Question 34 of 68intermediate🔧 ApplyNumerical4 marks

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (-1, 0, 0), (0, 1, 2), respectively, then find ∠ABC. [∠ABC is the angle between the vectors BA and BC].

Correct Answer

∠ABC = cos⁻¹(10/102\sqrt{102})

Exercise: EXERCISE 10.3 | Q: 15 | (Page 25)
For More Understanding

Explanation

This question tests the application of the dot product formula to find the angle between two vectors. The angle ∠ABC is formed at vertex B by vectors BA and BC. Using the formula cos θ = (BA · BC)/(|BA||BC|), we can determine the angle. The context shows similar examples of finding angles between vectors and calculating vector magnitudes from coordinates.

Solution Steps

  1. Step 1: Find vectors BA and BC. Given A(1, 2, 3), B(-1, 0, 0), C(0, 1, 2).

  2. Step 2: BA = A - B = (1-(-1), 2-0, 3-0) = (2, 2, 3)

  3. Step 3: BC = C - B = (0-(-1), 1-0, 2-0) = (1, 1, 2)

  4. Step 4: Calculate dot product: BA · BC = 2(1) + 2(1) + 3(2) = 2 + 2 + 6 = 10

  5. Step 5: Calculate magnitudes: |BA| = 22+22+32\sqrt{2² + 2² + 3²} = 4+4+9\sqrt{4 + 4 + 9} = 17\sqrt{17}

  6. Step 6: |BC| = 12+12+22\sqrt{1² + 1² + 2²} = 1+1+4\sqrt{1 + 1 + 4} = 6\sqrt{6}

  7. Step 7: Apply formula: cos θ = (BA · BC)/(|BA||BC|) = 10/(17\sqrt{17} × 6\sqrt{6}) = 10/102\sqrt{102}

  8. Step 8: Therefore, ∠ABC = cos⁻¹(10/102\sqrt{102})

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