Question

Let a,b,c be real numbers with `a^2 + b^2 + c^2 =1`. Show that the equation `|[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+c]|=0` represents a straight line.

Answer 1

Given, `|{:(ax-by-c," "bx+ay, cx+a),(bx+ay, -ax+by-c, cy+b),(cx+a," " cy+b, -ax-by+c):}| = 0`
`rArr (1)/(a)|{:(a^(2)x-aby-ac," "bx+ay, cx+a),(abx+a^(2)y, -ax+by-c, cy+b),(acx+a^(2)," " cy+b, -ax-by+c):}| = 0`
Applying `C_(1) to C_(1) + bC_(2) + cC_(3)`
`rArr (1)/(a)|{:((a^(2) + b^(2) + c^(2))x," "ay+bx, cx+a),((a^(2) + b^(2) + c^(2))y, by-c-ax, b+cy),(a^(2) + b^(2) + c^(2)," " b+cy, c-ax-by):}| = 0`
`rArr (1)/(a) |{:(x," "ay+bx, cx+a),(y, by-c-ax, b+cy),(1," " b+cy, c-ax-by):}| = 0 " " [because a^(2) + b^(2) + c^(2) = 1]`
Applying `C_(1) to C_(2) -bC_(1) " and " C_(3) to C_(3) -cC_(1)`
`rArr (1)/(a) |{:(x," "ay, cx+a),(y, -c-ax, " "b),(1," " cy, -ax-by):}| = 0`
`rArr (1)/(ax) |{:(x^(2)," "axy," " ax),(y, -c-ax, " "b),(1," " cy, -ax-by):}| = 0`
Applying `R_(1) to R_(1) + yR_(2) + R_(3)`
`rArr (1)/(ax) |{:(x^(2) +y^(2) +1," "0," " 0),(" "y, -c-ax, " "b),(" "1," " cy, -ax-by):}| = 0`
`rArr (1)/(ax) [(x^(2) + y^(2) + 1){(-c-ax)(-ax-by)-b(cy)}] = 0`
`rArr (1)/(ax) [(x^(2) + y^(2) + 1) (acx + bcy + a^(2)x^(2) + abxy -bcy)] = 0`
`rArr (1)/(ax) [(x^(2) + y^(2) + 1)(acx + a^(2)x^(2) + abxy)] = 0`d
`rArr (1)/(ax) [ax(x^(2) + y^(2) + 1)(c + ax + by)] = 0`
`rArr (x^(2) + y^(2) + 1)(ax + by +c) = 0`
`rArr ax +by +c = 0`
which represents a straight line.