Vectors Questions - JEE

Let $OPQR$ is a tetrahedon such that O is origin and $\vec{p}, \vec{q}, \vec{r}$ are position vectors of P, Q, R respectively and $\alpha$ is the angle which OP makes with face PQR then:

If $\vec{r} = l(\vec{b} \times \vec{c}) + m(\vec{c} \times \vec{a}) + n(\vec{a} \times \vec{b})$ and $[\vec{b}\vec{c}\vec{c}] = 2$, then $l + m + n$ is equal to:

If vectors $\vec{b} = (\tan\alpha, -1, 2\sqrt{\sin\frac{\alpha}{2}})$ and $\vec{c} = (\tan\alpha, \tan\alpha, -\frac{3}{\sqrt{\sin\alpha/2}})$ are orthogonal and vector $\vec{a} = (1, 3, \sin2\alpha)$ makes an obtuse angle with the z-axis then:

Let position vector of point $A$ be $\vec{i} + \vec{j} + \vec{k}$ and that of point $B$ be $-\vec{i} + \vec{k}$, then the position vector of point $R(\vec{r})$ such that $AR$ is perpendicular to $BR$ and $\vec{r}$ is not perpendicular to $\vec{r} - (\vec{j} + 2\vec{k})$ is:

If three coterminous edges of a tetrahedron are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 4$, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$, $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{4}$ and $\vec{c}$ and $\vec{a}$ is $\frac{\pi}{6}$. The area of the base is $2$ sq. units, then the height of the tetrahedron is:

If $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) \cdot (\vec{c} \times \vec{b}) = 0$ then which of the following is always true:

If $\vec{\alpha}$ and $\vec{\beta}$ be two perpendicular unit vectors such that $\vec{x} = \vec{\beta} - (\vec{a} \times \vec{x})$, then $|\vec{x}|$ is equal to:

If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are on a circle of radius $R$ whose centre is at origin and $\vec{c} - \vec{a}$ is perpendicular to $\vec{d} - \vec{b}$, then $|\vec{d} - \vec{a}|^2 + |\vec{b} - \vec{c}|^2$ (AC is diameter)

The vector, directed along the internal bisector of the angle between the vectors $\vec{a} = 7\vec{i} - 4\vec{j} - 4\vec{k}$ & $\vec{b} = -2\vec{i} - \vec{j} + 2\vec{k}$ with $|\vec{c}| = 5\sqrt{6}$ is:

Let $ABC$ be a triangle and $\vec{a},\vec{b},\vec{c}$ be the position vectors of the point $A, B, C$ respectively. External bisectors of $\angle B$ and $\angle C$ meet at $P$ with the sides of the triangle as $a, b, c$, the position vector of $P$ becomes :

If the interior and exterior bisectors of the angle $A$ of a triangle $ABC$ meet the side $BC$ at $D$ and $E$, then :

If $ABC$ be a triangle of sides $a, b, c$ with position vectors of $A, B, C$ as $\vec{a}, \vec{b}$ and $\vec{c}$ respectively, then the position vector of its incentre is:

The position vectors of the points $A, B, C$ are respectively $(1,1,1), (1,-1,2), (0,2,-1)$. The unit vector parallel to the plane determined by $A, B, C$ and perpendicular to the vector $(1,0,1)$ is/are:

A line passes through the points whose position vectors are $\vec{r} + \vec{j} - 2\vec{k}$ and $\vec{r} - 3\vec{j} + \vec{k}$. The position vector of a point on it at a unit distance from the first point is:

The volume of a right triangular prism $ABC A_1 B_1 C_1$ is equal to 3. If the position vectors of the vertices of the base $ABC$ are $A(1, 0, 1); B(2,0, 0)$ and $C(0, 1, 0)$ the position vectors of the vertex $A_1$ can be:

If non-zero vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ satisfy $(\vec{a} \times \vec{b}) \cdot \vec{c} = |\vec{a}||\vec{b}||\vec{c}|$ holds then:

Three points having position vectors $\vec{a}, \vec{b}$ and $\vec{c}$ will be collinear if:

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are any four vectors, then $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d})$ is a vector:

$\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $\alpha(\alpha \in [0, \pi])$ to each other and $|\vec{a} + \vec{b}| < 1$ then $\alpha$ can lie in:

The equation of the plane $P_1$ parallel to line $L: \frac{x-1}{2}=\frac{y-0}{3}=\frac{z-1}{1}$, at distance 6 from the intersection of $L$ with plane $P: 2x+3y+z-18=0$, and having maximum distance from $Q(5,4,-4)$, is $ax+by+cz+d=0$. The value of $\dfrac{bd}{ac}$ is:

Let $\gamma\in\mathbb{R}$ such that $L_1: \frac{x+1}{11}=\frac{y+2}{21}=\frac{z+3}{29}$ and $L_2: \frac{x+3}{16}=\frac{y+2}{11}=\frac{z+4}{\gamma}$ intersect at $R_1$. Match P)$\gamma$; Q)unit normal $\hat{n}$; R)$\overrightarrow{OR_1}$; S)$\overrightarrow{OR_1}\cdot\hat{n}$ with List-II: 1)$-\hat{i}-\hat{j}+\hat{k}$; 2)$\frac{3}{2}$; 3)$1$; 4)$\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$; 5)$\frac{2}{3}$

There is a plane $P$ passing through the line $\dfrac{x+1}{2}=\dfrac{y-1}{1}=\dfrac{z+3}{2}$ and parallel to the $x$-axis. Let $\vec{v}$ be a vector parallel to plane $P$ such that the angle between $\vec{v}$ and $\vec{u}=\hat{i}+\hat{j}+\hat{k}$ is least. There is a circle lying on plane $P$ with centre $C(0,0,-5)$ and radius $\sqrt{10}$. A line $L$ on plane $P$ is tangent to the circle at a point $Q(x,y,z)$ with the least possible value of $z$. Then which of the following is/are true?

Let $\vec{a},\vec{b},\vec{c},\vec{d}$ be position vectors of $A,B,C,D$ with $|\vec{a}|=1$, $|\vec{b}|=2$, $|\vec{c}|=3$, $|\vec{d}|=4$, $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=24$, $\vec{a}\cdot\vec{c}=\frac{3}{2}$, $\vec{b}\cdot\vec{c}>0$. Area of $\triangle ABC=\frac{1}{4}(2\alpha+\beta\sqrt{3})$ and area of $\triangle BCD=\frac{1}{2}(m-n\sqrt{3})$ where $\alpha,\beta,m,n\in\mathbb{N}$. Match P)$\alpha$, Q)$\beta$, R)$m$, S)$n$ with 1)1, 2)3, 3)4, 4)9.