Let a = 2i - 5j + 5k and b = i - j + 3k. If c is a vector such that 2(a × c) + 3(b × c) = 0 and (a -

Q: Why can we just factor out $\vec{c}$ in the cross product equation? Is the cross product distributive?

Yes, absolutely! The cross product is distributive over vector addition. That is, $\vec{u} \times (\vec{v} + \vec{w}) = (\vec{u} \times \vec{v}) + (\vec{u} \times \vec{w})$. In our case, we are essentially reversing this: $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = (2\vec{a} + 3\vec{b}) \times \vec{c} = \vec{0}$. It is a powerful tool that simplifies complex expressions into manageable ones.

Q: How do I know for sure that $\vec{c}$ is parallel to $(2\vec{a} + 3\vec{b})$?

Great question! The cross product of two vectors $\vec{u}$ and $\vec{v}$ is the zero vector if and only if the vectors are parallel (or one of them is the zero vector). Since $(2\vec{a} + 3\vec{b}) \times \vec{c} = \vec{0}$, it forces the vector $\vec{c}$ to lie along the same line as the vector $(2\vec{a} + 3\vec{b})$. This is the key geometric insight that unlocks the problem.

Q: What if the dot product condition $(\vec{a} - \vec{b}) \cdot \vec{c} = -97$ had resulted in $\lambda = 0$? What would that mean?

If $\lambda$ were $0$, then $\vec{c}$ would be the zero vector. However, in this problem, the dot product condition is a constraint that fixes the magnitude and direction of $\vec{c}$. If the dot product were zero, it would imply that $\vec{c}$ is orthogonal to $(\vec{a} - \vec{b})$. Always check your scalar value; it is the bridge between the direction we found and the specific vector we need.

Question

Let $\vec{a} = 2\hat{i} - 5\hat{j} + 5\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 3\hat{k}$. If $\vec{c}$ is a vector such that $2(\vec{a} 	imes \vec{c}) + 3(\vec{b} 	imes \vec{c}) = \vec{0}$ and $(\vec{a} - \vec{b}) \cdot \vec{c} = -97$, then $|\vec{c} 	imes \hat{k}|^2$ is equal to

Options:

(A) 233

(B) 218

(C) 193

(D) 205

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors, $\vec{a} = 2\hat{i} - 5\hat{j} + 5\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 3\hat{k}$. We seek the squared magnitude of $\vec{c} 	imes \hat{k}$, where $\vec{c}$ satisfies two specific conditions.

### The Power of Distributivity

Our first clue is the equation $2(\vec{a} 	imes \vec{c}) + 3(\vec{b} 	imes \vec{c}) = \vec{0}$. Because the cross product is distributive, we can factor $\vec{c}$ out of the expression.

This allows us to rewrite the equation as:
$$(2\vec{a} + 3\vec{b}) 	imes \vec{c} = \vec{0}$$

### The Geometric Bridge

When the cross product of two vectors is the zero vector, it implies that the vectors are parallel. Therefore, our mystery vector $\vec{c}$ must be parallel to the vector $(2\vec{a} + 3\vec{b})$.

We can express $\vec{c}$ as a scalar multiple of this resultant vector:
$$\vec{c} = \lambda(2\vec{a} + 3\vec{b})$$

### The Scalar Hunt

First, let us calculate the vector $(2\vec{a} + 3\vec{b})$:
$$2\vec{a} = 4\hat{i} - 10\hat{j} + 10\hat{k}$$
$$3\vec{b} = 3\hat{i} - 3\hat{j} + 9\hat{k}$$
$$2\vec{a} + 3\vec{b} = 7\hat{i} - 13\hat{j} + 19\hat{k}$$

Thus, we have $\vec{c} = \lambda(7\hat{i} - 13\hat{j} + 19\hat{k})$. We now use the second clue, $(\vec{a} - \vec{b}) \cdot \vec{c} = -97$, to solve for $\lambda$.

### The Dot Product Constraint

First, we find $\vec{a} - \vec{b}$:
$$\vec{a} - \vec{b} = (2-1)\hat{i} + (-5 - (-1))\hat{j} + (5-3)\hat{k} = \hat{i} - 4\hat{j} + 2\hat{k}$$

Substituting this and our expression for $\vec{c}$ into the dot product equation:
$$(\hat{i} - 4\hat{j} + 2\hat{k}) \cdot \lambda(7\hat{i} - 13\hat{j} + 19\hat{k}) = -97$$

Performing the dot product calculation:
$$\lambda [ (1)(7) + (-4)(-13) + (2)(19) ] = -97$$
$$\lambda (7 + 52 + 38) = -97$$
$$97\lambda = -97 \implies \lambda = -1$$

With $\lambda = -1$, we identify the vector $\vec{c}$:
$$\vec{c} = -1(7\hat{i} - 13\hat{j} + 19\hat{k}) = -7\hat{i} + 13\hat{j} - 19\hat{k}$$

### Final Calculation

We now compute the cross product $\vec{c} 	imes \hat{k}$:
$$\vec{c} 	imes \hat{k} = (-7\hat{i} + 13\hat{j} - 19\hat{k}) 	imes \hat{k}$$
$$\vec{c} 	imes \hat{k} = -7(\hat{i} 	imes \hat{k}) + 13(\hat{j} 	imes \hat{k}) - 19(\hat{k} 	imes \hat{k})$$
$$\vec{c} 	imes \hat{k} = -7(-\hat{j}) + 13(\hat{i}) - 0 = 13\hat{i} + 7\hat{j}$$

Finally, the squared magnitude is the sum of the squares of the components:
$$|\vec{c} 	imes \hat{k}|^2 = 13^2 + 7^2 = 169 + 49 = \mathbf{218}$$

Animated Solution for Mathematics - Vector Algebra: Let $a = 2 \hat{i} - 5 \hat{j} + 5 \hat{k}$ and $b = \hat{i} - \hat{j} + 3 \hat{k}$ . If $c$ is a vector such that $2 (a \times c) + 3 (b \times c) = 0$ and $(a - b) \cdot c = - 97$ , then $∣ c \times \hat{k} ∣^{2}$ is equal to

Select Answer:

Visualized Solution

$Analyze the Given Vectors$

$Simplify the Cross Product Equation$

$Parallel Vectors Concept$

$Calculate 2 a + 3 b$

$Express c with a Scalar Parameter$

$Calculate a - b$

$Apply the Dot Product Condition$

$Solve for λ$

$Determine Vector c$

$Calculate c \times \hat{k}$

$Find the Squared Magnitude$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of Distributivity

The Geometric Bridge

The Scalar Hunt

The Dot Product Constraint

Final Calculation

Similar Questions

Let $λ \in Z$ , $a = λ \hat{i} + \hat{j} - \hat{k}$ and $b = 3 \hat{i} - \hat{j} + 2 \hat{k}$ . Let $c$ be a vector such that $(a + b + c) \times c = 0$ , $a \cdot c = - 17$ and $b \cdot c = - 20$ . Then $∣ c \times (λ \hat{i} + \hat{j} + \hat{k}) ∣^{2}$ is equal to

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = 5 \hat{i} - 3 \hat{j} + 3 \hat{k}$ be there vectors. If $r$ is a vector such that, $r \times b = c \times b$ and $r \cdot a = 0$ . Then $25∣ r ∣^{2}$ 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 :

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 $a = \hat{j} - \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ . Then the vector $b$ satisfying $a \times b + c = 0$ and $a \cdot b = 3$

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 $a = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}, b = \hat{i} + \hat{k}$ and $c = \hat{i} + 2 \hat{j} - 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ and $r \cdot b = 0$ , then $∣ r ∣$ 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 $a = \hat{i} + 4 \hat{j} + 2 \hat{k}$ , $b = 3 \hat{i} - \hat{j} + 7 \hat{k}$ 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

Let $a = 4 \hat{i} - \hat{j} + \hat{k}, b = 11 \hat{i} - \hat{j} + \hat{k}$ and $c$ be a vector such that $(a + b) \times c = c \times (- 2 a + 3 b)$ . If $(2 a + 3 b) \cdot c = 1670$ , then $∣ c ∣^{2}$ is equal to :

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyze the Given Vectors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplify the Cross Product Equation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculate 2a+3b

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Express c with a Scalar Parameter

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Apply the Dot Product Condition

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculate c×k^

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the Squared Magnitude