Consider two vectors u = 3i - j and v = 2i + j - λk, λ 0. The angle between them is given by cos⁻¹

Q: Wait, why do we reject the negative root for $\lambda$?

The problem explicitly states $\lambda > 0$. Even if it didn't, notice that the final expression we need is $|\vec{v}|^2 = 5 + \lambda^2$. Since we are squaring $\lambda$, the sign of $\lambda$ actually doesn't affect the final magnitude! The condition $\lambda > 0$ is just to ensure the vector $\vec{v}$ is uniquely defined.

Q: Why can't we just find $\vec{v}_1$ and $\vec{v}_2$ explicitly? Is there a shortcut?

You absolutely could find them, but that is the long road! $\vec{v}_1$ is the projection of $\vec{v}$ onto $\vec{u}$, which is $\left(\frac{\vec{v} \cdot \vec{u}}{|\vec{u}|^2}\right)\vec{u}$. Then $\vec{v}_2 = \vec{v} - \vec{v}_1$. While this works, it is computationally heavy. The Pythagorean shortcut $|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$ is much faster and less prone to calculation errors.

Q: Is the angle between $\vec{v}_1$ and $\vec{v}_2$ always 90 degrees?

Yes, always! Because $\vec{v}_1$ is defined to be parallel to $\vec{u}$ and $\vec{v}_2$ is defined to be perpendicular to $\vec{u}$, the angle between $\vec{v}_1$ and $\vec{v}_2$ must be the same as the angle between $\vec{u}$ and any vector perpendicular to it, which is $90^\circ$. This orthogonality is the foundation of the Pythagorean theorem in vector space.

Question

Consider two vectors $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}, \lambda > 0$. The angle between them is given by $\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to

Options:

(A) $23/2$

(B) $14$

(C) $25/2$

(D) $10$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are working with two vectors in three-dimensional space: $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$. These vectors are inclined at an angle $	heta$, where $\cos 	heta = \frac{\sqrt{5}}{2\sqrt{7}}$.

Our objective is to determine the value of $|\vec{v}_1|^2 + |\vec{v}_2|^2$, where $\vec{v}_1$ is the component of $\vec{v}$ parallel to $\vec{u}$, and $\vec{v}_2$ is the component perpendicular to $\vec{u}$.

### The Bridge: The Dot Product

The dot product serves as the fundamental bridge between algebraic components and geometric orientation. We use the definition:

$$\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$$

First, we calculate the dot product:
$\vec{u} \cdot \vec{v} = (3)(2) + (-1)(1) + (0)(-\lambda) = 5$.

Next, we determine the magnitudes of the vectors:
$|\vec{u}| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$
$|\vec{v}| = \sqrt{2^2 + 1^2 + (-\lambda)^2} = \sqrt{5 + \lambda^2}$

### The Mystery of $\lambda$

Substituting these values into our cosine formula, we obtain:

$$\frac{\sqrt{5}}{2\sqrt{7}} = \frac{5}{\sqrt{10}\sqrt{5 + \lambda^2}}$$

To solve for $\lambda$, we square both sides of the equation:

$$\frac{5}{4 	imes 7} = \frac{25}{10(5 + \lambda^2)}$$

This simplifies to:

$$\frac{5}{28} = \frac{5}{2(5 + \lambda^2)}$$

Cross-multiplying yields $2(5 + \lambda^2) = 28$, which simplifies to $5 + \lambda^2 = 14$, or $\lambda^2 = 9$.

### The Geometric Insight

We are given that $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1 \parallel \vec{u}$ and $\vec{v}_2 \perp \vec{u}$. Because $\vec{v}_1$ and $\vec{v}_2$ are orthogonal, they form a right-angled triangle with $\vec{v}$ as the hypotenuse.

By the **Pythagorean theorem**, we have:

$$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$$

We do not need to calculate the individual vectors. We simply evaluate the square of the magnitude of $\vec{v}$ using our previously derived expression:

$$|\vec{v}|^2 = 5 + \lambda^2$$

Substituting $\lambda^2 = 9$ into the equation, we find:

$$|\vec{v}|^2 = 5 + 9 = 14$$

Thus, the final result is **14**.

Select Answer:

Visualized Solution

Visualizing the Vectors $u$ and $v$

The Dot Product Formula

Calculating $u \cdot v$

Calculating Magnitude $∣ u ∣$

Calculating Magnitude $∣ v ∣$

Setting up the Equation

Squaring Both Sides

Solving for $λ^{2}$

The Orthogonal Decomposition

Applying Pythagoras Theorem

Final Calculation

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Bridge

The Mystery of $λ$

The Geometric Insight

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Select Answer:

Visualized Solution

Visualizing the Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

The Dot Product Formula

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

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

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