Class 12 Vector Algebra · Long Answer3 Marksorthogonality
Class 12 Maths · Vector Algebra · Long Answer #122
Find a vector of magnitude 3 perpendicular to (3,1,−4) and (6,5,−2).
✅ Step-by-Step Solution
Final Answer: ±(2,−2,1)
- Compute (3,1,−4)×(6,5,−2)=(18,−18,9)=9(2,−2,1).
- The vector (2,−2,1) has norm √(4+4+1)=3.
- Its dot products with the given vectors are 6−2−4=0 and 12−10−2=0.
- Thus either (2,−2,1) or its negative has the prescribed magnitude and perpendicularity.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Compute (3,1,−4)×(6,5,−2)=(18,−18,9)=9(2,−2,1).
- [1 mark] The vector (2,−2,1) has norm √(4+4+1)=3.
- [1 mark] Its dot products with the given vectors are 6−2−4=0 and 12−10−2=0.
- [0.5 marks] Thus either (2,−2,1) or its negative has the prescribed magnitude and perpendicularity.
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