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`        Please help me solve this asap.... the correct answer is (x+y-5)  Question is attatched`
one year ago

```							Given : a + xb + yc = 0   ...(1)  and  a x b + b x c + c x a = 6(b x c)         ...(2)To solve this question, we’ll represent a in terms of b and c since RHS in second equation is in terms of (b x c).We will use the facts that cross product of a vector with itself is a zero vector and negative of cross productof a vector with another vector is same as cross product of second vector with the first i.e.a x a = 0  and  – (a x b) = (b x a) a + xb + yc = 0 ,  a = – (xb + yc)Now, a x b = – (xb + yc) x b  =  – (x(b x b) + y(c x b))  =  – y(c x b)  =  y(b x c)  [ b x b = 0]&  c x a = c x {– (xb + yc)} = – c x (xb + yc)  =  – (x (c x b) + y(c x c)) = – x(c x b) = x(c x b)[ c x c = 0](2) becomes  y(b x c) + b x c + x(c x b)  =  6(b x c)  i.e.  (x + y +1)(b x c)  =  6(b x c)Comparing coefficients of b x c on both sides, we get(x + y +1)  =  6    x + y – 5 = 0.
```
one year ago
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### Course Features

• 19 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions