The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat { \mathrm { i } } - 2 \hat { \mathrm { j } } + 2 \widehat { \mathrm { k } }$ and $\alpha \hat { \mathrm { i } } + 2 \alpha \hat { \mathrm { j } } - 2 \widehat { \mathrm { k } }$ is acute, is $\_\_\_\_$.
The set of all $\alpha$, for which the vectors $\vec { a } = \alpha t \hat { i } + 6 \hat { j } - 3 \hat { k }$ and $\vec { b } = t \hat { i } - 2 \hat { j } - 2 \alpha t \hat { k }$ are inclined at an obtuse angle for all $t \in \mathbb { R }$, is (1) $\left( - \frac { 4 } { 3 } , 1 \right)$ (2) $[ 0,1 )$ (3) $\left( - \frac { 4 } { 3 } , 0 \right]$ (4) $( - 2,0 ]$
Let $P ( x , y , z )$ be a point in the first octant, whose projection in the $x y$-plane is the point $Q$. Let $O P = \gamma$; the angle between $O Q$ and the positive $x$-axis be $\theta$; and the angle between $O P$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is (1) $\gamma \sqrt { 1 - \sin ^ { 2 } \phi \cos ^ { 2 } \theta }$ (2) $\gamma \sqrt { 1 - \sin ^ { 2 } \theta \cos ^ { 2 } \phi }$ (3) $\gamma \sqrt { 1 + \cos ^ { 2 } \phi \sin ^ { 2 } \theta }$ (4) $\gamma \sqrt { 1 + \cos ^ { 2 } \theta \sin ^ { 2 } \phi }$
Q21. The resultant of two vectors $\vec { A }$ and $\vec { B }$ is perpendicular to $\vec { A }$ and its magnitude is half that of $\vec { B }$. The angle between vectors $\vec { A }$ and $\vec { B }$ is $\_\_\_\_$ ○.
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1$, $|\vec{b}| = 2$ and the angle formed by $\vec{a}$ and $\vec{b}$ is $60^\circ$. Set $\vec{u} = x\vec{a} + \vec{b}$ and $\vec{v} = x\vec{a} - \vec{b}$ for a real number $x$. When $x > 1$, we are to find the value of $x$ such that the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$. In the following, $\vec{u} \cdot \vec{v}$ denotes the inner product of $\vec{u}$ and $\vec{v}$, and $\vec{a} \cdot \vec{b}$ denotes the inner product of $\vec{a}$ and $\vec{b}$. First of all, since the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$, we obtain $$(\vec{u} \cdot \vec{v})^2 = \frac{\mathbf{A}}{\mathbf{B}}|\vec{u}|^2|\vec{v}|^2.$$ When we express this equation in terms of $x$, noting $\vec{a} \cdot \vec{b} = \mathbf{C}$, we have $$x^4 - \mathbf{DE}x^2 + \mathbf{FG} = 0.$$ By transforming this, we also have $$\left(x^2 - \mathbf{H}\right)^2 = (\mathbf{I}x)^2.$$ When this is solved for $x$, we obtain $$x = \mathbf{J} + \sqrt{\mathbf{KL}},$$ noting $x > 1$.
Let $a , b$ be real numbers, and let $O$ be the origin of the coordinate plane. It is known that the graph of the quadratic function $f ( x ) = a x ^ { 2 }$ and the circle $\Omega : x ^ { 2 } + y ^ { 2 } - 3 y + b = 0$ both pass through point $P \left( 1 , \frac { 1 } { 2 } \right)$, and let point $C$ be the center of $\Omega$. Find the cosine of the angle between vectors $\overrightarrow { C O }$ and $\overrightarrow { C P }$.
Let $\mathrm{A}(0,\ -1,\ 1)$ be a point in coordinate space. Suppose a point $\mathrm{P}$ in the $xy$-plane satisfies all of the following conditions (i), (ii), (iii).
[(i)] $\mathrm{P}$ is different from the origin $\mathrm{O}$.
[(ii)] $\angle \mathrm{AOP} \geq \dfrac{2}{3}\pi$
[(iii)] $\angle \mathrm{OAP} \leq \dfrac{\pi}{6}$
Sketch the region that $\mathrm{P}$ can occupy in the $xy$-plane. %% Page 2