = (1)(x + 2) + (1)(6) + (1)(-2) = x + 6$$ Now, we calculate the magnitude $|\vec{b} + \vec{c}|$: $$|\vec{b} + \vec{c}| = \sqrt{(x + 2)^2 + 6^2 + (-2)^2} = \sqrt{x^2 + 4x + 4 + 36 + 4} = \sqrt{x^2 + 4x + 44}$$ Equating the two expressions, we have: $$x + 6 = \sqrt{x^2 + 4x + 44}$$ Squaring both sides yields: $$(x + 6)^2 = x^2 + 4x + 44$$ $$x^2 + 12x + 36 = x^2 + 4x + 44$$ Subtracting $x^2$ from both sides and simplifying, we find: $$8x = 8 \Rightarrow x = 1$$ ### Final Calculation With $x = 1$, our vector $\vec{c}$ is defined as $\vec{c} = \hat{i} + 2\hat{j} + 3\hat{k}$. The volume of the parallelepiped is given by the scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$, which is calculated via the determinant: $$\begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & -5 \\ 1 & 2 & 3 \end{vmatrix}$$ Expanding along the first row: $$\text{Volume} = 1(12 - (-10)) - 1(6 - (-5)) + 1(4 - 4)$$ $$\text{Volume} = 1(22) - 1(11) + 0$$ $$\text{Volume} = 22 - 11 = 11$$ The final volume of the parallelepiped is **11 cubic units**.

Wait, why can we just say $\vec{a} \cdot (\vec{b} + \vec{c}) = |\vec{b} + \vec{c}|$? Isn't there a cosine involved?

That is a brilliant observation! The definition is indeed $\vec{a} \cdot \vec{d} = |\vec{a}| |\vec{d}| \cos \theta$. However, since $\vec{d}$ is a unit vector, $|\vec{d}| = 1$. The problem states $\vec{a} \cdot \vec{d} = 1$. If you look at the components, you will see that the projection of $\vec{a}$ onto the direction of $(\vec{b} + \vec{c})$ happens to be exactly the magnitude of the vector $(\vec{b} + \vec{c})$ normalized. By substituting $\vec{d} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|}$, the denominator moves to the other side, leaving us with the dot product equal to the magnitude.

JEE Main 2024 (04 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $c = x \hat{i} + 2 \hat{j} + 3 \hat{k}, x \in R$ . If $d$ is the unit vector in the direction of $b + c$ such that $a \cdot d = 1$ , then $(a \times b) \cdot c$ is equal to

Q: Why do we use a determinant for $(\vec{a} \times \vec{b}) \cdot \vec{c}$?

The expression $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is the definition of the Scalar Triple Product. Geometrically, it represents the signed volume of a parallelepiped defined by these three vectors. Algebraically, the most efficient way to compute this is to arrange the components of $\vec{a}, \vec{b},$ and $\vec{c}$ as rows in a $3 \times 3$ matrix and calculate its determinant. It is a beautiful shortcut that saves you from performing the cross product first!

Q: Is it possible for $x$ to have two values? I'm worried about the squaring step.

When you square both sides of an equation like $A = \sqrt{B}$, you must always check for extraneous roots. In this specific case, squaring $(x+6) = \sqrt{x^2 + 4x + 44}$ leads to a linear equation because the $x^2$ terms cancel out. Since it reduces to a linear equation, there is only one unique solution for $x$. Always keep an eye on the domain of the square root, but here, the math remains elegant and consistent.

Q: What if I calculated the cross product first? Would it be wrong?

Not at all! You are absolutely free to calculate $\vec{a} \times \vec{b}$ first and then take the dot product with $\vec{c}$. However, the determinant method is the 'JEE way'—it is faster, less prone to arithmetic errors, and demonstrates a deeper understanding of linear algebra. Both paths lead to the same destination, but one is a scenic hike and the other is a high-speed train.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the condition

a \cdot d = 1

. Since

d

is a unit vector, this implies the projection of

a

onto the direction of

(b + c)

is equal to the magnitude of

a

's projection, which only happens if the vectors are aligned or if the geometry forces a specific constraint. Remember that

a \cdot d = ∣ a ∣∣ d ∣ cos θ = 1

Hint 2

Hint 2: Use the definition of the unit vector

d = \frac{b + c}{∣ b + c ∣}

. Substitute this into the given dot product equation to get

a \cdot (b + c) = ∣ b + c ∣

. This transforms a vector problem into a simple algebraic equation involving

x

Hint 3

Hint 3: Once you have the equation

x + 6 = x^{2} + 4 x + 44

, square both sides to eliminate the radical. Solve for

x

, and then use the scalar triple product property, which can be calculated as the determinant of the matrix formed by the components of

a, b,

and

c

Visualized Solution

Given Vectors $a, b, c$

$a = \hat{i} + \hat{j} + \hat{k}$
$b = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$
$c = x \hat{i} + 2 \hat{j} + 3 \hat{k}$

Vector Sum $b + c$

$b + c = (2 + x) \hat{i} + (4 + 2) \hat{j} + (- 5 + 3) \hat{k}$
$b + c = (x + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}$

Unit Vector $d$

$d$ is the unit vector along $b + c$ .
$d = \frac{b + c}{∣ b + c ∣}$

Condition $a \cdot d = 1$

$a \cdot (\frac{b + c}{∣ b + c ∣}) = 1$
$a \cdot (b + c) = ∣ b + c ∣$

Evaluating $a \cdot (b + c)$

$a \cdot (b + c) = (1) (x + 2) + (1) (6) + (1) (- 2)$
$a \cdot (b + c) = x + 2 + 6 - 2$
$a \cdot (b + c) = x + 6$

Magnitude $∣ b + c ∣$

$∣ b + c ∣ = (x + 2)^{2} + 6^{2} + (- 2)^{2}$
$∣ b + c ∣ = x^{2} + 4 x + 4 + 36 + 4$
$∣ b + c ∣ = x^{2} + 4 x + 44$

Solving for $x$

$x + 6 = x^{2} + 4 x + 44$
$(x + 6)^{2} = x^{2} + 4 x + 44$
$x^{2} + 12 x + 36 = x^{2} + 4 x + 44$

Value of $x$

$12 x - 4 x = 44 - 36$
$8 x = 8$
$x = 1$

Scalar Triple Product Setup

We need to find $(a \times b) \cdot c$
$(a \times b) \cdot c = 121142 1 - 5 3$

Evaluating the Determinant

$= 1 (12 - (- 10)) - 1 (6 - (- 5)) + 1 (4 - 4)$
$= 1 (22) - 1 (11) + 1 (0)$
$= 22 - 11$

Final Answer

$(a \times b) \cdot c = 11$
The scalar triple product represents the volume of the parallelepiped formed by $a, b, c$ .

00:00 / 00:00

The Sigma Insight: Scalar Triple Product

Solution Diagram

Analyzing the Setup

We are given three vectors:

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

b = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}

, and

c = x \hat{i} + 2 \hat{j} + 3 \hat{k}

. These vectors represent the edges of a parallelepiped in three-dimensional space.

The Bridge of the Unit Vector

We are introduced to a unit vector

d

, defined as the direction of the sum

b + c

. First, we calculate the sum:

b + c = (x + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}

The problem provides the condition

a \cdot d = 1

. Since

d

is the unit vector along

b + c

, we define it as:

d = \frac{b + c}{∣ b + c ∣}

Substituting this into the condition, we obtain:

a \cdot (\frac{b + c}{∣ b + c ∣}) = 1

By multiplying both sides by the magnitude, we arrive at the fundamental relation:

a \cdot (b + c) = ∣ b + c ∣

The Algebraic Unfolding

Next, we compute the dot product

a \cdot (b + c)

using the components:

a \cdot (b + c) = (1) (x + 2) + (1) (6) + (1) (- 2) = x + 6

Now, we calculate the magnitude

∣ b + c ∣

∣ b + c ∣ = (x + 2)^{2} + 6^{2} + (- 2)^{2} = x^{2} + 4 x + 4 + 36 + 4 = x^{2} + 4 x + 44

Equating the two expressions, we have:

x + 6 = x^{2} + 4 x + 44

Squaring both sides yields:

(x + 6)^{2} = x^{2} + 4 x + 44

x^{2} + 12 x + 36 = x^{2} + 4 x + 44

Subtracting

x^{2}

from both sides and simplifying, we find:

8 x = 8 \Rightarrow x = 1

Final Calculation

With

x = 1

, our vector

c

is defined as

c = \hat{i} + 2 \hat{j} + 3 \hat{k}

. The volume of the parallelepiped is given by the scalar triple product

(a \times b) \cdot c

, which is calculated via the determinant:

121142 1 - 5 3

Expanding along the first row:

Volume = 1 (12 - (- 10)) - 1 (6 - (- 5)) + 1 (4 - 4)

Volume = 1 (22) - 1 (11) + 0

Volume = 22 - 11 = 11

The final volume of the parallelepiped is 11 cubic units.

Common Traps & FAQs

Wait, why can we just say

a \cdot (b + c) = ∣ b + c ∣

? Isn't there a cosine involved?

That is a brilliant observation! The definition is indeed

a \cdot d = ∣ a ∣∣ d ∣ cos θ

. However, since

d

is a unit vector,

∣ d ∣ = 1

. The problem states

a \cdot d = 1

. If you look at the components, you will see that the projection of

a

onto the direction of

(b + c)

happens to be exactly the magnitude of the vector

(b + c)

normalized. By substituting

d = \frac{b + c}{∣ b + c ∣}

, the denominator moves to the other side, leaving us with the dot product equal to the magnitude.

Why do we use a determinant for

(a \times b) \cdot c

The expression

(a \times b) \cdot c

is the definition of the Scalar Triple Product. Geometrically, it represents the signed volume of a parallelepiped defined by these three vectors. Algebraically, the most efficient way to compute this is to arrange the components of

a, b,

and

c

as rows in a

3 \times 3

matrix and calculate its determinant. It is a beautiful shortcut that saves you from performing the cross product first!

Is it possible for

x

to have two values? I'm worried about the squaring step.

When you square both sides of an equation like

A = B

, you must always check for extraneous roots. In this specific case, squaring

(x + 6) = x^{2} + 4 x + 44

leads to a linear equation because the

x^{2}

terms cancel out. Since it reduces to a linear equation, there is only one unique solution for

x

. Always keep an eye on the domain of the square root, but here, the math remains elegant and consistent.

What if I calculated the cross product first? Would it be wrong?

Not at all! You are absolutely free to calculate

a \times b

first and then take the dot product with

c

. However, the determinant method is the 'JEE way'—it is faster, less prone to arithmetic errors, and demonstrates a deeper understanding of linear algebra. Both paths lead to the same destination, but one is a scenic hike and the other is a high-speed train.

Does the order of vectors in the scalar triple product matter?

Yes, it does! The scalar triple product follows cyclic permutation rules:

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

. If you swap any two vectors, the sign of the result flips. Always be careful with the order, but as long as you maintain the cyclic order, the volume remains invariant.

Select Answer:

Visualized Solution

Given Vectors $a, b, c$

Vector Sum $b + c$

Unit Vector $d$

Condition $a \cdot d = 1$

Evaluating $a \cdot (b + c)$

Magnitude $∣ b + c ∣$

Solving for $x$

Value of $x$

Scalar Triple Product Setup

Evaluating the Determinant

Final Answer

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Bridge of the Unit Vector

The Algebraic Unfolding

Final Calculation

Similar Questions

Let $a = \hat{i} - \hat{j}$ , $b = \hat{j} - \hat{k}$ , $c = \hat{k} - \hat{i}$ . If $d$ is a unit vector such that $a \cdot d = 0 = [b c d]$ , then $d$ equals

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

Let $a = \hat{i} - α \hat{j} + β \hat{k}, b = 3 \hat{i} + β \hat{j} - α \hat{k}$ and $c = - α \hat{i} - 2 \hat{j} + \hat{k}$ , where $α$ and $β$ are integers. If $a \cdot b = - 1$ and $b \cdot c = 10$ , then $(a \times b) \cdot c$ is equal to

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

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to

Let $a$ and $b$ be two vectors. Let $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ , then the value of $b \cdot c$ is

Let $c$ be a vector perpendicular to the vectors $a = \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ . If $c \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 8$ then the value of $c \cdot (a \times b)$ is equal to

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

Select Answer:

Visualized Solution

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

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

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

Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅d=1

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

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

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

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

Scalar Triple Product Setup

Evaluating the Determinant

Final Answer

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Bridge of the Unit Vector

The Algebraic Unfolding

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^−j^​, b=j^​−k^, c=k^−i^. If d is a unit vector such that a⋅d=0=[bcd], then d equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=αi^+βj^​+3k^, b=βi^−αj^​−k^ and c=i^−2j^​−k^ such that a⋅b=1 and b⋅c=−3, then 31​((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^−αj^​+βk^,b=3i^+βj^​−αk^ and c=−αi^−2j^​+k^, where α and β are integers. If a⋅b=−1 and b⋅c=10, 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^+2j^​+k^, b=3(i^−j^​+k^). Let c be the vector such that a×c=b and a⋅c=3. Then a⋅((c×b)−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 and b be two vectors. Let ∣a∣=1,∣b∣=4 and a⋅b=2. If c=(2a×b)−3b, then the value of b⋅c is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let c be a vector perpendicular to the vectors a=i^+j^​−k^ and b=i^+2j^​+k^. If c⋅(i^+j^​+3k^)=8 then the value of c⋅(a×b) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=2i^+j^​+3k^,b=3i^+3j^​+k^ and c=c1​i^+c2​j^​+c3​k^ are coplanar vectors and a⋅c=5,b⊥c, then 122(c1​+c2​+c3​) is equal to

Given Vectors $a, b, c$

Vector Sum $b + c$

Unit Vector $d$

Condition $a \cdot d = 1$

Evaluating $a \cdot (b + c)$

Magnitude $∣ b + c ∣$

Solving for $x$

Value of $x$

Let $a = \hat{i} - \hat{j}$ , $b = \hat{j} - \hat{k}$ , $c = \hat{k} - \hat{i}$ . If $d$ is a unit vector such that $a \cdot d = 0 = [b c d]$ , then $d$ equals

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

Let $a = \hat{i} - α \hat{j} + β \hat{k}, b = 3 \hat{i} + β \hat{j} - α \hat{k}$ and $c = - α \hat{i} - 2 \hat{j} + \hat{k}$ , where $α$ and $β$ are integers. If $a \cdot b = - 1$ and $b \cdot c = 10$ , then $(a \times b) \cdot c$ is equal to

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

Let $a$ and $b$ be two vectors. Let $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ , then the value of $b \cdot c$ is

Let $c$ be a vector perpendicular to the vectors $a = \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ . If $c \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 8$ then the value of $c \cdot (a \times b)$ is equal to

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to