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

Q: Wait, why can't we just say $\vec{c} = \vec{a}$? If $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$, doesn't that imply $\vec{c} = \vec{a}$?

That is a classic trap! The cross product is not injective. It only tells us about the components perpendicular to $\vec{b}$. There is an entire line of vectors that satisfy this, which is why we introduce the scalar $k$ to account for the component parallel to $\vec{b}$.

Q: Why do we use $\vec{c} - \vec{a} = k\vec{b}$ instead of $\vec{c} = k\vec{b} + \vec{a}$?

They are mathematically identical! Using $\vec{c} = \vec{a} + k\vec{b}$ is just a way of saying that $\vec{c}$ is the vector $\vec{a}$ shifted by some amount along the direction of $\vec{b}$. It helps us keep track of the geometry, but you are free to rearrange it however makes sense to you.

Q: Is there a faster way to find $\vec{b} \cdot \vec{c}$ without finding $k$?

While you could try to manipulate the dot product directly, finding $k$ is the most robust and error-proof method. In JEE Advanced, reliability is just as important as speed. Stick to the systematic approach!

Question

Let $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ 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 $\vec{b} \cdot \vec{c}$ is equal to:

Options:

(A) $1/2$

(B) $-3/2$

(C) $-1/2$

(D) $-1$

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 aim to determine the properties of a third vector, $\vec{c}$, using the principles of vector algebra.

### The Cross Product Trap

The problem begins with the equation $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{a}$. We must avoid the common mistake of "canceling" $\vec{b}$, as the cross product is not a simple algebraic multiplication.

Instead, we rearrange the equation to:
$$\vec{b} 	imes \vec{c} - \vec{b} 	imes \vec{a} = \vec{0}$$

By the distributive property, this becomes:
$$\vec{b} 	imes (\vec{c} - \vec{a}) = \vec{0}$$

When the cross product of two vectors is the zero vector, they are collinear. Thus, $(\vec{c} - \vec{a})$ must be parallel to $\vec{b}$, which we express as:
$$\vec{c} - \vec{a} = k\vec{b} \implies \vec{c} = \vec{a} + k\vec{b}$$

### The Dot Product Constraint

We are given the condition $\vec{c} \cdot \vec{a} = 0$. Substituting our expression for $\vec{c}$ into this condition yields:
$$(\vec{a} + k\vec{b}) \cdot \vec{a} = 0$$

Distributing the dot product, we obtain:
$$\vec{a} \cdot \vec{a} + k(\vec{b} \cdot \vec{a}) = 0$$
$$|\vec{a}|^2 + k(\vec{b} \cdot \vec{a}) = 0$$

Calculating the necessary values:
$$|\vec{a}|^2 = (1)^2 + (-2)^2 + (1)^2 = 6$$
$$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(-2) + (1)(1) = 4$$

Substituting these into the equation:
$$6 + 4k = 0 \implies k = -\frac{3}{2}$$

### The Final Computation

We now calculate $\vec{b} \cdot \vec{c}$ using the relation $\vec{c} = \vec{a} + k\vec{b}$:
$$\vec{b} \cdot \vec{c} = \vec{b} \cdot (\vec{a} + k\vec{b}) = \vec{b} \cdot \vec{a} + k|\vec{b}|^2$$

We know $\vec{b} \cdot \vec{a} = 4$ and calculate $|\vec{b}|^2$:
$$|\vec{b}|^2 = (1)^2 + (-1)^2 + (1)^2 = 3$$

Assembling the final expression:
$$\vec{b} \cdot \vec{c} = 4 + \left(-\frac{3}{2}ight)(3)$$
$$\vec{b} \cdot \vec{c} = 4 - \frac{9}{2} = \frac{8-9}{2}$$

The final result is:
$$\vec{b} \cdot \vec{c} = -\frac{1}{2}$$

Animated Solution for Mathematics - Vector Algebra: 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:

Select Answer:

Visualized Solution

Visualizing Vectors $a$ and $b$

The Cross Product Equation

Rearranging the Equation

Collinearity of Vectors

Expressing $c$

The Dot Product Condition

Substituting $c$

Calculating $∣ a ∣^{2}$

Calculating $b \cdot a$

Solving for $k$

Target: $b \cdot c$

Calculating $∣ b ∣^{2}$

Final Computation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Trap

The Dot Product Constraint

The Final Computation

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 $a = - \hat{i} - \hat{k}$ , $b = - \hat{i} + \hat{j}$ 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 $a = 3 \hat{i} + \hat{j} - 2 \hat{k}$ , $b = 4 \hat{i} + \hat{j} + 7 \hat{k}$ 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 $a = \hat{j} - \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ . Then the vector $b$ satisfying $a \times b + c = 0$ and $a \cdot b = 3$

Let $a = 6 \hat{i} + 9 \hat{j} + 12 \hat{k}$ , $b = α \hat{i} + 11 \hat{j} - 2 \hat{k}$ and $c$ be vectors such that $a \times c = a \times b$ . If $a \cdot c = - 12$ , and $c \cdot (\hat{i} - 2 \hat{j} + \hat{k}) = 5$ then $c \cdot (\hat{i} + \hat{j} + \hat{k})$ 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 :

The vectors $a$ and $b$ are not perpendicular and $c$ and $d$ are two vectors satisfying $b \times c = b \times d$ and $a \cdot d = 0$ . Then the vector $d$ 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:

Select Answer:

Visualized Solution

Visualizing 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 Equation

Rearranging the Equation

Collinearity of Vectors

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

The Dot Product Condition

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

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

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

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

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

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