Question 20 of 68intermediate🔧 ApplyLong Answer4 marks

Show that the points A, B and C with position vectors,

a⃗=3i^−4j^−4k^,b⃗=2i^−j^+k^andc⃗=i^−3j^−5k^,\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}, \quad \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \quad \text{and} \quad \vec{c} = \hat{i} - 3\hat{j} - 5\hat{k},

respectively form the vertices of a right angled triangle.

Correct Answer

The points A, B and C form the vertices of a right angled triangle, right-angled at A.

Exercise: EXERCISE 10.2 | Q: 17 | (Page 18)
For More Understanding

Explanation

This question tests the concept of position vectors and the condition for a right-angled triangle. The textbook demonstrates in Example 12 that when the sum of squares of two sides equals the square of the third side (Pythagoras theorem), the triangle is right-angled.

Alternatively, if the dot product of two sides is zero, those sides are perpendicular, confirming a right angle at that vertex.

Solution Steps

  1. Step 1: Find the vectors representing the sides of the triangle.

  2. AB→=b⃗−a⃗=(2−3)i^+(−1−(−4))j^+(1−(−4))k^=−i^+3j^+5k^\overrightarrow{AB} = \vec{b} - \vec{a} = (2-3)\hat{i} + (-1-(-4))\hat{j} + (1-(-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}
  3. BC→=c⃗−b⃗=(1−2)i^+(−3−(−1))j^+(−5−1)k^=−i^−2j^−6k^\overrightarrow{BC} = \vec{c} - \vec{b} = (1-2)\hat{i} + (-3-(-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}
  4. CA→=a⃗−c⃗=(3−1)i^+(−4−(−3))j^+(−4−(−5))k^=2i^−j^+k^\overrightarrow{CA} = \vec{a} - \vec{c} = (3-1)\hat{i} + (-4-(-3))\hat{j} + (-4-(-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}

  5. Step 2: Calculate the magnitudes squared of each side.

  6. ∣AB→∣2=(−1)2+32+52=1+9+25=35|\overrightarrow{AB}|^2 = (-1)^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35
  7. ∣BC→∣2=(−1)2+(−2)2+(−6)2=1+4+36=41|\overrightarrow{BC}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41
  8. ∣CA→∣2=22+(−1)2+12=4+1+1=6|\overrightarrow{CA}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6

  9. Step 3: Verify Pythagoras theorem.

  10. ∣BC→∣2=41=35+6=∣AB→∣2+∣CA→∣2|\overrightarrow{BC}|^2 = 41 = 35 + 6 = |\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2

  11. Step 4: Conclusion.

  12. Since ∣BC→∣2=∣AB→∣2+∣CA→∣2|\overrightarrow{BC}|^2 = |\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2, by Pythagoras theorem, the triangle is right-angled at A.

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