Class 12 Vector Algebra · Long Answer3 Markscross_product_area
Class 12 Maths · Vector Algebra · Long Answer #77
Vectors a=(p,q,r), b=(s,3,4), c=(3,1,−2) satisfy a=b+c and form a triangle of area 5√6. Find p,q,r and possible s.
✅ Step-by-Step Solution
Final Answer: (s,p,q,r)=(5,8,4,2) or (−11,−8,4,2)
- From a=b+c, obtain p=s+3, q=4, r=2.
- Compute b×c=(−10,2s+12,s−9). Its squared norm is 5s²+30s+325.
- Since ½|b×c|=5√6, obtain 5s²+30s+325=600, or s²+6s−55=0. Thus s=5 or −11.
- For s=5: (p,q,r)=(8,4,2). For s=−11: (p,q,r)=(−8,4,2); both satisfy the area condition.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] From a=b+c, obtain p=s+3, q=4, r=2.
- [1 mark] Compute b×c=(−10,2s+12,s−9). Its squared norm is 5s²+30s+325.
- [1 mark] Since ½|b×c|=5√6, obtain 5s²+30s+325=600, or s²+6s−55=0. Thus s=5 or −11.
- [0.5 marks] For s=5: (p,q,r)=(8,4,2). For s=−11: (p,q,r)=(−8,4,2); both satisfy the area condition.
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