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Samyak Jain
333 Points
2 years ago
Given : a + xb + yc = 0   ...(1)  and  a x b + c + c x a = 6(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 (c).
We will use the facts that cross product of a vector with itself is a zero vector and negative of cross product
of 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)
$\dpi{100} \because$ 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)  [$\dpi{80} \because$ b x b = 0]
&  c x a = x {– (xb + yc)} = – x (xb + yc)  =  – (x (c x b) + y(c x c)) = – x(c x b) = x(c x b)
[$\dpi{80} \because$ c x c = 0]
(2) becomes  y(b x c) + c + x(c x b)  =  6(c)  i.e.
(x + y +1)(c)  =  6(c)
Comparing coefficients of c on both sides, we get
(x + y +1)  =  6  $\dpi{100} \Rightarrow$  x + y – 5 = 0.