Find the direction cosines of the sides of the triangle whose vertice…

Question 20 of 25intermediate🔧 ApplyShort Answer4 marks

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4), (-1, 1, 2)$ and $(-5, -5, -2)$ .

Correct Answer

Let the vertices of the triangle be $A(3, 5, -4)$ , $B(-1, 1, 2)$ , and $C(-5, -5, -2)$ . The direction cosines of the line passing through two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are given by $\frac{x_2 - x_1}{PQ}, \frac{y_2 - y_1}{PQ}, \frac{z_2 - z_1}{PQ}$ .

For side AB: The direction ratios are $(-1-3), (1-5), (2-(-4))$ which are $-4, -4, 6$ . Distance $AB = \sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{68}$ . Thus, the direction cosines are $\frac{-4}{\sqrt{68}}, \frac{-4}{\sqrt{68}}, \frac{6}{\sqrt{68}}$ or $\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$ .

For side BC: The direction ratios are $(-5-(-1)), (-5-1), (-2-2)$ which are $-4, -6, -4$ . Distance $BC = \sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{68}$ . Thus, the direction cosines are $\frac{-4}{\sqrt{68}}, \frac{-6}{\sqrt{68}}, \frac{-4}{\sqrt{68}}$ or $\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}$ .

For side CA: The direction ratios are $(3-(-5)), (5-(-5)), (-4-(-2))$ which are $8, 10, -2$ . Distance $CA = \sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{168}$ . Thus, the direction cosines are $\frac{8}{\sqrt{168}}, \frac{10}{\sqrt{168}}, \frac{-2}{\sqrt{168}}$ or $\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}$ .

Exercise: EXERCISE 11.1 | Q: 5 | (Chapter: Page 5)

For More Understanding

Explanation

To find the direction cosines of the sides of a triangle, we treat each side as a line segment joining two vertices. Using the formula provided in the context from Example 3, we calculate the direction ratios (difference in coordinates) and the distance between the vertices for each side. The direction cosines are then found by dividing the direction ratios by the distance. We perform this calculation for all three sides: AB, BC, and CA.

Solution Steps

Step 1: Identify the vertices as A(3, 5, -4), B(-1, 1, 2), and C(-5, -5, -2).
Step 2: For side AB, calculate direction ratios: -4, -4, 6. Calculate distance AB = $\sqrt{16+16+36}$ = $\sqrt{68}$ . Direction cosines are -4/ $\sqrt{68}$ , -4/ $\sqrt{68}$ , 6/ $\sqrt{68}$ , which simplifies to -2/ $\sqrt{17}$ , -2/ $\sqrt{17}$ , 3/ $\sqrt{17}$ .
Step 3: For side BC, calculate direction ratios: -4, -6, -4. Calculate distance BC = $\sqrt{16+36+16}$ = $\sqrt{68}$ . Direction cosines are -4/ $\sqrt{68}$ , -6/ $\sqrt{68}$ , -4/ $\sqrt{68}$ , which simplifies to -2/ $\sqrt{17}$ , -3/ $\sqrt{17}$ , -2/ $\sqrt{17}$ .
Step 4: For side CA, calculate direction ratios: 8, 10, -2. Calculate distance CA = $\sqrt{64+100+4}$ = $\sqrt{168}$ . Direction cosines are 8/ $\sqrt{168}$ , 10/ $\sqrt{168}$ , -2/ $\sqrt{168}$ , which simplifies to 4/ $\sqrt{42}$ , 5/ $\sqrt{42}$ , -1/ $\sqrt{42}$ .

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