Let O be the origin and OA = 2i + 2j + k, OB = i - 2j + 2k and OC = 1/2(OB - λOA) for some λ 0. If

Q: Why does $\vec{OB} \times \vec{OB}$ become zero? Isn't it just a vector?

That is a fantastic question! Remember the definition of the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta)$. When you cross a vector with itself, the angle $\theta$ is $0^\circ$. Since $\sin(0^\circ) = 0$, the entire cross product vanishes into the zero vector. It is one of the most powerful simplification tools in your vector toolkit!

Q: Why is the area of $\triangle ABC$ equal to the area of $\triangle OAB$?

It is a beautiful geometric coincidence in this specific problem! When you expand the cross product $\vec{AB} \times \vec{AC}$, the terms simplify exactly to $\vec{OA} \times \vec{OB}$. This means the 'effective base and height' of the triangle $ABC$ are identical to those of triangle $OAB$. It is a great reminder to trust your algebra when the geometry seems complex.

Q: How do I know if I should use the dot product or cross product for the angle between diagonals?

Always use the dot product formula $\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$. The dot product is the scalar projection of one vector onto another, which is directly related to the cosine of the angle. The cross product would give you the sine of the angle, which is usually more work to convert back to cosine.

Question

Let $O$ be the origin and $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$, $\vec{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{OC} = \frac{1}{2}(\vec{OB} - \lambda \vec{OA})$ for some $\lambda > 0$. If $|\vec{OB} 	imes \vec{OC}| = \frac{9}{2}$, then which of the following statements is (are) TRUE?

Options:

(A) Projection of $\vec{OC}$ on $\vec{OA}$ is $-\frac{3}{2}$

(B) Area of the triangle $OAB$ is $\frac{9}{2}$

(C) Area of the triangle $ABC$ is $\frac{9}{2}$

(D) The acute angle between the diagonals of the parallelogram with adjacent sides $\vec{OA}$ and $\vec{OC}$ is $\frac{\pi}{3}$

Sigma Percentile · Accepted Answer

### The Vector Playground: A Journey into 3D Space

Welcome, fellow traveler of the JEE Advanced landscape. Today, we are not just solving a problem; we are exploring the elegant architecture of 3D space.

We have vectors $\vec{OA}$, $\vec{OB}$, and a mysterious third vector $\vec{OC}$ defined by a parameter $\lambda$. This problem is a classic test of your ability to manipulate vector algebra with precision and intuition.

### Phase 1: The Foundation of Orthogonality

Before we touch the complex definitions, we must understand our building blocks. We are given $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$.

First, we calculate their magnitudes. Using the standard formula $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$, we find:

$$|\vec{OA}| = \sqrt{2^2 + 2^2 + 1^2} = 3$$

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

Now, here is the spark of genius. Let us calculate the dot product:

$$\vec{OA} \cdot \vec{OB} = (2)(1) + (2)(-2) + (1)(2) = 2 - 4 + 2 = 0$$

Stop and breathe. A dot product of zero is a gift from the problem setter. It tells us that $\vec{OA}$ and $\vec{OB}$ are perfectly perpendicular, forming the axes of our local coordinate system.

### Phase 2: Unmasking the Mystery of $\lambda$

We are given $\vec{OC} = \frac{1}{2}(\vec{OB} - \lambda \vec{OA})$ and the condition $|\vec{OB} 	imes \vec{OC}| = \frac{9}{2}$. Do not substitute coordinates immediately; stay in the vector domain.

Let us expand the cross product:

$$\vec{OB} 	imes \vec{OC} = \vec{OB} 	imes \frac{1}{2}(\vec{OB} - \lambda \vec{OA})$$

Using the distributive property of the cross product, we get:

$$\frac{1}{2}(\vec{OB} 	imes \vec{OB}) - \frac{\lambda}{2}(\vec{OB} 	imes \vec{OA})$$

As we discussed, $\vec{OB} 	imes \vec{OB} = \vec{0}$. The expression simplifies to $-\frac{\lambda}{2}(\vec{OB} 	imes \vec{OA})$. Since $\vec{OB} 	imes \vec{OA} = -(\vec{OA} 	imes \vec{OB})$, the negative signs cancel, leaving us with $\frac{\lambda}{2}(\vec{OA} 	imes \vec{OB})$.

We know $|\vec{OA} 	imes \vec{OB}| = |\vec{OA}||\vec{OB}|\sin(90^{\circ}) = 3 	imes 3 	imes 1 = 9$. Equating the magnitudes:

$$\frac{\lambda}{2}(9) = \frac{9}{2} \Rightarrow \lambda = 1$$

The mystery is solved! We now have $\vec{OC} = \frac{1}{2}(\vec{OB} - \vec{OA})$.

### Phase 3: Evaluating the Truth

Now that we have the full picture, we can evaluate the statements.

**Statement A: Projection of $\vec{OC}$ on $\vec{OA}$**

The projection is given by $\frac{\vec{OC} \cdot \vec{OA}}{|\vec{OA}|}$. Substituting $\vec{OC} = \frac{1}{2}(\vec{OB} - \vec{OA})$:

$$	ext{Proj} = \frac{\frac{1}{2}(\vec{OB} - \vec{OA}) \cdot \vec{OA}}{3} = \frac{\frac{1}{2}(\vec{OB} \cdot \vec{OA} - |\vec{OA}|^2)}{3}$$

Since $\vec{OB} \cdot \vec{OA} = 0$ and $|\vec{OA}|^2 = 9$, we get $\frac{\frac{1}{2}(0 - 9)}{3} = -\frac{3}{2}$. **Statement A is TRUE.**

**Statement B: Area of $	riangle OAB$**

The area is $\frac{1}{2} |\vec{OA} 	imes \vec{OB}| = \frac{1}{2}(9) = \frac{9}{2}$. **Statement B is TRUE.**

**Statement C: Area of $	riangle ABC$**

We define the sides as $\vec{AB} = \vec{OB} - \vec{OA}$ and $\vec{AC} = \vec{OC} - \vec{OA} = \frac{1}{2}\vec{OB} - \frac{3}{2}\vec{OA}$. Calculating the cross product $\vec{AB} 	imes \vec{AC}$:

$$(\vec{OB} - \vec{OA}) 	imes (\frac{1}{2}\vec{OB} - \frac{3}{2}\vec{OA})$$

Expanding this, the self-cross products vanish, leaving $-\frac{3}{2}(\vec{OB} 	imes \vec{OA}) - \frac{1}{2}(\vec{OA} 	imes \vec{OB})$. This simplifies to $\vec{OA} 	imes \vec{OB}$. The area is $\frac{1}{2} |\vec{OA} 	imes \vec{OB}| = \frac{9}{2}$. **Statement C is TRUE.**

**Statement D: The Angle Between Diagonals**

For the parallelogram with sides $\vec{OA}$ and $\vec{OC}$, the diagonals are $\vec{d_1} = \vec{OA} + \vec{OC}$ and $\vec{d_2} = \vec{OA} - \vec{OC}$.

After calculating the dot product $\vec{d_1} \cdot \vec{d_2}$ and the magnitudes, we find $\cos 	heta = \frac{1}{\sqrt{5}}$. Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, **Statement D is FALSE.**

Select Answer:

Visualized Solution

Magnitudes and Dot Product

Magnitude of $O A \times O B$

Simplifying $O B \times O C$

Finding $λ$ and $O C$

Option A: Projection of $O C$ on $O A$

Option B: Area of $△ O A B$

Option C: Vectors $A B$ and $A C$

Option C: Area of $△ A B C$

Option D: Diagonals of Parallelogram

Dot Product of Diagonals

Magnitudes of Diagonals

Angle Between Diagonals

The Sigma Insight: Vector (Cross) Product

The Vector Playground

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Magnitudes and Dot Product

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

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OB×OC

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

Option A: Projection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OC on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA

Option B: Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Option C: Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Option C: Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC