A vector $\vec a=(x,y,z)$ makes an obtuse angle with Y-axis, and make equal angles with $\vec b=(y,-2z,3x)$ and $\vec c=(2z,3x,-y)$ and $\vec a$ is perpendicular to $\vec d=(1,-1,2)$ if $|\vec a|=2\sqrt{3}$ then vector $\vec a$ is:
Answer & explanation
Correct answer: option 2
$\frac{\vec a.\vec b}{|\vec b|}=\frac{\vec a.\vec c}{|\vec c|}⇒\frac{xy-2zy+3xz}{\sqrt{y^2+4z^2+9z^2}}=\frac{2xz+3xy-yz}{\sqrt{4z^2+9x^2+y^2}}$
⇒ 2xy - yz - zx = 0 …… (i)
$\vec a.\vec d$ ⇒ x - y + z = 0 …… (ii)
Putting x = y - 2z in (i)
2(y - 2z)y + yz - (y - 2z)z = 0
$⇒2y^2+4zy+2z^2=0⇒y^2+z^2-2zy=0⇒(z-y)^2=0⇒z=y$
So, x = -z
$\vec a≡(x,-x,-x)\&\,\sqrt{3x^2}=2\sqrt{3}⇒3x^2=12⇒x^2=4⇒x±2$
$\vec a≡(2,-2,-2)$ [obtuse angle with y-axis]