7.3 Q-4

Question Statement

Find the number $z$ such that the triangle with vertices $A(1, -1, 0)$, $B(-2, 2, 1)$, and $C(0, 2, z)$ forms a right triangle with the right angle at $C$.

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Background and Explanation

In a right triangle, the two vectors representing the sides that meet at the right angle are perpendicular to each other. For the triangle with vertices $A$, $B$, and $C$, we need to find the value of $z$ such that the vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$ are perpendicular. The condition for two vectors to be perpendicular is that their dot product equals zero:

$$\overrightarrow{AC} \cdot \overrightarrow{BC} = 0$$

We will calculate the vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$, and then solve for $z$.

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Solution

1. Find the vector $\overrightarrow{AC}$:

The vector $\overrightarrow{AC}$ is given by the difference between the coordinates of points $C(0, 2, z)$ and $A(1, -1, 0)$:

$$\overrightarrow{AC} = (0 - 1, 2 + 1, z - 0) = (-1, 3, z)$$

Thus, $\overrightarrow{AC} = -\underline{\mathbf{i}} + 3 \underline{\mathbf{j}} + z \underline{\mathbf{k}}$.

2. Find the vector $\overrightarrow{BC}$:

The vector $\overrightarrow{BC}$ is given by the difference between the coordinates of points $C(0, 2, z)$ and $B(-2, 2, 1)$:

$$\overrightarrow{BC} = (0 + 2, 2 - 2, z - 1) = (2, 0, z - 1)$$

Thus, $\overrightarrow{BC} = 2 \underline{\mathbf{i}} + 0 \underline{\mathbf{j}} + (z - 1) \underline{\mathbf{k}}$.

3. Set up the condition for perpendicularity:

For $\overrightarrow{AC}$ and $\overrightarrow{BC}$ to be perpendicular, their dot product must be zero:

$$\overrightarrow{AC} \cdot \overrightarrow{BC} = 0$$

Now, compute the dot product:

$$(-1)(2) + (3)(0) + z(z - 1) = 0$$ $$-2 + 0 + z^2 - z = 0$$ $$z^2 - z - 2 = 0$$

4. Solve the quadratic equation:

The quadratic equation $z^2 - z - 2 = 0$ can be factored as:

$$z^2 - z - 2 = (z - 2)(z + 1) = 0$$

Thus, the solutions are:

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

Therefore, the possible values of $z$ are $z = 2$ and $z = -1$.

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Key Formulas or Methods Used

• Dot product of vectors: The dot product of two vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ is:

$$\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3$$

• Condition for perpendicularity: Two vectors are perpendicular if their dot product is zero:

$$\mathbf{u} \cdot \mathbf{v} = 0$$

• Quadratic equation: The quadratic equation $ax^2 + bx + c = 0$ can be solved by factoring, completing the square, or using the quadratic formula.

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Summary of Steps

1. Calculate the vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$.
2. Set up the dot product equation for perpendicularity: $\overrightarrow{AC} \cdot \overrightarrow{BC} = 0$.
3. Solve the resulting quadratic equation for $z$.
4. Find the solutions for $z$, which are $z = 2$ and $z = -1$.