Let a = 6i + 9j + 12k, b = αi + 11j - 2k and c be vectors such that a × c = a × b. If a · c = -12, a

Q: Wait, if $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$, does that mean $\vec{c} = \vec{b}$?

That is the most common trap in vector algebra! Remember, the cross product only captures the component of a vector perpendicular to the reference vector. If $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$, it only implies that the components of $\vec{c}$ and $\vec{b}$ perpendicular to $\vec{a}$ are identical. The parallel components can differ by any scalar multiple of $\vec{a}$. That is why we introduce the parameter $\lambda$!

Q: Why did the $\lambda$ terms cancel out in the second dot product calculation?

That wasn't a coincidence; it was the beauty of the problem design! When we substituted $\vec{c} = \vec{b} + \lambda \vec{a}$ into the dot product $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$, the coefficient of $\lambda$ became the dot product of $\vec{a}$ and the vector $(\hat{i} - 2\hat{j} + \hat{k})$. Since $\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$, calculating $(6)(1) + (9)(-2) + (12)(1)$ gives $6 - 18 + 12 = 0$. The math is telling us that $\vec{a}$ is perpendicular to that specific vector, which is why $\lambda$ vanished!

Q: Is there a geometric way to solve this without components?

Absolutely! You could project $\vec{c}$ onto $\vec{a}$ and use the perpendicular components. However, in JEE Advanced, when you have unknown components like $\alpha$, the algebraic approach using the parametric form $\vec{c} = \vec{b} + \lambda \vec{a}$ is usually the most robust and least prone to error. Always choose the path that minimizes the chance of a sign error!

Question

Let $\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$, $\vec{b} = \alpha\hat{i} + 11\hat{j} - 2\hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} 	imes \vec{c} = \vec{a} 	imes \vec{b}$. If $\vec{a} \cdot \vec{c} = -12$, and $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$ then $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given vectors $\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$ and $\vec{b} = \alpha\hat{i} + 11\hat{j} - 2\hat{k}$, along with a vector $\vec{c}$. The governing condition is $\vec{a} 	imes \vec{c} = \vec{a} 	imes \vec{b}$.

### The Cross Product Trap

Many students instinctively want to cancel $\vec{a}$ from both sides. However, we must treat this as a vector equation: $\vec{a} 	imes \vec{c} - \vec{a} 	imes \vec{b} = \vec{0}$.

By the distributive property of the cross product, this simplifies to:
$$\vec{a} 	imes (\vec{c} - \vec{b}) = \vec{0}$$

This implies that the vector $(\vec{c} - \vec{b})$ is parallel to $\vec{a}$. Mathematically, we express this as:
$$\vec{c} - \vec{b} = \lambda \vec{a} \implies \vec{c} = \vec{b} + \lambda \vec{a}$$

### The Parametric Bridge

By defining $\vec{c} = \vec{b} + \lambda \vec{a}$, we transform the problem into one involving two scalars: $\alpha$ and $\lambda$. Substituting the components of $\vec{a}$ and $\vec{b}$, we obtain:
$$\vec{c} = (\alpha + 6\lambda)\hat{i} + (11 + 9\lambda)\hat{j} + (-2 + 12\lambda)\hat{k}$$

### The Dance of Dot Products

First, we use the constraint $\vec{a} \cdot \vec{c} = -12$. Expanding this, we get $\vec{a} \cdot \vec{b} + \lambda |\vec{a}|^2 = -12$.

Calculating the components:
$$\vec{a} \cdot \vec{b} = (6)(\alpha) + (9)(11) + (12)(-2) = 6\alpha + 75$$
$$|\vec{a}|^2 = 6^2 + 9^2 + 12^2 = 36 + 81 + 144 = 261$$

Substituting these into our equation:
$$6\alpha + 75 + 261\lambda = -12 \implies 6\alpha + 261\lambda = -87 \implies 2\alpha + 87\lambda = -29$$

Next, we use the second constraint: $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$. Substituting the component form of $\vec{c}$:
$$(\alpha + 6\lambda)(1) + (11 + 9\lambda)(-2) + (-2 + 12\lambda)(1) = 5$$

Simplifying the expression:
$$\alpha + 6\lambda - 22 - 18\lambda - 2 + 12\lambda = 5$$
$$\alpha + (6 - 18 + 12)\lambda - 24 = 5$$
$$\alpha - 24 = 5 \implies \alpha = 29$$

### Final Calculation

With $\alpha = 29$, we substitute back into $2\alpha + 87\lambda = -29$:
$$2(29) + 87\lambda = -29 \implies 58 + 87\lambda = -29 \implies 87\lambda = -87 \implies \lambda = -1$$

Now we determine $\vec{c}$:
$$\vec{c} = (29 - 6)\hat{i} + (11 - 9)\hat{j} + (-2 - 12)\hat{k} = 23\hat{i} + 2\hat{j} - 14\hat{k}$$

The final requirement is to calculate $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})$:
$$23 + 2 - 14 = 11$$

The final answer is **11**.

Enter Numerical Value:

Visualized Solution

Given Vectors $a$ and $b$

The Cross Product Condition

Geometric Interpretation

Component Form of $c$

First Dot Product Condition

Calculating $a \cdot b$ and $∣ a ∣^{2}$

Forming Equation 1

Second Dot Product Condition

Solving for $α$

Solving for $λ$

Finding Vector $c$

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Trap

The Parametric Bridge

The Dance of Dot Products

Final Calculation

Similar Questions

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

Let $a = 2 \hat{i} + 5 \hat{j} - \hat{k}$ , $b = 2 \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $c$ be three vectors such that $(c + \hat{i}) \times (a + b + \hat{i}) = a \times (c + \hat{i})$ . If $a \cdot c = - 29$ , then $c \cdot (- 2 \hat{i} + \hat{j} + \hat{k})$ 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 :

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

Let $a = - \hat{i} - \hat{k}$ , $b = - \hat{i} + \hat{j}$ and $c = \hat{i} + 2 \hat{j} + 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times b = c \times b$ and $r \cdot a = 0$ , then the value of $r \cdot b$ is .........

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

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 $a = 3 \hat{i} + \hat{j} - 2 \hat{k}$ , $b = 4 \hat{i} + \hat{j} + 7 \hat{k}$ and $c = \hat{i} - 3 \hat{j} + 4 \hat{k}$ be three vectors. If a vector $p$ satisfies $p \times b = c \times b$ and $p \cdot a = 0$ , then $p \cdot (\hat{i} - \hat{j} - \hat{k})$ is equal to

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to

Enter Numerical Value:

Visualized Solution

Given 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

The Cross Product Condition

Geometric Interpretation

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

First Dot Product Condition

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

Forming Equation 1

Second Dot Product Condition

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

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

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

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Trap

The Parametric Bridge

The Dance of Dot Products

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^​−k^,b=i^−j^​ and c=i^−j^​−k^ be three given vectors. If r is a vector such that r×a=c×a and r⋅b=0, then r⋅a is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+5j^​−k^, b=2i^−2j^​+2k^ and c be three vectors such that (c+i^)×(a+b+i^)=a×(c+i^). If a⋅c=−29, then c⋅(−2i^+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=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; } Let a=i^−2j^​+k^, b=i^−j^​+k^ be two vectors. If c is a vector such that b×c=b×a and c⋅a=0 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=−i^−k^, b=−i^+j^​ and c=i^+2j^​+3k^ be three given vectors. If r is a vector such that r×b=c×b and r⋅a=0, then the value of r⋅b is .........

.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

.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^+j^​−2k^,b=i^+j^​ and c=a×b. Let d be a vector such that ∣d−a∣=11​,∣c×d∣=3 and the angle between c and d is 4π​. Then a⋅d is equal to

Given Vectors $a$ and $b$

Component Form of $c$

Calculating $a \cdot b$ and $∣ a ∣^{2}$

Solving for $α$

Solving for $λ$

Finding Vector $c$

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

Let $a = 2 \hat{i} + 5 \hat{j} - \hat{k}$ , $b = 2 \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $c$ be three vectors such that $(c + \hat{i}) \times (a + b + \hat{i}) = a \times (c + \hat{i})$ . If $a \cdot c = - 29$ , then $c \cdot (- 2 \hat{i} + \hat{j} + \hat{k})$ 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 :

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

Let $a = - \hat{i} - \hat{k}$ , $b = - \hat{i} + \hat{j}$ and $c = \hat{i} + 2 \hat{j} + 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times b = c \times b$ and $r \cdot a = 0$ , then the value of $r \cdot b$ is .........

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

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 $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to