Let $\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{q} = \hat{i} - \hat{j} + \hat{k}$. If for some real numbers $\alpha, \beta$, and $\gamma$, we have $15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha (2\vec{p} + \vec{q}) + \beta (\vec{p} - 2\vec{q}) + \gamma (\vec{p} \times \vec{q})$, then the value of $\gamma$ is ________.

### Analyzing the Setup We are given two vectors: $\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}$ $\vec{q} = \hat{i} - \hat{j} + \hat{k}$ The master equation is defined as: $$15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha (2\vec{p} + \vec{q}) + \beta (\vec{p} - 2\vec{q}) + \gamma (\vec{p} \times \vec{q})$$ Our objective is to determine the value of the scalar $\gamma$. ### Phase 1: The Pre-flight Check First, we compute the linear combinations of the vectors: $2\vec{p} + \vec{q} = 2(2\hat{i} + \hat{j} + 3\vec{k}) + (\hat{i} - \hat{j} + \hat{k}) = 5\hat{i} + \hat{j} + 7\hat{k}$ $\vec{p} - 2\vec{q} = (2\hat{i} + \hat{j} + 3\hat{k}) - 2(\hat{i} - \hat{j} + \hat{k}) = 0\hat{i} + 3\hat{j} + \hat{k}$ Next, we calculate the cross product $\vec{p} \times \vec{q}$ using the determinant method: $$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 + 3) - \hat{j}(2 - 3) + \hat{k}(-2 - 1) = 4\hat{i} + \hat{j} - 3\hat{k}$$ ### Phase 2: The Master Equation and the System Substituting these results into the master equation: $$15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha (5\hat{i} + \hat{j} + 7\hat{k}) + \beta (3\hat{j} + \hat{k}) + \gamma (4\hat{i} + \hat{j} - 3\hat{k})$$ By equating the components of $\hat{i}, \hat{j},$ and $\hat{k}$, we obtain the following system of linear equations: (1) $5\alpha + 4\gamma = 15$ (2) $\alpha + 3\beta + \gamma = 10$ (3) $7\alpha + \beta - 3\gamma = 6$ ### Phase 3: The Surgical Strike To isolate $\gamma$, we eliminate $\beta$ by manipulating equations (2) and (3). Multiply equation (3) by $3$: $$21\alpha + 3\beta - 9\gamma = 18$$ Subtract equation (2) from this result: $$(21\alpha + 3\beta - 9\gamma) - (\alpha + 3\beta + \gamma) = 18 - 10$$ $$20\alpha - 10\gamma = 8 \Rightarrow 5\alpha - 2.5\gamma = 2$$ Now, we solve the system consisting of equation (1) and our new derived equation: $5\alpha + 4\gamma = 15$ $5\alpha - 2.5\gamma = 2$ Subtracting the second from the first: $$(5\alpha + 4\gamma) - (5\alpha - 2.5\gamma) = 15 - 2$$ $$6.5\gamma = 13$$ Solving for $\gamma$: $$\gamma = \frac{13}{6.5} = 2$$ The final value is **$\gamma = 2$**.

Wait, why do we need to compute the cross product $\vec{p} \times \vec{q}$? Can't we just use dot products?

Great question! While dot products are excellent for finding angles or projections, the cross product is essential here because it generates a vector perpendicular to both $\vec{p}$ and $\vec{q}$. Since the master equation is a linear combination of $\vec{p}$, $\vec{q}$, and their cross product, we are essentially working in a basis formed by these three vectors. The cross product is a required component of this basis.

Is there a faster way to find $\gamma$ without solving the full system of equations?

Yes, there is! You could take the dot product of the entire master equation with a vector perpendicular to both $(2\vec{p} + \vec{q})$ and $(\vec{p} - 2\vec{q})$. This would eliminate $\alpha$ and $\beta$ in one step. However, for most students, the systematic approach of setting up the linear system is less prone to calculation errors and is highly reliable in a high-pressure exam environment.

I keep getting a different value for $\gamma$. Where could I be going wrong?

The most common trap is a sign error in the cross product calculation. Remember that when expanding the determinant $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix}$, the $\hat{j}$ component must be multiplied by a negative sign. Double-check your determinant expansion: $\hat{i}(1 - (-3)) - \hat{j}(2 - 3) + \hat{k}(-2 - 1)$. If your signs are correct, re-verify your arithmetic in the system of equations.

JEE(ADVANCED)-202

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: Let $p = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and $q = \hat{i} - \hat{j} + \hat{k}$ . If for some real numbers $α, β$ , and $γ$ , we have $15 \hat{i} + 10 \hat{j} + 6 \hat{k} = α (2 p + q) + β (p - 2 q) + γ (p \times q)$ , then the value of $γ$ is ________.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the problem as a system of linear equations. You are given a target vector and a linear combination of three other vectors. Instead of solving for all variables, focus on the structure of the equation and identify which variables are truly necessary to find the value of

γ

Hint 2

Hint 2 (The Arsenal): Use the property that if two vectors are equal, their corresponding components must be equal. This will allow you to set up a system of three linear equations. Look for ways to eliminate

α

and

β

strategically to isolate

γ

without solving the entire system.

Hint 3

Hint 3 (The Execution): Start by setting up the system of equations:

5 α + 4 γ = 15

α + 3 β + γ = 10

, and

7 α + β - 3 γ = 6

. Notice that the first equation already links

α

and

γ

. Use the other two equations to eliminate

β

and create a second equation involving only

α

and

γ

Visualized Solution

Given Vectors & Equation

Given vectors:
$p = 2 \hat{i} + \hat{j} + 3 \hat{k}$
$q = \hat{i} - \hat{j} + \hat{k}$
Master Equation:
$15 \hat{i} + 10 \hat{j} + 6 \hat{k} = α (2 p + q) + β (p - 2 q) + γ (p \times q)$

Compute $2 p + q$

First, evaluate the term $(2 p + q)$ :
$2 p + q = 2 (2 \hat{i} + \hat{j} + 3 \hat{k}) + (\hat{i} - \hat{j} + \hat{k})$
$= (4 \hat{i} + 2 \hat{j} + 6 \hat{k}) + (\hat{i} - \hat{j} + \hat{k})$
$= 5 \hat{i} + \hat{j} + 7 \hat{k}$

Compute $p - 2 q$

Next, evaluate the term $(p - 2 q)$ :
$p - 2 q = (2 \hat{i} + \hat{j} + 3 \hat{k}) - 2 (\hat{i} - \hat{j} + \hat{k})$
$= (2 \hat{i} + \hat{j} + 3 \hat{k}) - (2 \hat{i} - 2 \hat{j} + 2 \hat{k})$
$= 0 \hat{i} + 3 \hat{j} + \hat{k}$

Setup & Evaluate $p \times q$

Set up the cross product $(p \times q)$ using a determinant:
$p \times q = \hat{i} 21 \hat{j} 1 - 1 \hat{k} 31$
Expanding along the first row:
$= \hat{i} (1 - (- 3)) - \hat{j} (2 - 3) + \hat{k} (- 2 - 1)$
$= 4 \hat{i} + \hat{j} - 3 \hat{k}$

Substitute into Master Equation

Substitute the evaluated terms back into the master equation:
$15 \hat{i} + 10 \hat{j} + 6 \hat{k} = α (5 \hat{i} + \hat{j} + 7 \hat{k}) + β (3 \hat{j} + \hat{k}) + γ (4 \hat{i} + \hat{j} - 3 \hat{k})$

Group Components

Group the coefficients of $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ on the right side:
$15 \hat{i} + 10 \hat{j} + 6 \hat{k} = (5 α + 4 γ) \hat{i} + (α + 3 β + γ) \hat{j} + (7 α + β - 3 γ) \hat{k}$

Equate Coefficients

Equating corresponding components from both sides yields a system of 3 equations:
Eq 1: $5 α + 4 γ = 15$
Eq 2: $α + 3 β + γ = 10$
Eq 3: $7 α + β - 3 γ = 6$

Strategy to find $γ$

Goal: Find $γ$ .
Notice that Eq 1 only contains $α$ and $γ$ .
Strategy: Eliminate $β$ using Eq 2 and Eq 3 to get another equation in terms of $α$ and $γ$ .

Eliminate $β$

Multiply Eq 3 by $3$ to match the $β$ coefficient in Eq 2:
$3 \times (7 α + β - 3 γ) = 3 \times 6$
Eq 3': $21 α + 3 β - 9 γ = 18$
Subtract Eq 2 from Eq 3':
$(21 α + 3 β - 9 γ) - (α + 3 β + γ) = 18 - 10$
$20 α - 10 γ = 8$

Solve for $γ$

Divide the new equation by $4$ to match the $α$ coefficient in Eq 1:
$\frac{20 α - 10 γ}{4} = \frac{8}{4} ⟹ 5 α - \frac{5}{2} γ = 2$ (Eq 4)
Subtract Eq 4 from Eq 1 ( $5 α + 4 γ = 15$ ):
$(5 α + 4 γ) - (5 α - \frac{5}{2} γ) = 15 - 2$
$4 γ + \frac{5}{2} γ = 13$
$\frac{13}{2} γ = 13 ⟹ γ = 2$

00:00 / 00:00

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

We are given two vectors:

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

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

The master equation is defined as:

15 \hat{i} + 10 \hat{j} + 6 \hat{k} = α (2 p + q) + β (p - 2 q) + γ (p \times q)

Our objective is to determine the value of the scalar

γ

Phase 1

The Pre-flight Check

First, we compute the linear combinations of the vectors:

2 p + q = 2 (2 \hat{i} + \hat{j} + 3 k) + (\hat{i} - \hat{j} + \hat{k}) = 5 \hat{i} + \hat{j} + 7 \hat{k}

p - 2 q = (2 \hat{i} + \hat{j} + 3 \hat{k}) - 2 (\hat{i} - \hat{j} + \hat{k}) = 0 \hat{i} + 3 \hat{j} + \hat{k}

Next, we calculate the cross product

p \times q

using the determinant method:

p \times q = \hat{i} 21 \hat{j} 1 - 1 \hat{k} 31 = \hat{i} (1 + 3) - \hat{j} (2 - 3) + \hat{k} (- 2 - 1) = 4 \hat{i} + \hat{j} - 3 \hat{k}

Phase 2

The Master Equation and the System

Substituting these results into the master equation:

15 \hat{i} + 10 \hat{j} + 6 \hat{k} = α (5 \hat{i} + \hat{j} + 7 \hat{k}) + β (3 \hat{j} + \hat{k}) + γ (4 \hat{i} + \hat{j} - 3 \hat{k})

By equating the components of

\hat{i}, \hat{j},

and

\hat{k}

, we obtain the following system of linear equations: (1)

5 α + 4 γ = 15

(2)

α + 3 β + γ = 10

(3)

7 α + β - 3 γ = 6

Phase 3

The Surgical Strike

To isolate

γ

, we eliminate

β

by manipulating equations (2) and (3). Multiply equation (3) by

3

21 α + 3 β - 9 γ = 18

Subtract equation (2) from this result:

(21 α + 3 β - 9 γ) - (α + 3 β + γ) = 18 - 10

20 α - 10 γ = 8 \Rightarrow 5 α - 2.5 γ = 2

Now, we solve the system consisting of equation (1) and our new derived equation:

5 α + 4 γ = 15

5 α - 2.5 γ = 2

Subtracting the second from the first:

(5 α + 4 γ) - (5 α - 2.5 γ) = 15 - 2

6.5 γ = 13

Solving for

γ

γ = \frac{13}{6.5} = 2

The final value is $γ = 2$ .

Common Traps & FAQs

Wait, why do we need to compute the cross product

p \times q

? Can't we just use dot products?

Great question! While dot products are excellent for finding angles or projections, the cross product is essential here because it generates a vector perpendicular to both

p

and

q

. Since the master equation is a linear combination of

p

q

, and their cross product, we are essentially working in a basis formed by these three vectors. The cross product is a required component of this basis.

Is there a faster way to find

γ

without solving the full system of equations?

Yes, there is! You could take the dot product of the entire master equation with a vector perpendicular to both

(2 p + q)

and

(p - 2 q)

. This would eliminate

α

and

β

in one step. However, for most students, the systematic approach of setting up the linear system is less prone to calculation errors and is highly reliable in a high-pressure exam environment.

I keep getting a different value for

γ

. Where could I be going wrong?

The most common trap is a sign error in the cross product calculation. Remember that when expanding the determinant

\hat{i} 21 \hat{j} 1 - 1 \hat{k} 31

, the

\hat{j}

component must be multiplied by a negative sign. Double-check your determinant expansion:

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

. If your signs are correct, re-verify your arithmetic in the system of equations.

Enter Numerical Value:

Visualized Solution

Given Vectors & Equation

Compute $2 p + q$

Compute $p - 2 q$

Setup & Evaluate $p \times q$

Substitute into Master Equation

Group Components

Equate Coefficients

Strategy to find $γ$

Eliminate $β$

Solve for $γ$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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 $a = \hat{i} - 3 \hat{j} + 7 \hat{k}$ , $b = 2 \hat{i} - \hat{j} + \hat{k}$ and $c$ be a vector such that $(a + 2 b) \times c = 3 (c \times a)$ . If $a \cdot c = 130$ , then $b \cdot c$ is equal to _______

Let $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $b = 7 \hat{i} + \hat{j} - 6 \hat{k}$ . If $r \times a = r \times b$ , $r \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = - 3$ , then $r \cdot (2 \hat{i} - 3 \hat{j} + \hat{k})$ is equal to

Let $α = 3 \hat{i} + \hat{j}$ and $β = 2 \hat{i} - \hat{j} + 3 \hat{k}$ . If $β = β_{1} - β_{2}$ , where $β_{1}$ is parallel to $α$ and $β_{2}$ is perpendicular to $α$ , then $β_{1} \times β_{2}$ is equal to

Let $a = 2 \hat{i} - 3 \hat{j} + \hat{k}, b = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$ and a vector $c$ be such that $(a - c) \times b = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}$ and $a \cdot c = 3$ . If $b \times c = d$ , then $∣ a \cdot d ∣$ is equal to :

Given that $a = (1, 1, 1)$ , $c = (0, 1, - 1)$ , $a \cdot b = 3$ and $a \times b = c$ , then $b = .........$

If $a = \hat{i} + 2 \hat{k}, b = \hat{i} + \hat{j} + \hat{k}, c = 7 \hat{i} - 3 \hat{j} + 4 \hat{k}$ , $r \times b + b \times c = 0$ and $r \cdot a = 0$ then $r \cdot c$ is equal to

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

Enter Numerical Value:

Visualized Solution

Given Vectors & Equation

Compute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2p​+q​

Compute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​−2q​

Setup & Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​×q​

Substitute into Master Equation

Group Components

Equate Coefficients

Strategy to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ

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

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

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

.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^−3j^​+7k^, b=2i^−j^​+k^ and c be a vector such that (a+2b)×c=3(c×a). If a⋅c=130, then 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=2i^−3j^​+4k^ and b=7i^+j^​−6k^. If r×a=r×b, r⋅(i^+2j^​+k^)=−3, then r⋅(2i^−3j^​+k^) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^−3j^​+k^,b=3i^+2j^​+5k^ and a vector c be such that (a−c)×b=−18i^−3j^​+12k^ and a⋅c=3. If b×c=d, then ∣a⋅d∣ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given that a=(1,1,1), c=(0,1,−1), a⋅b=3 and a×b=c, then b=.........

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+2j^​+λk^, b=3i^−5j^​−λk^, a⋅c=7, 2b⋅c+43=0, a×c=b×c. Then ∣a⋅b∣ is equal to

Compute $2 p + q$

Compute $p - 2 q$

Setup & Evaluate $p \times q$

Strategy to find $γ$

Eliminate $β$

Solve for $γ$

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 $a = \hat{i} - 3 \hat{j} + 7 \hat{k}$ , $b = 2 \hat{i} - \hat{j} + \hat{k}$ and $c$ be a vector such that $(a + 2 b) \times c = 3 (c \times a)$ . If $a \cdot c = 130$ , then $b \cdot c$ is equal to _______

Let $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $b = 7 \hat{i} + \hat{j} - 6 \hat{k}$ . If $r \times a = r \times b$ , $r \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = - 3$ , then $r \cdot (2 \hat{i} - 3 \hat{j} + \hat{k})$ is equal to

Let $a = 2 \hat{i} - 3 \hat{j} + \hat{k}, b = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$ and a vector $c$ be such that $(a - c) \times b = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}$ and $a \cdot c = 3$ . If $b \times c = d$ , then $∣ a \cdot d ∣$ is equal to :

Given that $a = (1, 1, 1)$ , $c = (0, 1, - 1)$ , $a \cdot b = 3$ and $a \times b = c$ , then $b = .........$

If $a = \hat{i} + 2 \hat{k}, b = \hat{i} + \hat{j} + \hat{k}, c = 7 \hat{i} - 3 \hat{j} + 4 \hat{k}$ , $r \times b + b \times c = 0$ and $r \cdot a = 0$ then $r \cdot c$ is equal to

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