Let $\vec{b}$ and $\vec{c}$ be non-collinear vectors satisfying $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b})\vec{b} = (4 - 2x - \sin y)\vec{b} + (x^2 - 1)\vec{c}$ and $(\vec{c} \cdot \vec{c})\vec{a} = \vec{c}$,then $x$ is equal to

For any two vectors $\vec{a}$ and $\vec{b}$,we always have $|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ (Cauchy-Schwarz inequality). Is this statement true or false?

Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $7 \hat{i}-\hat{j}+8 \hat{k}$.

If $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$ and $\vec{u} = (q-r)\hat{i} + (r-p)\hat{j} + (p-q)\hat{k}$ and $\vec{v} = \frac{\hat{i}}{a} + \frac{\hat{j}}{b} + \frac{\hat{k}}{c}$,then:

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is:

If $A, B, C$ and $D$ are points whose position vectors are $\hat{i}+\hat{j}+\hat{k}, 4 \hat{i}-\hat{j}+2 \hat{k}, 5 \hat{i}+\hat{j}$ and $7 \hat{i}+2 \hat{j}+3 \hat{k}$ respectively,then the projection of $\vec{AB}$ on $\vec{CD}$ is