Question 46 of 68intermediate🔧 ApplyNumerical4 marks

Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Correct Answer

The area of the triangle is 61\sqrt{61}/2 square units.

Exercise: EXERCISE 10.4 | Q: 9 | (Page 31)
For More Understanding

Explanation

This question tests the application of vector cross product to find the area of a triangle in 3D space. The context from Example 24 demonstrates the method: Area = 1/2|AB × AC|. Students need to first find the vectors AB and AC using the given vertices, then compute their cross product using the determinant method, find its magnitude, and finally multiply by 1/2.

Solution Steps

  1. Step 1: Find vectors AB and AC. AB = B - A = (2-1)î + (3-1)ĵ + (5-2)k̂ = î + 2ĵ + 3k̂ AC = C - A = (1-1)î + (5-1)ĵ + (5-2)k̂ = 4ĵ + 3k̂

  2. Step 2: Calculate AB × AC using determinant. AB × AC = |î ĵ k̂| |1 2 3| |0 4 3| = î(6-12) - ĵ(3-0) + k̂(4-0) = -6î - 3ĵ + 4k̂

  3. Step 3: Find the magnitude. |AB × AC| = (−6\sqrt{(-6}² + (-3)² + 4²) = 36+9+16\sqrt{36 + 9 + 16} = 61\sqrt{61}

  4. Step 4: Calculate the area. Area = 1/2 × |AB × AC| = 1/2 × 61\sqrt{61} = 61\sqrt{61}/2 square units

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