7.3 Q-3

Question Statement

Find a real number $\alpha$ such that the vectors $\underline{\mathbf{U}}$ and $\underline{\mathbf{V}}$ are perpendicular.

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Background and Explanation

Two vectors are perpendicular if and only if their dot product is zero. The dot product of two vectors $\underline{\mathbf{U}} = [u_1, u_2, u_3]$ and $\underline{\mathbf{V}} = [v_1, v_2, v_3]$ is given by:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = u_1v_1 + u_2v_2 + u_3v_3$$

For vectors to be perpendicular, we must have:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = 0$$

Given that the vectors contain the variable $\alpha$, we will solve for $\alpha$ such that the dot product equals zero.

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Solution

(i) Given vectors:

$$\underline{\mathbf{U}} = 2\alpha \underline{\mathbf{i}} + \underline{\mathbf{j}} - \underline{\mathbf{k}}, \quad \underline{\mathbf{V}} = \underline{\mathbf{i}} + \alpha \underline{\mathbf{j}} + 4\underline{\mathbf{k}}$$

1. Dot product formula: Since the vectors are perpendicular, their dot product should be zero:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = 0$$

2. Compute the dot product:

$$(2\alpha)(1) + (1)(\alpha) + (-1)(4) = 0$$ $$2\alpha + \alpha - 4 = 0$$ $$3\alpha - 4 = 0$$

3. Solve for $\alpha$:

$$\alpha = \frac{4}{3}$$

Thus, the value of $\alpha$ that makes $\underline{\mathbf{U}}$ and $\underline{\mathbf{V}}$ perpendicular is $\alpha = \frac{4}{3}$.

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(ii) Given vectors:

$$\underline{\mathbf{U}} = \alpha \underline{\mathbf{i}} + 2\alpha \underline{\mathbf{j}} - \underline{\mathbf{k}}, \quad \underline{\mathbf{V}} = \underline{\mathbf{i}} + \alpha \underline{\mathbf{j}} + 3\underline{\mathbf{k}}$$

1. Dot product formula: Again, the vectors are perpendicular, so their dot product equals zero:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = 0$$

2. Compute the dot product:

$$(\alpha)(1) + (2\alpha)(\alpha) + (-1)(3) = 0$$ $$\alpha + 2\alpha^2 - 3 = 0$$ $$2\alpha^2 + \alpha - 3 = 0$$

3. Solve the quadratic equation $2\alpha^2 + \alpha - 3 = 0$ using factoring:

$$\alpha(2\alpha + 3) - 1(2\alpha + 3) = 0$$ $$(\alpha - 1)(2\alpha + 3) = 0$$

4. Find the roots:

$$\alpha - 1 = 0 \quad \text{or} \quad 2\alpha + 3 = 0$$ $$\alpha = 1 \quad \text{or} \quad \alpha = \frac{-3}{2}$$

Thus, the possible values of $\alpha$ are $\alpha = 1$ and $\alpha = \frac{-3}{2}$.

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Key Formulas or Methods Used

• Dot product:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = u_1v_1 + u_2v_2 + u_3v_3$$

• Condition for perpendicularity:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = 0$$

• Solving quadratic equations: The quadratic equation can be solved using factoring or the quadratic formula.

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Summary of Steps

1. Write down the dot product formula for the given vectors.
2. Calculate the dot product and set it equal to zero.
3. Solve for $\alpha$ by simplifying the equation.
4. Find the possible values of $\alpha$ that satisfy the condition for perpendicularity.