Let a = 3i + j - 2k, b = 4i + j + 7k and c = i - 3j + 4k be three vectors. If a vector p satisfies p

Q: Wait, why can't I just assume $\vec{p} = \vec{c}$? The equation $\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$ is satisfied if $\vec{p} = \vec{c}$, right?

You are absolutely right that $\vec{p} = \vec{c}$ is a solution! However, it is not the *only* solution. The cross product equation defines a line in 3D space, not a single point. By assuming $\vec{p} = \vec{c}$, you are missing the entire family of vectors that satisfy the condition. We need the general form $\vec{p} = \vec{c} + \lambda \vec{b}$ to capture every possibility, allowing the orthogonality condition to pick the specific vector we need.

Q: Why does $\vec{X} \times \vec{Y} = \vec{0}$ imply $\vec{X} = \lambda \vec{Y}$? I always forget this property.

Think about the magnitude of the cross product: $|\vec{X} \times \vec{Y}| = |\vec{X}||\vec{Y}| \sin(\theta)$. For this to be zero, either the magnitude of one vector is zero, or $\sin(\theta)$ must be zero. If $\sin(\theta) = 0$, then $\theta$ is $0^\circ$ or $180^\circ$, meaning the vectors are parallel. Thus, one must be a scalar multiple of the other! It is a fundamental property of vector geometry.

Q: Is there a faster way than finding $\vec{p}$ explicitly? This feels like a lot of arithmetic.

In some problems, you can use properties of the scalar triple product, but here, finding $\vec{p}$ explicitly is the most robust and error-proof method. JEE Advanced problems often test your patience and precision with arithmetic. Embrace the calculation; it is the final gatekeeper between you and the correct answer!

Question

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

Options:

(A) 24

(B) 36

(C) 28

(D) 32

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a quest to uncover the identity of a mysterious vector, $\vec{p}$.

We are given three known vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, and we are told that $\vec{p}$ is bound by two distinct laws. This is the beauty of vector algebra: it is the language of constraints.

### The Geometry of the Cross Product

Our first clue is the equation $\vec{p} 	imes \vec{b} = \vec{c} 	imes \vec{b}$. Let us bring everything to one side:
$$\vec{p} 	imes \vec{b} - \vec{c} 	imes \vec{b} = \vec{0}$$

By the distributive property of the cross product, we can factor out $\vec{b}$ to get:
$$(\vec{p} - \vec{c}) 	imes \vec{b} = \vec{0}$$

In the realm of vectors, if the cross product of two entities is the zero vector, they must be parallel. This means the vector $(\vec{p} - \vec{c})$ is parallel to $\vec{b}$.

Geometrically, this implies that $\vec{p}$ lies on a line passing through the tip of $\vec{c}$ and running parallel to $\vec{b}$. We can express this as:
$$\vec{p} = \vec{c} + \lambda \vec{b}$$

### The Orthogonality Anchor

Now, we have a line of potential vectors, but we need the specific one that satisfies our second condition: $\vec{p} \cdot \vec{a} = 0$. This is the orthogonality condition, which tells us that $\vec{p}$ is perpendicular to $\vec{a}$.

Substituting our general form for $\vec{p}$ into this constraint, we obtain:
$$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$$

Using the distributive property of the dot product, we expand this into:
$$\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$$

### The Calculation

First, let us calculate $\vec{c} \cdot \vec{a}$. Given $\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{c} = \hat{i} - 3\hat{j} + 4\hat{k}$:
$$\vec{c} \cdot \vec{a} = (1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8$$

Next, we calculate $\vec{b} \cdot \vec{a}$ using $\vec{b} = 4\hat{i} + \hat{j} + 7\hat{k}$ and $\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$:
$$\vec{b} \cdot \vec{a} = (4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1$$

Substituting these values back into our linear equation:
$$-8 + \lambda(-1) = 0 \Rightarrow \lambda = -8$$

### The Final Reveal

We reconstruct $\vec{p}$ using $\vec{p} = \vec{c} - 8\vec{b}$:
$$\vec{p} = (\hat{i} - 3\hat{j} + 4\hat{k}) - 8(4\hat{i} + \hat{j} + 7\hat{k})$$

Distributing the $-8$ and combining like terms:
$$\vec{p} = (\hat{i} - 32\hat{i}) + (-3\hat{j} - 8\hat{j}) + (4\hat{k} - 56\hat{k}) = -31\hat{i} - 11\hat{j} - 52\hat{k}$$

Finally, the question asks for the dot product of $\vec{p}$ with the vector $\vec{v} = \hat{i} - \hat{j} - \hat{k}$:
$$\vec{p} \cdot \vec{v} = (-31)(1) + (-11)(-1) + (-52)(-1)$$
$$\vec{p} \cdot \vec{v} = -31 + 11 + 52 = 32$$

We have arrived at the destination. The **final answer is 32**.

Select Answer:

Visualized Solution

The Given Vectors

The Cross Product Condition

Collinearity of Vectors

The Orthogonality Condition

Expanding the Dot Product

Calculate $c \cdot a$

Calculate $b \cdot a$

Solve for $λ$

Construct Vector $p$

Calculate Components of $p$

Final Dot Product

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Geometry of the Cross Product

The Orthogonality Anchor

The Calculation

The Final Reveal

Similar Questions

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Select Answer:

Visualized Solution

The Given Vectors

The Cross Product Condition

Collinearity of Vectors

The Orthogonality Condition

Expanding the Dot Product

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

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

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

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

Calculate Components of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​

Final Dot Product

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Geometry of the Cross Product

The Orthogonality Anchor

The Calculation

The Final Reveal

Similar Questions

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