Question

x, y, z are distinct scalars such that

$\lbrack x\mathbf{a} + y\mathbf{b} + z\mathbf{c},x\mathbf{b} + y\mathbf{c} + z\mathbf{a},x\mathbf{c} + y\mathbf{a} + z\mathbf{b}\rbrack = 0$ where a, b, c are non-coplanar vectors then

• A. $x + y + z = 0$
• B. $xy + yz + zx = 0$
• C. $x^{3} + y^{3} + z^{3} = 0$
• D. $x^{2} + y^{2} + z^{2} = 0$

Answer 1

Answer: (A)

$x + y + z = 0$

Answer 2

$\mathbf{a},\mathbf{b},\mathbf{c}$ are non-coplanar

$\therefore$ $\lbrack\mathbf{abc}\rbrack \neq 0$

Now,$\lbrack x\mathbf{a} + y\mathbf{b} + z\mathbf{c},x\mathbf{b} + y\mathbf{c} + z\mathbf{a},x\mathbf{c} + y\mathbf{a} + z\mathbf{b}\rbrack = 0$

⇒$(x\mathbf{a} + y\mathbf{b} + z\mathbf{c}).\{(x\mathbf{b} + y\mathbf{c} + z\mathbf{a}) \times (x\mathbf{c} + y\mathbf{a} + z\mathbf{b})\} = 0$

⇒$(x\mathbf{a} + y\mathbf{b} + z\mathbf{c}).\{(x^{2} - yz)(\mathbf{b} \times \mathbf{c}) + (z^{2} - xy)(\mathbf{a} \times \mathbf{b}) + (y^{2} - zx)(\mathbf{c} \times \mathbf{a})\} = 0$

⇒$x(x^{2} - yz)\lbrack\mathbf{abc}\rbrack + y(y^{2} - zx)\lbrack\mathbf{bca}\rbrack + z(z^{2} - xy)\lbrack\mathbf{cab}\rbrack = 0$

⇒$(x^{3} - xyz)\lbrack\mathbf{abc}\rbrack + (y^{3} - xyz)\lbrack\mathbf{abc}\rbrack + (z^{3} - xyz)\lbrack\mathbf{abc}\rbrack = 0$

As $\lbrack\mathbf{abc}\rbrack \neq 0$, $x^{3} + y^{3} + z^{3} - 3xyz = 0$

⇒$(x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - zx) = 0$

⇒ $\frac{1}{2}(x + y + z)\{(x - y)^{2} + (y - z)^{2} + (z - x)^{2}\} = 0$

⇒ $x + y + z = 0$ or $x = y = z$

But x, y, z are distinct. $\therefore$ $x + y + z = 0$