Key Concepts and Formulas

• Perpendicular Vectors: Two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product is zero: $\vec{u} \cdot \vec{v} = 0$.
• Dot Product Definition: For vectors $\vec{x}$ and $\vec{y}$ with angle $\theta$ between them, $\vec{x} \cdot \vec{y} = |\vec{x}| |\vec{y}| \cos \theta$.
• Unit Vector Property: If $\vec{u}$ is a unit vector, then $|\vec{u}| = 1$, and $\vec{u} \cdot \vec{u} = |\vec{u}|^2 = 1$.
• Distributive Property of Dot Product: $\vec{x} \cdot (\vec{y} + \vec{z}) = \vec{x} \cdot \vec{y} + \vec{x} \cdot \vec{z}$.
• Quadratic Formula: For a quadratic equation $A\lambda^2 + B\lambda + C = 0$, the solutions are $\lambda = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

Step-by-Step Solution

Step 1: Understand the Given Information and Calculate Essential Dot Products

We are given that $\vec{a}$ and $\vec{b}$ are unit vectors, which means $|\vec{a}| = 1$ and $|\vec{b}| = 1$. The angle between them is $\frac{\pi}{3}$. We need to find the values of $\lambda$ for which the vectors $\lambda \vec{a}+2 \vec{b}$ and $3 \vec{a}-\lambda \vec{b}$ are perpendicular.

First, let's calculate the dot products of $\vec{a}$ and $\vec{b}$ that will be needed:

• $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1^2 = 1$ (Since $\vec{a}$ is a unit vector)
• $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1^2 = 1$ (Since $\vec{b}$ is a unit vector)
• $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\left(\frac{\pi}{3}\right) = (1)(1)\left(\frac{1}{2}\right) = \frac{1}{2}$
  • Why: This uses the definition of the dot product with the given magnitudes and angle. Since the dot product is commutative, $\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b} = \frac{1}{2}$.

Step 2: Apply the Perpendicularity Condition

Two vectors are perpendicular if their dot product is zero. Let $\vec{u} = \lambda \vec{a}+2 \vec{b}$ and $\vec{v} = 3 \vec{a}-\lambda \vec{b}$. The condition for perpendicularity is $\vec{u} \cdot \vec{v} = 0$. $(\lambda \vec{a}+2 \vec{b}) \cdot (3 \vec{a}-\lambda \vec{b}) = 0$ * Why: This is the direct application of the definition of perpendicular vectors.

Step 3: Expand the Dot Product

We use the distributive property of the dot product to expand the expression: $(\lambda \vec{a}) \cdot (3 \vec{a}) + (\lambda \vec{a}) \cdot (-\lambda \vec{b}) + (2 \vec{b}) \cdot (3 \vec{a}) + (2 \vec{b}) \cdot (-\lambda \vec{b}) = 0$ Rearranging the scalar coefficients: $3\lambda (\vec{a} \cdot \vec{a}) - \lambda^2 (\vec{a} \cdot \vec{b}) + 6 (\vec{b} \cdot \vec{a}) - 2\lambda (\vec{b} \cdot \vec{b}) = 0$ * Why: Expanding converts the vector equation into a form where we can substitute the pre-calculated dot product values, leading to an equation solely in terms of $\lambda$.

Step 4: Substitute Known Dot Product Values and Form a Quadratic Equation

Substitute the values calculated in Step 1 into the expanded equation: $3\lambda (1) - \lambda^2 \left(\frac{1}{2}\right) + 6 \left(\frac{1}{2}\right) - 2\lambda (1) = 0$ Simplify the equation: $3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda = 0$ Combine like terms: $\lambda - \frac{\lambda^2}{2} + 3 = 0$ To obtain a standard quadratic form, multiply the entire equation by 2: $2\lambda - \lambda^2 + 6 = 0$ Rearrange into the standard quadratic form $A\lambda^2 + B\lambda + C = 0$: $-\lambda^2 + 2\lambda + 6 = 0$ Multiplying by -1 to make the leading coefficient positive: $\lambda^2 - 2\lambda - 6 = 0$ * Why: This step transforms the vector equation into a solvable algebraic equation for $\lambda$.

Step 5: Solve the Quadratic Equation for $\lambda$

We use the quadratic formula $\lambda = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ for the equation $\lambda^2 - 2\lambda - 6 = 0$, where $A=1$, $B=-2$, and $C=-6$. $\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-6)}}{2(1)}$ $\lambda = \frac{2 \pm \sqrt{4 + 24}}{2}$ $\lambda = \frac{2 \pm \sqrt{28}}{2}$ Simplify the square root: $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$. $\lambda = \frac{2 \pm 2\sqrt{7}}{2}$ Divide by 2: $\lambda = 1 \pm \sqrt{7}$ * Why: Solving the quadratic equation provides the potential values of $\lambda$ that satisfy the perpendicularity condition.

This yields two possible values for $\lambda$: $\lambda_1 = 1 + \sqrt{7}$ $\lambda_2 = 1 - \sqrt{7}$

Step 6: Check the Values of $\lambda$ Against the Given Interval

The problem requires us to find the number of values of $\lambda$ in the interval $[-1, 3]$. We need to approximate the values of $\sqrt{7}$. We know that $2^2 = 4$ and $3^2 = 9$, so $2 < \sqrt{7} < 3$. A good approximation is $\sqrt{7} \approx 2.65$.

• For $\lambda_1 = 1 + \sqrt{7}$: $\lambda_1 \approx 1 + 2.65 = 3.65$. Since $3.65 > 3$, this value of $\lambda$ is not in the interval $[-1, 3]$.

• For $\lambda_2 = 1 - \sqrt{7}$: $\lambda_2 \approx 1 - 2.65 = -1.65$. Since $-1.65 < -1$, this value of $\lambda$ is not in the interval $[-1, 3]$.

  • Why: The question specifies a range for $\lambda$. It's essential to filter the solutions obtained from the algebraic equation to only include those that satisfy this condition.

Since neither of the calculated values of $\lambda$ falls within the interval $[-1, 3]$, there are no such values of $\lambda$ in the given range.

Common Mistakes & Tips

• Forgetting the Interval: Always check if the obtained values of $\lambda$ lie within the specified interval. This is a common oversight that can lead to an incorrect answer.
• Algebraic Errors: Be meticulous when expanding dot products and solving the quadratic equation. Small arithmetic mistakes can significantly alter the results.
• Approximation Accuracy: For interval checks, ensure your approximations of square roots are accurate enough to distinguish whether a value is inside or outside the boundary. For example, knowing $\sqrt{7}$ is between 2 and 3 is crucial.

Summary

The problem involved finding values of $\lambda$ for which two vector expressions are perpendicular. This was achieved by utilizing the property that the dot product of perpendicular vectors is zero. After calculating the necessary dot products of the given unit vectors, the perpendicularity condition was expanded and simplified into a quadratic equation in $\lambda$. Solving this quadratic equation yielded two possible values for $\lambda$, $1+\sqrt{7}$ and $1-\sqrt{7}$. However, upon checking these values against the given interval $[-1, 3]$, neither value was found to be within this range. Therefore, there are no values of $\lambda$ in the specified interval that satisfy the problem's conditions.

The final answer is $\boxed{0}$.