Let a = i - 2j + k and b = i - j + k be two vectors. If c is a vector such that b×c = b×a and c·a =

Q: Wait, if $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$, why can't I just say $\vec{c} = \vec{a}$?

That is the classic JEE trap! If $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$, then $\vec{b} \times (\vec{c} - \vec{a}) = 0$. This means the vector $(\vec{c} - \vec{a})$ is parallel to $\vec{b}$, not necessarily zero. So, $\vec{c} = \vec{a} + k\vec{b}$ for some scalar $k$. You must solve for $k$ using the other condition, $\vec{c} \cdot \vec{a} = 0$.

Q: Why did we take the cross product with $\vec{a}$? Could we have done something else?

Excellent question. You could have substituted $\vec{c} = \vec{a} + k\vec{b}$ directly into $\vec{c} \cdot \vec{a} = 0$. That is actually a faster method! But the triple product method is a fundamental technique that works even when you don't have a simple linear relationship. Mastering it is essential for JEE Advanced.

Q: Is there a physical interpretation of $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$?

Yes! Geometrically, it means that the projection of $\vec{c}$ onto the plane perpendicular to $\vec{b}$ is the same as the projection of $\vec{a}$ onto that same plane. In other words, $\vec{c}$ and $\vec{a}$ share the same 'shadow' when viewed along the direction of $\vec{b}$.

Question

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

Options:

(A) $\frac{1}{2}$

(B) $\frac{1}{3}$

(C) $-\frac{1}{2}$

(D) $-\frac{1}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors, $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. We are tasked with finding the properties of an unknown vector $\vec{c}$ given the conditions $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{a}$ and $\vec{c} \cdot \vec{a} = 0$.

### The Trap of Intuition

The first instinct might be to assume $\vec{c} = \vec{a}$. However, this is a common trap in vector algebra.

If $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{a}$, it implies $\vec{b} 	imes (\vec{c} - \vec{a}) = 0$. This means the vector $(\vec{c} - \vec{a})$ is parallel to $\vec{b}$, leading to the general form:
$$\vec{c} = \vec{a} + k\vec{b}$$
where $k$ is a scalar constant.

### The Triple Product Weapon

To solve for $\vec{c}$, we take the cross product with $\vec{a}$ on both sides of the equation $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{a}$:
$$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \vec{a} 	imes (\vec{b} 	imes \vec{a})$$

We now invoke the Vector Triple Product identity, known as the "BAC-CAB" rule:
$$\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$$

Applying this to the left-hand side, we get:
$$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$

Since we are given $\vec{c} \cdot \vec{a} = 0$, the first term vanishes. We are left with the simplified expression:
$$-(\vec{a} \cdot \vec{b})\vec{c}$$

### The Calculation

Now, we evaluate the right-hand side, $\vec{a} 	imes (\vec{b} 	imes \vec{a})$, using the same identity:
$$\vec{a} 	imes (\vec{b} 	imes \vec{a}) = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$$

First, we calculate the necessary dot products:
$$\vec{a} \cdot \vec{b} = (1)(1) + (-2)(-1) + (1)(1) = 1 + 2 + 1 = 4$$
$$\vec{a} \cdot \vec{a} = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6$$

Substituting these values back into our equation, we obtain:
$$-4\vec{c} = 6\vec{b} - 4\vec{a}$$

### The Final Stretch

Substituting the components of $\vec{a}$ and $\vec{b}$ into the equation:
$$6\vec{b} - 4\vec{a} = 6(\hat{i} - \hat{j} + \hat{k}) - 4(\hat{i} - 2\hat{j} + \hat{k})$$
$$= (6-4)\hat{i} + (-6+8)\hat{j} + (6-4)\hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

Thus, $-4\vec{c} = 2\hat{i} + 2\hat{j} + 2\hat{k}$, which simplifies to:
$$\vec{c} = -\frac{1}{2}(\hat{i} + \hat{j} + \hat{k})$$

Finally, we calculate the dot product $\vec{c} \cdot \vec{b}$:
$$\vec{c} \cdot \vec{b} = -\frac{1}{2}(\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k})$$
$$\vec{c} \cdot \vec{b} = -\frac{1}{2}( (1)(1) + (1)(-1) + (1)(1) ) = -\frac{1}{2}(1 - 1 + 1) = -\frac{1}{2}$$

The final answer is **$-\frac{1}{2}$**.

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

Select Answer:

Visualized Solution

Identify Given Vectors $a$ and $b$

Analyze the Given Conditions

The Strategic Move: Cross Product with $a$

Recalling the Vector Triple Product Identity

Applying Identity to the Left Hand Side

Applying Identity to the Right Hand Side

Calculating the Dot Product $a \cdot b$

Calculating the Dot Product $a \cdot a$

Constructing the Vector Equation

Simplifying the Vector Expression

Solving for Vector $c$

Final Step: Calculating $c \cdot b$

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Trap of Intuition

The Triple Product Weapon

The Calculation

The Final Stretch

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If $A, B$ and $C$ are vectors such that $∣ B ∣ = ∣ C ∣$ . Prove that $[(A + B) \times (A + C)] \times (B \times C) \cdot (B + C) = 0$ .

Select Answer:

Visualized Solution

Identify 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

Analyze the Given Conditions

The Strategic Move: Cross Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Recalling the Vector Triple Product Identity

Applying Identity to the Left Hand Side

Applying Identity to the Right Hand Side

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

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