Let a = i + 2j + k, b = 3i - j + 3k, c = 2i - j + 2k and d be a vector such that b × d = c × d and a

Q: Why can't we just solve for $\vec{d}$ by setting the cross product components to zero?

The equation $(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}$ only tells us that $\vec{d}$ is parallel to $(\vec{b} - \vec{c})$. It does not uniquely determine $\vec{d}$ because any scalar multiple $\lambda(\vec{b} - \vec{c})$ will satisfy the cross product condition. We need the second condition, $\vec{a} \cdot \vec{d} = 4$, to fix the magnitude and sign of $\vec{d}$.

Q: Is Lagrange's Identity really necessary? Can't I just calculate the cross product $\vec{a} \times \vec{d}$ and then find its magnitude?

You absolutely can! However, in competitive exams like JEE, time is your most precious resource. Lagrange's Identity allows you to find the squared magnitude directly using the dot product and the individual squared magnitudes, which are much easier to compute and less prone to arithmetic errors than the cross product determinant.

Q: Why did we get a negative value for $\lambda$? Does that mean the magnitude is negative?

Great observation! The scalar $\lambda$ being negative simply means that $\vec{d}$ points in the opposite direction of the vector $(\vec{b} - \vec{c})$. The magnitude of a vector is always non-negative, and when we square the components to find the magnitude, the negative sign disappears. The geometry remains perfectly consistent.

Question

Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}, \vec{b} = 3\hat{i} - \hat{j} + 3\hat{k}, \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{d}$ be a vector such that $\vec{b} 	imes \vec{d} = \vec{c} 	imes \vec{d}$ and $\vec{a} \cdot \vec{d} = 4$. Then $|(\vec{a} 	imes \vec{d})|^2$ is equal to ______ .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to unravel a beautiful puzzle involving vectors. Imagine you are standing in a three-dimensional coordinate system with three known vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, and one mysterious vector, $\vec{d}$.

We are given two clues about $\vec{d}$: a cross product relationship and a dot product constraint. Our mission is to find the squared magnitude of the cross product of $\vec{a}$ and $\vec{d}$.

Let us begin by examining the first clue: $\vec{b} 	imes \vec{d} = \vec{c} 	imes \vec{d}$. If we bring everything to one side, we get $\vec{b} 	imes \vec{d} - \vec{c} 	imes \vec{d} = \vec{0}$.

By the distributive property of the cross product, this simplifies to $(\vec{b} - \vec{c}) 	imes \vec{d} = \vec{0}$. This is a profound geometric statement! When the cross product of two vectors is the zero vector, it implies that the vectors are parallel.

Thus, our mysterious vector $\vec{d}$ must be parallel to the vector $(\vec{b} - \vec{c})$. We can express this as $\vec{d} = \lambda(\vec{b} - \vec{c})$, where $\lambda$ is a scalar constant.

### Calculating the Direction

First, let us find the vector $(\vec{b} - \vec{c})$. Following the logic provided in the problem steps, we utilize the vector difference:
$$(\vec{b} - \vec{c}) = \hat{i} - 2\hat{j} + \hat{k}$$

Consequently, we define our vector $\vec{d}$ in terms of the scalar $\lambda$:
$$\vec{d} = \lambda(\hat{i} - 2\hat{j} + \hat{k})$$

### The Dot Product Constraint

Now, we use the second clue: $\vec{a} \cdot \vec{d} = 4$. With $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$, we substitute our expression for $\vec{d}$:
$$(\hat{i} + 2\hat{j} + \hat{k}) \cdot \lambda(\hat{i} - 2\hat{j} + \hat{k}) = 4$$

This expands to the following scalar equation:
$$\lambda [ (1)(1) + (2)(-2) + (1)(1) ] = 4$$

Simplifying the terms inside the bracket, we get:
$$\lambda [ 1 - 4 + 1 ] = 4 \Rightarrow -2\lambda = 4$$

Thus, we find the value of the scalar constant:
$$\lambda = -2$$

Substituting this back, our vector $\vec{d}$ is:
$$\vec{d} = -2(\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$$

### The Elegant Shortcut

We need to calculate $|\vec{a} 	imes \vec{d}|^2$. Instead of computing the cross product directly, we use **Lagrange's Identity**:
$$|\vec{a} 	imes \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2$$

First, we calculate the squared magnitudes:
$$|\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 6$$
$$|\vec{d}|^2 = (-2)^2 + 4^2 + (-2)^2 = 4 + 16 + 4 = 24$$

Finally, substituting these values into the identity:
$$|\vec{a} 	imes \vec{d}|^2 = (6)(24) - (4)^2$$
$$|\vec{a} 	imes \vec{d}|^2 = 144 - 16 = 128$$

And there we have it! The final result is **128**. Keep practicing, and remember that every vector problem is just a story waiting to be told.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + 2 \hat{j} + \hat{k}, b = 3 \hat{i} - \hat{j} + 3 \hat{k}, c = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $d$ be a vector such that $b \times d = c \times d$ and $a \cdot d = 4$ . Then $∣ (a \times d) ∣^{2}$ is equal to ______ .

Enter Numerical Value:

Visualized Solution

The Vector Space

Analyze the Cross Product Equation

Interpret the Parallel Condition

Calculate the Difference Vector $b - c$

Express $d$ in Terms of $λ$

Apply the Dot Product Condition

Substitute Vectors into Dot Product

Evaluate the Dot Product

Solve for $λ$

Find the Exact Vector $d$

Introduce Lagrange's Identity

Calculate Squared Magnitudes

Substitute into Lagrange's Identity

Final Computation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Calculating the Direction

The Dot Product Constraint

The Elegant Shortcut

Similar Questions

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