7.2 Q-4

Question Statement

Find the value of $\alpha$ such that the magnitude of the vector $\alpha \underline{\mathbf{i}} + (\alpha+1) \underline{\mathbf{j}} + 2 \underline{\mathbf{k}}$ is equal to 3.

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Background and Explanation

This problem involves finding the value of $\alpha$ that satisfies a given condition on the magnitude of a vector. The magnitude of a vector $\mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k}$ is calculated using the following formula: $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

In this case, the vector components depend on the variable $\alpha$, and we are asked to solve for $\alpha$ such that the magnitude of the vector is 3. We will follow standard algebraic steps to isolate $\alpha$.

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Solution

We are given the vector: $\mathbf{v} = \alpha \underline{\mathbf{i}} + (\alpha + 1) \underline{\mathbf{j}} + 2 \underline{\mathbf{k}}$

1. Write the equation for the magnitude of the vector: The magnitude of $\mathbf{v}$ is given by:

$$|\mathbf{v}| = \sqrt{(\alpha)^2 + (\alpha+1)^2 + (2)^2}$$

2. Substitute the given magnitude: We are told that the magnitude is 3, so we set the equation equal to 3:

$$\sqrt{\alpha^2 + (\alpha + 1)^2 + 4} = 3$$

3. Simplify the terms: Expand $(\alpha + 1)^2$:

$$\sqrt{\alpha^2 + (\alpha^2 + 2\alpha + 1) + 4} = 3$$

Combine like terms:

$$\sqrt{2\alpha^2 + 2\alpha + 5} = 3$$

4. Square both sides: To eliminate the square root, square both sides of the equation:

$$2\alpha^2 + 2\alpha + 5 = 9$$

5. Simplify the equation: Subtract 9 from both sides:

$$2\alpha^2 + 2\alpha + 5 - 9 = 0$$ $$2\alpha^2 + 2\alpha - 4 = 0$$

6. Divide by 2: To simplify, divide the entire equation by 2:

$$\alpha^2 + \alpha - 2 = 0$$

7. Factor the quadratic equation: Factor the quadratic equation:

$$\alpha(\alpha + 2) - 1(\alpha + 2) = 0$$

This gives:

$$(\alpha + 2)(\alpha - 1) = 0$$

8. Solve for $\alpha$: Set each factor equal to 0:

$$\alpha + 2 = 0 \quad \text{or} \quad \alpha - 1 = 0$$

So, we get two possible solutions:

$$\alpha = -2 \quad \text{or} \quad \alpha = 1$$

Thus, the two possible values of $\alpha$ are:

$$\boxed{\alpha = -2 \text{ or } \alpha = 1}$$

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Key Formulas or Methods Used

• Magnitude of a Vector: $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

• Solving Quadratic Equations: Factoring and using the zero product property.

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Summary of Steps

1. Set up the equation for the magnitude of the vector.
2. Substitute the given magnitude and simplify the terms.
3. Square both sides of the equation to eliminate the square root.
4. Simplify the resulting equation and solve the quadratic equation.
5. Factor the quadratic and solve for $\alpha$.
6. Conclude that $\alpha = -2$ or $\alpha = 1$.