Let a = 3i + 2j + 2k and b = i + 2j - 2k be two vectors. If a vector perpendicular to both the vecto

Q: Wait, why do we use the cross product here? Can't we just use dot products and solve a system of equations?

You absolutely could! Setting the dot product of your unknown vector (x, y, z) with both (a+b) and (a-b) to zero would give you two linear equations. However, that leads to a system with three variables and only two equations, meaning you'd have an infinite family of vectors. The cross product is the elegant 'shortcut' that gives you the direction immediately!

Q: Why does (a + b) x (a - b) simplify to -2(a x b)?

Great catch! When you expand the cross product, you get (a x a) - (a x b) + (b x a) - (b x b). Since the cross product of any vector with itself is the zero vector, and (b x a) is simply -(a x b), the expression becomes 0 - (a x b) - (a x b) - 0, which is exactly -2(a x b).

Q: Does the order of the cross product matter? Should I do (a - b) x (a + b) instead?

It matters for the sign! (a - b) x (a + b) would give you 2(a x b). Since the problem asks for 'one such vector' and provides options, you can check the sign by looking at the resulting components. Both the vector and its negative are valid perpendicular vectors.

Q: How do I know if I should normalize the vector first?

Always check the magnitude of your cross product result first. If the magnitude is already the value requested (in this case, 12), you don't need to divide by the magnitude and then multiply by 12. It saves time and prevents messy fractions!

Q: Is there a geometric interpretation for why the magnitude of the cross product is 12?

Yes! The magnitude of the cross product of two vectors is equal to the area of the parallelogram they span. Here, the cross product of (a+b) and (a-b) represents twice the area of the triangle formed by vectors a and b. It's a beautiful connection between algebra and geometry.

Question

Let $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$ be two vectors. If a vector perpendicular to both the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ has the magnitude 12 then one such vector is

Options:

(A) $4(2\hat{i} + 2\hat{j} - \hat{k})$

(B) $4(-2\hat{i} + 2\hat{j} + \hat{k})$

(C) $4(2\hat{i} - 2\hat{j} + \hat{k})$

(D) $4(2\hat{i} + 2\hat{j} + \hat{k})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the world of JEE Advanced mathematics. Today, we aren't just solving a problem; we are exploring the elegant architecture of 3D space.

We are given two vectors, $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$. Our mission is to find a vector that stands perfectly perpendicular to two other vectors: the sum, $\vec{a} + \vec{b}$, and the difference, $\vec{a} - \vec{b}$.

### The Power of the Cross Product

When you hear the word 'perpendicular' in the context of two vectors, your mind should immediately race to the cross product. The cross product is the ultimate tool for finding a normal vector—a vector that refuses to lie in the plane defined by its parents.

We need a vector $\vec{v}$ that is parallel to the result of $(\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b})$.

Now, you might be tempted to calculate $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ first. While that works, let's embrace the elegance of algebra. Let's expand this cross product using the distributive property:

$$(\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b}) = \vec{a} 	imes \vec{a} - \vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{a} - \vec{b} 	imes \vec{b}$$

Pause for a moment and look at the beauty here. The cross product of any vector with itself is the zero vector, $\vec{0}$, because the angle between them is zero, and $\sin(0) = 0$.

Furthermore, the cross product is anti-commutative, meaning $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$. Our expression collapses into something remarkably simple:

$$-(\vec{a} 	imes \vec{b}) - (\vec{a} 	imes \vec{b}) = -2(\vec{a} 	imes \vec{b})$$

### Calculating the Normal

Now, we only need to calculate $\vec{a} 	imes \vec{b}$. We use the determinant method, which is our reliable compass in 3D space:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 2 \ 1 & 2 & -2 \end{vmatrix}$$

Let's break this down component by component. For the $\hat{i}$ component, we have $(2)(-2) - (2)(2) = -4 - 4 = -8$.

For the $\hat{j}$ component, we must remember the negative sign in the expansion: $-[(3)(-2) - (2)(1)] = -[-6 - 2] = 8$.

Finally, for the $\hat{k}$ component, we have $(3)(2) - (2)(1) = 6 - 2 = 4$. Thus, our normal vector $\vec{n}$ is $-8\hat{i} + 8\hat{j} + 4\hat{k}$.

### The Final Verification

The problem demands a vector of magnitude 12. Let's check the magnitude of our calculated vector $\vec{n}$:

$$|\vec{n}| = \sqrt{(-8)^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$$

It is a perfect match! The magnitude is exactly 12.

Since our vector $\vec{v}$ must be parallel to $-2(\vec{a} 	imes \vec{b})$, we are looking for a vector that is a scalar multiple of our result. Given the options, we can see that **$4(-2\hat{i} + 2\hat{j} + \hat{k})$** is the correct direction.

### Conclusion

Mathematics is not about memorizing formulas; it is about recognizing patterns. By using the properties of the cross product, we avoided unnecessary arithmetic and arrived at the solution with clarity and confidence.

Keep practicing, keep visualizing, and remember: every vector is just a step toward understanding the geometry of the universe.

Animated Solution for Mathematics - Vector Algebra: Let $a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - 2 \hat{k}$ be two vectors. If a vector perpendicular to both the vectors $a + b$ and $a - b$ has the magnitude 12 then one such vector is

Select Answer:

Visualized Solution

Given Vectors $a$ and $b$

The Perpendicularity Condition

Using the Cross Product

Simplifying the Cross Product

Applying Cross Product Properties

Setting up $a \times b$

Expanding the Determinant

The Normal Vector $n$

Calculating Magnitude

Finalizing the Vector

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of the Cross Product

Calculating the Normal

The Final Verification

Conclusion

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