Let a unit vector C which makes an angle of 60° with 2i + 2j - k and angle 45° with i - k be C. Then

Q: Wait, why do we use the dot product instead of the cross product?

Great question! The dot product is defined as $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta$. It directly involves the cosine of the angle between two vectors, which is exactly the information given in the problem. The cross product involves $\sin\theta$ and results in a vector perpendicular to the plane, which is not what we need here.

Q: How do I know which root of the quadratic equation to choose?

In this specific problem, the quadratic equation $36x^2 - 36x + 1 = 0$ yields two possible values for $x$. However, the problem asks for a specific vector $\vec{C}$ that satisfies the conditions. Often, checking the options provided in a multiple-choice question will reveal which root is intended. If it were a subjective question, both vectors would technically satisfy the geometric conditions unless further constraints were provided.

Q: Is there a geometric way to solve this without all this algebra?

Yes, you could visualize this as the intersection of two cones in 3D space. The set of all vectors making a $60^\circ$ angle with $\vec{A}$ forms a cone, and the set of all vectors making a $45^\circ$ angle with $\vec{B}$ forms another cone. The intersection of these two cones gives you the possible directions for $\vec{C}$. However, for JEE Advanced, the algebraic method is the most reliable and efficient way to arrive at the exact components.

Question

Let a unit vector $\vec{C}$ which makes an angle of $60^\circ$ with $2\hat{i} + 2\hat{j} - \hat{k}$ and angle $45^\circ$ with $\hat{i} - \hat{k}$ be $\vec{C}$. Then $\vec{C} + \left(-\frac{1}{2}\hat{i} + \frac{1}{3\sqrt{2}}\hat{j} - \frac{\sqrt{2}}{3}\hat{k}\right)$ is :

Options:

(A) $\frac{\sqrt{2}}{3}\hat{i} - \frac{1}{2}\hat{k}$

(B) $\left(\frac{1}{\sqrt{3}} + \frac{1}{2}\right)\hat{i} + \left(\frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{2}}\right)\hat{j} + \left(\frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{3}\right)\hat{k}$

(C) $\frac{\sqrt{2}}{3}\hat{i} + \frac{1}{3\sqrt{2}}\hat{j} - \frac{1}{2}\hat{k}$

(D) $-\frac{\sqrt{2}}{3}\hat{i} + \frac{\sqrt{2}}{3}\hat{j} + \left(\frac{1}{2} + \frac{2\sqrt{2}}{3}\right)\hat{k}$

Sigma Percentile · Accepted Answer

### The Geometry of Direction: Unveiling the Unit Vector

Imagine you are standing in the center of a three-dimensional coordinate system. You have a vector $\vec{C}$, a mysterious arrow of unit length pointing somewhere into the vastness of this space.

You don't know where it points, but you have two clues: its relationship with two other known vectors, $\vec{A} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{B} = \hat{i} - \hat{k}$. This is not just a math problem; it is a detective story where we use the language of vectors to pin down the exact orientation of $\vec{C}$.

### Phase 1: The Foundation

We define our vector as $\vec{C} = x\hat{i} + y\hat{j} + z\hat{k}$. The term "unit vector" is our most powerful constraint. It tells us that the magnitude of $\vec{C}$ is exactly one.

Mathematically, this translates to the foundational equation:

$$x^2 + y^2 + z^2 = 1$$

This equation represents a sphere of radius one centered at the origin. Our vector $\vec{C}$ must terminate somewhere on the surface of this sphere.

### Phase 2: The Bridge of the Dot Product

We are given that $\vec{C}$ makes an angle of $60^\circ$ with $\vec{A}$ and $45^\circ$ with $\vec{B}$. The dot product is the bridge between the abstract angle and the concrete components.

Recall the definition: $\vec{C} \cdot \vec{A} = |\vec{C}||\vec{A}|\cos	heta$. Since $|\vec{C}| = 1$, this simplifies to $\vec{C} \cdot \vec{A} = |\vec{A}|\cos	heta$.

First, let us calculate the magnitude of $\vec{A}$:

$$|\vec{A}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = 3$$

Now, the dot product $\vec{C} \cdot \vec{A} = 2x + 2y - z$. Setting this equal to $|\vec{A}|\cos 60^\circ$, we get:

$$2x + 2y - z = 3 	imes \frac{1}{2} = \frac{3}{2}$$

We repeat this for $\vec{B} = \hat{i} - \hat{k}$. The magnitude $|\vec{B}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. The dot product $\vec{C} \cdot \vec{B} = x - z$. Setting this equal to $|\vec{B}|\cos 45^\circ$:

$$x - z = \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 1$$

### Phase 3: The Algebraic Transformation

We now have a system of three equations. From $x - z = 1$, we find $z = x - 1$. Substituting this into our first dot product equation:

$$2x + 2y - (x - 1) = \frac{3}{2} \Rightarrow x + 2y + 1 = \frac{3}{2} \Rightarrow 2y = \frac{1}{2} - x \Rightarrow y = \frac{1 - 2x}{4}$$

Now, we have $y$ and $z$ expressed entirely in terms of $x$. We substitute these into our sphere equation, $x^2 + y^2 + z^2 = 1$:

$$x^2 + \left( \frac{1 - 2x}{4} ight)^2 + (x - 1)^2 = 1$$

Expanding this, we get:

$$x^2 + \frac{1 - 4x + 4x^2}{16} + (x^2 - 2x + 1) = 1$$

Multiplying the entire equation by $16$ to clear the denominator:

$$16x^2 + 1 - 4x + 4x^2 + 16x^2 - 32x + 16 = 16$$

Combining like terms leads us to the quadratic equation:

$$36x^2 - 36x + 1 = 0$$

### Phase 4: The Climax and Resolution

Solving for $x$ using the quadratic formula, we find:

$$x = \frac{36 \pm \sqrt{36^2 - 4(36)(1)}}{2(36)} = \frac{36 \pm \sqrt{1296 - 144}}{72} = \frac{36 \pm \sqrt{1152}}{72} = \frac{36 \pm 24\sqrt{2}}{72} = \frac{1}{2} \pm \frac{\sqrt{2}}{3}$$

Choosing the roots, we find the corresponding $y$ and $z$ components. Finally, the problem asks us to add a specific vector to $\vec{C}$.

When we perform this addition, the components align in such a way that the complex terms cancel out, leaving us with a clean, elegant result. This is the beauty of physics and mathematics—the complexity of the journey often leads to a surprisingly simple and harmonious destination.

Animated Solution for Mathematics - Vector Algebra: Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Select Answer:

Visualized Solution

Define the Unit Vector $C$

Angle with Vector $A$

Evaluate Dot Product with $A$

Angle with Vector $B$

Express $y$ and $z$ in terms of $x$

Substitute into Magnitude Equation

Expand and Simplify

Solve the Quadratic Equation

Calculate $y$ and $z$ Components

Final Vector Addition

The Sigma Insight: Scalar (Dot) Product

The Geometry of Direction

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Define the Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Angle with Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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