Two adjacent sides of a parallelogram ABCD are given by AB = 2i + 10j + 11k and AD = i + 2j + 2k. Th

Q: Why do we need to calculate the magnitude of the vectors first?

The dot product formula $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$ requires the magnitudes. Without them, the dot product is just a scalar without angular context. We need the lengths to isolate $\cos \theta$.

Q: Can we just use the cross product?

You could, but it is overkill. The dot product is the most direct path to the cosine of the angle. In JEE, efficiency is as important as accuracy.

Q: What if the rotation was not in the plane?

The problem explicitly states 'in the plane of the parallelogram'. If it were out of the plane, the geometry would be much more complex, involving 3D rotation matrices. Always pay attention to these constraints!

Question

Two adjacent sides of a parallelogram $ABCD$ are given by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$, then the cosine of the angle $\alpha$ is given by

Options:

(A) $\frac{8}{9}$

(B) $\frac{\sqrt{17}}{9}$

(C) $\frac{1}{9}$

(D) $\frac{4\sqrt{5}}{9}$

Sigma Percentile · Accepted Answer

### The Geometric Dance of Vectors

Welcome, future engineers. Today, we are not just solving a problem; we are choreographing a dance between vectors.

Imagine a parallelogram resting on a flat, infinite plane in 3D space. We are given two adjacent sides, $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + 2\hat{k}$.

These vectors define the foundation of our shape. Our goal is to understand how $\vec{AD}$ transforms when it is rotated by an acute angle $\alpha$ to a new position, $\vec{AD'}$, such that it becomes perfectly perpendicular to $\vec{AB}$.

### Phase 1: The Language of Angles

Before we can rotate anything, we must understand the initial state. What is the angle $	heta$ between $\vec{AB}$ and $\vec{AD}$?

In the world of vectors, the dot product is our bridge between algebra and geometry. We know that:

$$\vec{AB} \cdot \vec{AD} = |\vec{AB}| |\vec{AD}| \cos 	heta$$

To use this, we first need the magnitudes. The magnitude of $\vec{AB}$ is:

$$|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$$

Similarly, for $\vec{AD}$, we have:

$$|\vec{AD}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

Now, we compute the dot product:

$$\vec{AB} \cdot \vec{AD} = (2)(1) + (10)(2) + (11)(2) = 2 + 20 + 22 = 44$$

Substituting these into our formula, we find:

$$\cos 	heta = \frac{44}{15 	imes 3} = \frac{44}{45}$$

This value, $\frac{44}{45}$, is our anchor.

### Phase 2: The Geometric Transformation

Now, let us visualize the rotation. We take $\vec{AD}$ and rotate it by an angle $\alpha$ within the plane until it hits a position $\vec{AD'}$ that is perpendicular to $\vec{AB}$.

This means the angle between $\vec{AB}$ and $\vec{AD'}$ is exactly $90^{\circ}$. Geometrically, this implies that the original angle $	heta$ plus the rotation angle $\alpha$ must sum to $90^{\circ}$.

Thus, $\alpha = 90^{\circ} - 	heta$. The question asks for $\cos \alpha$.

Using our trigonometric identity:

$$\cos \alpha = \cos(90^{\circ} - 	heta) = \sin 	heta$$

We have successfully reduced a complex rotation problem to a simple trigonometric identity.

### Phase 3: The Final Calculation

We know $\cos 	heta = \frac{44}{45}$. To find $\sin 	heta$, we use the identity $\sin 	heta = \sqrt{1 - \cos^2 	heta}$.

Substituting our value, we get:

$$\sin 	heta = \sqrt{1 - \left(\frac{44}{45}ight)^2} = \sqrt{\frac{45^2 - 44^2}{45^2}}$$

This is a difference of squares:

$$\frac{\sqrt{(45-44)(45+44)}}{45} = \frac{\sqrt{1 	imes 89}}{45} = \frac{\sqrt{89}}{45}$$

This is our mathematically rigorous result. If you find that this does not match the options, do not panic.

In the high-stakes environment of competitive exams, typos can occur. Trust your derivation, trust your logic, and always prioritize the process over the final option. You have mastered the geometry behind the rotation.

Select Answer:

Visualized Solution

Defining the Initial Vectors

The Angle Between Vectors

Magnitude of $A B$

Magnitude of $A D$

The Dot Product

Calculating $cos θ$

The Rotation to $A D^{'}$

Relating $α$ and $θ$

Finding $cos α$

Calculating $sin θ$

Final Computation

Final Conclusion

The Sigma Insight: Scalar (Dot) Product

The Geometric Dance of Vectors

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Defining the Initial Vectors

The Angle Between Vectors

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

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

The Dot Product

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

The Rotation to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD′

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

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

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