JEE Advanced 2014 · Paper 1, Mathematics question 48

Dot products. $|\vec x|^2=|\vec y|^2=|\vec z|^2=2$ and every mixed product is $2\cos\frac\pi3=1$.

Direction of $\vec a$. $\vec a$ is perpendicular to $\vec x$ and to $\vec y\times\vec z$, so it is parallel to
$$\vec x\times(\vec y\times\vec z)=(\vec x\cdot\vec z)\vec y-(\vec x\cdot\vec y)\vec z=\vec y-\vec z.$$
Write $\vec a=k(\vec y-\vec z)$. Then $\vec a\cdot\vec y=k(2-1)=k$, so $\vec a=(\vec a\cdot\vec y)(\vec y-\vec z)$.

Direction of $\vec b$. Parallel to $\vec y\times(\vec z\times\vec x)=(\vec y\cdot\vec x)\vec z-(\vec y\cdot\vec z)\vec x=\vec z-\vec x$. With $\vec b=m(\vec z-\vec x)$, $\vec b\cdot\vec z=m(2-1)=m$.

• (A) $\vec b=(\vec b\cdot\vec z)(\vec z-\vec x)$. True.
• (B) True.
• (C) $\vec a\cdot\vec b=km(\vec y-\vec z)\cdot(\vec z-\vec x)=km(1-1-2+1)=-km=-(\vec a\cdot\vec y)(\vec b\cdot\vec z)$. True.
• (D) Has the wrong sign. False.

(A), (B), (C)