7.3 Q-6

Question Statement

i. Show that the vectors $(3 \underline{i} - 2 \underline{j} + \underline{k}), (\underline{i} - 3 \underline{j} + 5 \underline{k})$ and $(2 \underline{i} + \underline{j} - 4 \underline{k})$ form a right angle.

ii. Show that the set of points $P = (1, 3, 2)$, $Q = (4, 1, 4)$, and $R = (6, 5, 5)$ form a right triangle.

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Background and Explanation

In part (i), we are tasked with showing that the given vectors are perpendicular (form a right angle). Two vectors are perpendicular if their dot product is zero:

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

In part (ii), we are given three points in space, and we need to prove that they form a right triangle. For three points to form a right triangle, the vectors corresponding to the sides meeting at the right angle must be perpendicular to each other.

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Solution

(i) Proving that the vectors form a right angle:

We are given the following vectors:

• $\underline{\mathbf{a}} = 3 \underline{i} - 2 \underline{j} + \underline{k}$
• $\underline{\mathbf{b}} = \underline{i} - 3 \underline{j} + 5 \underline{k}$
• $\underline{\mathbf{c}} = 2 \underline{i} + \underline{j} - 4 \underline{k}$

We need to show that these vectors form a right angle, which means we need to prove that at least two of the vectors are perpendicular. To do this, we will compute the dot product between $\underline{\mathbf{a}}$ and $\underline{\mathbf{c}}$.

1. Dot product of $\underline{\mathbf{a}}$ and $\underline{\mathbf{c}}$:

$$\underline{\mathbf{a}} \cdot \underline{\mathbf{c}} = (3 \underline{i} - 2 \underline{j} + \underline{k}) \cdot (2 \underline{i} + \underline{j} - 4 \underline{k})$$ $$= 3(2) + (-2)(1) + 1(-4) = 6 - 2 - 4 = 0$$

Since the dot product is $0$, the vectors $\underline{\mathbf{a}}$ and $\underline{\mathbf{c}}$ are perpendicular, and thus form a right angle.

$$\underline{\mathbf{a}} \perp \underline{\mathbf{c}}$$

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(ii) Proving that the points $P$, $Q$, and $R$ form a right triangle:

We are given the points:

• $P = (1, 3, 2)$
• $Q = (4, 1, 4)$
• $R = (6, 5, 5)$

We need to show that these points form a right triangle. To do this, we will compute the vectors corresponding to the sides of the triangle and check if any two vectors are perpendicular.

1. Find the vectors:

   • $\overrightarrow{PQ} = Q - P = (4, 1, 4) - (1, 3, 2) = 3 \underline{i} - 2 \underline{j} + 2 \underline{k}$
   • $\overrightarrow{QR} = R - Q = (6, 5, 5) - (4, 1, 4) = 2 \underline{i} + 4 \underline{j} + \underline{k}$
   • $\overrightarrow{PR} = R - P = (6, 5, 5) - (1, 3, 2) = 5 \underline{i} + 2 \underline{j} + 3 \underline{k}$

2. Check if $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ are perpendicular:

$$\overrightarrow{PQ} \cdot \overrightarrow{QR} = (3 \underline{i} - 2 \underline{j} + 2 \underline{k}) \cdot (2 \underline{i} + 4 \underline{j} + \underline{k})$$ $$= 3(2) + (-2)(4) + 2(1) = 6 - 8 + 2 = 0$$

Since the dot product is $0$, the vectors $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ are perpendicular, and thus the points $P$, $Q$, and $R$ form a right triangle.

$$P, Q, R \text{ are the vertices of a right triangle.}$$

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Key Formulas or Methods Used

• Dot product of vectors:

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

Two vectors are perpendicular if their dot product is zero:

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

• Vectors between points: The vector between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by:

$$\overrightarrow{PQ} = (x_2 - x_1) \underline{i} + (y_2 - y_1) \underline{j} + (z_2 - z_1) \underline{k}$$

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Summary of Steps

1. For part (i), compute the dot product of $\underline{\mathbf{a}}$ and $\underline{\mathbf{c}}$ to check if they are perpendicular.
2. For part (ii), find the vectors corresponding to the sides of the triangle $PQR$.
3. Compute the dot product of vectors $\overrightarrow{PQ}$ and $\overrightarrow{QR}$.
4. Conclude that the vectors are perpendicular, so the points $P$, $Q$, and $R$ form a right triangle.