Key Concepts and Formulas

• Dot Product: For vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, their dot product is $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$.
• Magnitude of a Vector: The magnitude of a vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ is $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$. The square of the magnitude is $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$.
• Angle Between Vectors: The cosine of the angle $\theta$ between two non-zero vectors $\vec{a}$ and $\vec{b}$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$.
• Vector Decomposition and Orthogonality: If a vector $\vec{v}$ is decomposed into $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$, then $\vec{v}_1$ and $\vec{v}_2$ are orthogonal to each other. Consequently, $|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$.

Step-by-Step Solution

Step 1: Determine the value of $\lambda$ using the given angle between the vectors.

We are given $\vec{u}=3 \hat{i}-\hat{j}$ and $\vec{v}=2 \hat{i}+\hat{j}-\lambda \hat{k}$ with $\lambda > 0$. The angle $\theta$ between them satisfies $\cos \theta = \frac{\sqrt{5}}{2 \sqrt{7}}$.

First, calculate the dot product of $\vec{u}$ and $\vec{v}$: $\vec{u} \cdot \vec{v} = (3)(2) + (-1)(1) + (0)(-\lambda) = 6 - 1 + 0 = 5$

Next, calculate the magnitudes of $\vec{u}$ and $\vec{v}$: $|\vec{u}| = \sqrt{3^2 + (-1)^2 + 0^2} = \sqrt{9 + 1} = \sqrt{10}$ $|\vec{v}| = \sqrt{2^2 + 1^2 + (-\lambda)^2} = \sqrt{4 + 1 + \lambda^2} = \sqrt{5 + \lambda^2}$

Now, use the formula for the angle between vectors: $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$ Substitute the known values: $\frac{\sqrt{5}}{2 \sqrt{7}} = \frac{5}{\sqrt{10} \sqrt{5 + \lambda^2}}$

To solve for $\lambda$, square both sides of the equation: $\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)^2 = \left(\frac{5}{\sqrt{10} \sqrt{5 + \lambda^2}}\right)^2$ $\frac{5}{4 \times 7} = \frac{25}{10 (5 + \lambda^2)}$ $\frac{5}{28} = \frac{25}{10 (5 + \lambda^2)}$ Simplify the right side by dividing the numerator and denominator by 5: $\frac{5}{28} = \frac{5}{2 (5 + \lambda^2)}$ Since the numerators are equal, the denominators must be equal: $28 = 2 (5 + \lambda^2)$ Divide by 2: $14 = 5 + \lambda^2$ Solve for $\lambda^2$: $\lambda^2 = 14 - 5 = 9$ Since $\lambda > 0$, we have $\lambda = 3$.

With $\lambda=3$, the vector $\vec{v}$ is $2 \hat{i}+\hat{j}-3 \hat{k}$. The square of the magnitude of $\vec{v}$ is: $|\vec{v}|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$

Step 2: Utilize the vector decomposition and the Pythagorean theorem for orthogonal vectors.

We are given that $\vec{v}=\vec{v}_1+\overrightarrow{v_2}$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\overrightarrow{v_2}$ is perpendicular to $\vec{u}$. A key property of such a decomposition is that $\vec{v}_1$ and $\vec{v}_2$ are orthogonal to each other. This is because if $\vec{v}_1 = k\vec{u}$ for some scalar $k$, and $\vec{v}_2 \cdot \vec{u} = 0$, then $\vec{v}_1 \cdot \vec{v}_2 = (k\vec{u}) \cdot \vec{v}_2 = k(\vec{u} \cdot \vec{v}_2) = k(0) = 0$.

For orthogonal vectors $\vec{v}_1$ and $\vec{v}_2$, the Pythagorean theorem for vectors states: $|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$

In Step 1, we calculated $|\vec{v}|^2$ after finding $\lambda=3$. $|\vec{v}|^2 = 14$

Therefore, the value of $\left|\overrightarrow{v_1}\right|^2+\left|\overrightarrow{v_2}\right|^2$ is equal to $|\vec{v}|^2$. $\left|\overrightarrow{v_1}\right|^2+\left|\overrightarrow{v_2}\right|^2 = 14$

Common Mistakes & Tips

• Algebraic Errors: Squaring both sides of an equation can introduce extraneous solutions or lead to calculation errors. Double-check all algebraic manipulations.
• Ignoring $\lambda > 0$: The condition $\lambda > 0$ is crucial for selecting the correct value of $\lambda$ after solving $\lambda^2 = 9$.
• Unnecessary Calculation: Recognize that the problem directly asks for $|\vec{v}_1|^2 + |\vec{v}_2|^2$, which is equal to $|\vec{v}|^2$ due to the orthogonal decomposition. Calculating $\vec{v}_1$ and $\vec{v}_2$ explicitly is time-consuming and not required.

Summary

The problem involves determining an unknown scalar $\lambda$ using the angle between two vectors, and then finding the sum of the squares of the magnitudes of vector components. By first calculating $\lambda$ using the dot product and magnitude formulas, we found $\lambda=3$. Subsequently, we recognized that the decomposition of $\vec{v}$ into $\vec{v}_1$ (parallel to $\vec{u}$) and $\vec{v}_2$ (perpendicular to $\vec{u}$) implies that $\vec{v}_1$ and $\vec{v}_2$ are orthogonal. This allows us to directly use the Pythagorean theorem for vectors, $|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$. The required value is simply the square of the magnitude of $\vec{v}$, which was calculated to be 14.

The final answer is $\boxed{14}$.