Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \alpha \hat{k}$, $\alpha > 0$.
If the projection of $\vec{a} \times \vec{b}$ on the vector $-\hat{i} + 2\hat{j} - 2\hat{k}$ is $30$, then $\alpha$ is equal to:
Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$ and $\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$. Then the vector product $\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$ is equal to :
Let $P$ be the point of intersection of the lines
$\dfrac{x-2}{1}=\dfrac{y-4}{5}=\dfrac{z-2}{1}$ and $\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{2}$.
Then, the shortest distance of $P$ from the line $4x=2y=z$ is:
If $\vec{\alpha} = (\lambda - 2)\vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors $\vec{a}$ and $\vec{b}$ which are non-collinear, then the value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear, is –
Let $\vec a=5\hat{\imath}-\hat{\jmath}-3\hat{k}$ and $\vec b=\hat{\imath}+3\hat{\jmath}+5\hat{k}$ be two vectors. Then which one of the following statements is TRUE?
Let the point $A$ divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $r:1\ (r>0)$. If $O$ is the origin and $(\overrightarrow{OQ}\cdot\overrightarrow{OA})-\dfrac{1}{5}\lvert\overrightarrow{OP}\times\overrightarrow{OA}\rvert^{2}=10$, then the value of $r$ is:
If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{m}{n}$, where $m$, $n$ are coprime numbers, then $m+n$ is equal to :
The distance of the line
$\displaystyle \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$
from the point $(1,4,0)$ along the line
$\displaystyle \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:
Let \(\vec a = 2\hat i - 7\hat j + 5\hat k\), \(\vec b = \hat i + \hat k\) and \(\vec c = \hat i + 2\hat j - 3\hat k\) be three given vectors.
If \(\vec r\) is a vector such that \(\vec r \times \vec a = \vec c \times \vec a\) and \(\vec r \cdot \vec b = 0\), then \(|\vec r|\) is equal to:
Let $\vec a=\hat i+\hat j+\hat k,;\vec b=2\hat i+4\hat j-5\hat k$ and $\vec c=x\hat i+2\hat j+3\hat k,;x\in\mathbb R$.
If $\vec d$ is the unit vector in the direction of $(\vec b+\vec c)$ such that $\vec a\cdot\vec d=1$, then $(\vec a\times\vec b)\cdot\vec c$ equals:
Let the values of $p$, for which the shortest distance between the lines
$\dfrac{x+1}{3}=\dfrac{y}{4}=\dfrac{z}{5}$ and $\vec r=(p\hat i+2\hat j+\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$
is $\dfrac{1}{\sqrt6}$, be $a,b$ $(a
One vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along the $x$, $y$ and $z$ axes are $3,\,4$ and $5$ units respectively.
Let $P$ be the vertex $(3,4,5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to the $z$–axis, not passing through $O$ or $P$, is:
If the line $\dfrac{2-x}{3}=\dfrac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\dfrac{x+3}{3\mu}=\dfrac{1-2y}{6}=\dfrac{5-z}{7}$, then $4\lambda+9\mu$ equals:
Let $\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$ be two vectors, such that $\vec{a} \times \vec{b} = -\hat{i} + 9\hat{j} + 12\hat{k}$. Then the projection of $\vec{b} - 2\vec{a}$ on $\vec{b} + \vec{a}$ is equal to:
Let $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$ and $\vec{c}=-\hat{i}+4\hat{j}+3\hat{k}$.
If $\vec{d}$ is a vector perpendicular to both $\vec{b}$ and $\vec{c}$, and $\vec{a}\cdot\vec{d}=18$, then $\lvert \vec{a}\times \vec{d}\rvert^{2}$ is equal to:
If $\overrightarrow a ,\,\,\overrightarrow b ,$ and $\overrightarrow C $ are unit vectors such that $\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,$ then $\left| {\overrightarrow a \times \overrightarrow c } \right|$ is equal to :
Let $\vec{a}=\hat{i}+2\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+\lambda\hat{j}+4\hat{k}$ and $\vec{c}=2\hat{i}+4\hat{j}+(\lambda^{2}-1)\hat{k}$ be coplanar vectors. Then the non-zero vector $\vec{a}\times\vec{c}$ is :
If the vectors $\overrightarrow {AB} = 3\widehat i + 4\widehat k$ and $\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$ are the sides of a triangle $ABC,$ then the length of the median through $A$ is :
Let the angle $\theta,;0<\theta<\tfrac{\pi}{2}$ between two unit vectors $\hat a$ and $\hat b$ be $\sin^{-1}\left(\tfrac{\sqrt{65}}{9}\right)$. If the vector $\vec c=3\hat a+6\hat b+9(\hat a\times\hat b)$, then the value of $9(\vec c\cdot\hat a)-3(\vec c\cdot\hat b)$ is
The distance of the point $(7,-2,11)$ from the line
$\dfrac{x-6}{1}=\dfrac{y-4}{0}=\dfrac{z-8}{3}$
along the line
$\dfrac{x-5}{2}=\dfrac{y-1}{-3}=\dfrac{z-5}{6}$ is:
Let in a $\triangle ABC$, the length of the side $AC$ be $6$,
the vertex $B$ be $(1, 2, 3)$ and the vertices $A, C$ lie on the line
$\dfrac{x - 6}{3} = \dfrac{y - 7}{2} = \dfrac{z - 7}{-2}$.
Then the area (in sq. units) of $\triangle ABC$ is:
Let the line passing through the points $(-1, 2, 1)$ and parallel to the line $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z}{4}$ intersect the line $\dfrac{x + 2}{3} = \dfrac{y - 3}{2} = \dfrac{z - 4}{1}$ at the point $P$.
Then the distance of $P$ from the point $Q(4, -5, 1)$ is
Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$.
If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c} = 5$, then $|\vec{c}|$ is equal to
If the shortest distance between the lines
$\dfrac{x-4}{2}=\dfrac{y+1}{3}=\dfrac{z-\lambda}{2}$
and
$\dfrac{x-2}{1}=\dfrac{y+1}{4}=\dfrac{z-2}{-3}$
is $\dfrac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is:
Let $\widehat a$, $\widehat b$ be unit vectors. If $\overrightarrow c $ be a vector such that the angle between $\widehat a$ and $\overrightarrow c $ is ${\pi \over {12}}$, and $\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right)$, then ${\left| {6\overrightarrow c } \right|^2}$ is equal to :
Let the vectors $\vec a=(1+t)\hat i+(1-t)\hat j+\hat k$, $\vec b=(1-t)\hat i+(1+t)\hat j+2\hat k$ and $\vec c=t\hat i-t\hat j+\hat k$, $t\in\mathbb R$ be such that for $\alpha,\beta,\gamma\in\mathbb R$, $\alpha\vec a+\beta\vec b+\gamma\vec c=\vec 0\Rightarrow \alpha=\beta=\gamma=0$. Then, the set of all values of $t$ is:
If the constant term in the expansion of $\left(\dfrac{\sqrt{3}}{x}+\dfrac{2x}{\sqrt{5}}\right)^{12}$, $x\ne 0$, is $\alpha\times 2^{8}\times\sqrt{3}$, then $25\alpha$ is:
Let $\vec a=2\hat i+5\hat j-\hat k$, $\vec b=2\hat i-2\hat j+2\hat k$ and $\vec c$ be three vectors such that $(\vec c+\hat i)\times(\vec a+\vec b+\hat i)=\vec a\times(\vec c+\hat i)$. If $\vec a\cdot\vec c=-29$, then $\vec c\cdot(-2\hat i+\hat j+\hat k)$ is equal to:
Let a vector $\vec{a}$ has magnitude $9$. Let a vector $\vec{b}$ be such that for every $(x,y)\in\mathbb{R}\times\mathbb{R}-{(0,0)}$, the vector $(x\vec{a}+y\vec{b})$ is perpendicular to the vector $(6y\vec{a}-18x\vec{b})$. Then the value of $|\vec{a}\times\vec{b}|$ is equal to:
If the position vectors of the vertices $A$, $B$ and $C$ of a $\triangle ABC$ are respectively $4\hat{i}+7\hat{j}+8\hat{k}$, $2\hat{i}+3\hat{j}+4\hat{k}$ and $2\hat{i}+5\hat{j}+7\hat{k}$, then the position vector of the point where the bisector of $\angle A$ meets $BC$ is :
Let $\vec a=\hat i+2\hat j+\hat k$, $\quad \vec b=3(\hat i-\hat j+\hat k)$.
Let $\vec c$ be the vector such that $\vec a\times\vec c=\vec b$ and $\vec a\cdot\vec c=3$.
Then $\vec a\cdot\big((\vec c\times\vec b)-\vec b-\vec c\big)$ is equal to:
If the equation of the line passing through the point $ \left( 0, -\frac{1}{2}, 0 \right) $ and perpendicular to the lines $ \vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right) $ and $ \vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right) $ is $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, then $ a+b+c+d $ is equal to :
Let the position vectors of the vertices $A,B,$ and $C$ of a triangle be
$2\hat i+2\hat j+\hat k$, $\ \hat i+2\hat j+2\hat k$ and $2\hat i+\hat j+2\hat k$ respectively.
Let $l_1,l_2,l_3$ be the lengths of perpendiculars drawn from the orthocenter of the triangle on the sides $AB,BC,$ and $CA$ respectively, then $l_1^{2}+l_2^{2}+l_3^{2}$ equals:
$\text{Consider the lines } L_{1}: , x-1=y-2=z \quad \text{and} \quad L_{2}: , x-2=y=z-1.$
$\text{Let the feet of the perpendiculars from the point } P(5,1,-3) \text{ on } L_{1} \text{ and } L_{2} \text{ be } Q \text{ and } R \text{ respectively.}$
$\text{If the area of the triangle } PQR \text{ is } A, \text{ then } 4A^{2}\text{ is equal to:}$
Let $\vec a$ and $\vec b$ be vectors of the same magnitude such that
$\displaystyle \frac{\lvert\vec a+\vec b\rvert+\lvert\vec a-\vec b\rvert}{\lvert\vec a+\vec b\rvert-\lvert\vec a-\vec b\rvert}=\sqrt2+1.$
Then $\displaystyle \frac{\lvert\vec a+\vec b\rvert^{2}}{\lvert\vec a\rvert^{2}}$ is:
If the shortest distance between the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{\lambda}$ and $\dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-5}{5}$ is $\dfrac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is:
Consider three vectors $\vec a,\vec b,\vec c$. Let $|\vec a|=2$, $|\vec b|=3$ and $\vec a=\vec b\times\vec c$. If $\alpha\in[0,\tfrac{\pi}{3}]$ is the angle between $\vec b$ and $\vec c$, then the minimum value of $27,|\vec c-\vec a|^{2}$ is:
Let the position vectors of three vertices of a triangle be $4\vec p+\vec q-3\vec r$, $-5\vec p+\vec q+2\vec r$ and $2\vec p-\vec q+2\vec r$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\dfrac{\vec p+\vec q+\vec r}{4}$ and $\alpha \vec p+\beta \vec q+\gamma \vec r$ respectively, then $\alpha+2\beta+5\gamma$ is equal to:
If the shortest distance between the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{\lambda}$ and $\dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-5}{5}$ is $\dfrac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is:
Let $\vec a = 3\hat{i}-\hat{j}+2\hat{k}$, $\vec b=\vec a \times (\hat{i}-2\hat{k})$ and $\vec c=\vec b \times \hat{k}$. Then the projection of $\vec c-2\hat{j}$ on $\vec a$ is:
$ \text{If the points with position vectors } \alpha\hat{i}+10\hat{j}+13\hat{k},; 6\hat{i}+11\hat{j}+11\hat{k},; \dfrac{9}{2}\hat{i}+\beta\hat{j}-8\hat{k} \text{ are collinear, then } (19\alpha-6\beta)^2 \text{ is equal to:} $
Let $\sqrt{3}\,\hat{i}+\hat{j}$, $\ \hat{i}+\sqrt{3}\,\hat{j}$ and $\ \beta\,\hat{i}+(1-\beta)\,\hat{j}$ respectively be the position vectors of the points $A$, $B$ and $C$ with respect to the origin $O$. If the distance of $C$ from the bisector of the acute angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is $\dfrac{3}{\sqrt{2}}$, then the sum of all possible values of $\beta$ is:
$ \text{Let } S \text{ be the set of all } a \in \mathbb{R} \text{ for which the angle between the vectors } \vec{u}=a(\log_{e} b),\hat{i}-6\hat{j}+3\hat{k} \text{ and } \vec{v}=(\log_{e} b),\hat{i}+2\hat{j}+2a(\log_{e} b),\hat{k},\ (b>1), \text{ is acute. Then } S \text{ is equal to:} $
The position vectors of the vertices $A,B,C$ of a triangle are
$2\hat i-3\hat j+3\hat k$, $2\hat i+2\hat j+3\hat k$ and $-\hat i+\hat j+3\hat k$
respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$
(where $D$ is on the line segment $BC$). Then $2l^{2}$ equals:
et $A(x,y,z)$ be a point in $xy$-plane, which is equidistant from three points $(0,3,2)$, $(2,0,3)$ and $(0,0,1)$. Let $B=(1,4,-1)$ and $C=(2,0,-2)$. Then among the statements
(S1): $\triangle ABC$ is an isosceles right angled triangle, and
(S2): the area of $\triangle ABC$ is $\dfrac{9\sqrt{2}}{2}$,
Let $\vec{u}$ be a vector coplanar with the vectors $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$. If $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{u}\cdot\vec{b}=24$, then $\lvert\vec{u}\rvert^{2}$ is equal to :
Let $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ ${a_i} > 0$, $i = 1,2,3$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\overrightarrow a $ on the vector $3\widehat i + 4\widehat j$ be 7. Let $\overrightarrow b $ be a vector obtained by rotating $\overrightarrow a $ with 90$^\circ$. If $\overrightarrow a $, $\overrightarrow b $ and x-axis are coplanar, then projection of a vector $\overrightarrow b $ on $3\widehat i + 4\widehat j$ is equal to:
Let $\vec a,\vec b,\vec c$ be three non-zero vectors such that $\vec b$ and $\vec c$ are non-collinear.
If $\vec a+5\vec b$ is collinear with $\vec c$, $\ \vec b+6\vec c$ is collinear with $\vec a$ and
$\vec a+\alpha\vec b+\beta\vec c=\vec 0$, then $\alpha+\beta$ is equal to:
Let O be the origin and the position vector of the point P be $ - \widehat i - 2\widehat j + 3\widehat k$. If the position vectors of the points A, B and C are $ - 2\widehat i + \widehat j - 3\widehat k,2\widehat i + 4\widehat j - 2\widehat k$ and $ - 4\widehat i + 2\widehat j - \widehat k$ respectively, then the projection of the vector $\overrightarrow {OP} $ on a vector perpendicular to the vectors $\overrightarrow {AB} $ and $\overrightarrow {AC} $ is :
Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{4}$. If $\theta$ is the angle between the vectors $(\hat{a}+\hat{b})$ and $(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$, then the value of $164 \,\cos ^{2} \theta$ is equal to :
Let $O$ be the origin and the position vectors of $A$ and $B$ be $2\hat i+2\hat j+\hat k$ and $2\hat i+4\hat j+4\hat k$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is:
If the image of the point $(4,4,3)$ in the line $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-1}{3}$ is $(\alpha,\beta,\gamma)$, then $\alpha+\beta+\gamma$ is equal to
The arc $PQ$ of a circle subtends a right angle at its centre $O$. The midpoint of the arc $PQ$ is $R$. If $\overrightarrow{OP}=\vec{u}$, $\overrightarrow{OR}=\vec{v}$ and $\overrightarrow{OQ}=\alpha\vec{u}+\beta\vec{v}$, then $\alpha,\ \beta^{2}$ are the roots of the equation:
Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $\alpha$ when divided by $3$
and $\beta$ when divided by $7$.
Then $(\alpha^2 + \beta^2)$ is equal to:
Let a unit vector $\hat{\mathbf u}=x\hat i+y\hat j+z\hat k$ make angles
$\dfrac{\pi}{2},\ \dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$ with the vectors
$\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat k$,
$\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat k$ and
$\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j$ respectively.
If $\vec v=\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat k$,
then $|\hat{\mathbf u}-\vec v|^{2}$ is equal to:
If vectors $\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$ and $\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$ are collinear, then a possible unit vector parallel to the vector $x\widehat i + y\widehat j + z\widehat k$ is :
Let a vector $\alpha \widehat i + \beta \widehat j$ be obtained by rotating the vector $\sqrt 3 \widehat i + \widehat j$ by an angle 45$^\circ$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($\alpha$, $\beta$), (0, $\beta$) and (0, 0) is equal to :
Let $\vec a=2\hat i+\hat j-\hat k,\quad \vec b=\big((\vec a\times(\hat i+\hat j))\times\hat i\big)\times\hat i.$
Then the square of the projection of $\vec a$ on $\vec b$ is:
Let $\vec{a} = 2\hat{i} + 7\hat{j} - \hat{k}$, $\vec{b} = 3\hat{i} + 5\hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$.
Let $\vec{d}$ be a vector which is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c} \cdot \vec{d} = 12$.
Then $(-\hat{i} + \hat{j} - \hat{k}) \cdot (\vec{c} \times \vec{d})$ is equal to:
Let $\vec a=6\hat i+\hat j-\hat k$ and $\vec b=\hat i+\hat j$. If $\vec c$ is a vector such that $|\vec c|\ge 6$, $\ \vec a\cdot\vec c=6|\vec c|$, $|\vec c-\vec a|=2\sqrt2$ and the angle between $\vec a\times\vec b$ and $\vec c$ is $60^\circ$, then $|(,(\vec a\times\vec b)\times\vec c,)|$ equals:
Let
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$,
$\vec{c} = \hat{j} - \hat{k}$,
and a vector $\vec{b}$ be such that
$\vec{a} \times \vec{b} = \vec{c}$
and
$\vec{a} \cdot \vec{b} = 3$.
Then $|\vec{b}|$ equals:
Let the position vectors of two points P and Q be 3$\widehat i$ $-$ $\widehat j$ + 2$\widehat k$ and $\widehat i$ + 2$\widehat j$ $-$ 4$\widehat k$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $-$1, 2) and ($-$2, 1, $-$2), respectively. Let lines PR and QS intersect at T. If the vector $\overrightarrow {TA} $ is perpendicular to both $\overrightarrow {PR} $ and $\overrightarrow {QS} $ and the length of vector $\overrightarrow {TA} $ is $\sqrt 5 $ units, then the modulus of a position vector of A is :
Let $\vec{a},\vec{b},\vec{c}$ be three coplanar concurrent vectors such that the angles between any two of them are the same. If the product of their magnitudes is $14$ and
$
(\vec{a}\times\vec{b})\cdot(\vec{b}\times\vec{c})
+(\vec{b}\times\vec{c})\cdot(\vec{c}\times\vec{a})
+(\vec{c}\times\vec{a})\cdot(\vec{a}\times\vec{b})=168,
$
then $|\vec{a}|+|\vec{b}|+|\vec{c}|$ is equal to:
If the points $\mathbf{P}$ and $\mathbf{Q}$ are respectively the circumcenter and the orthocentre of a $\triangle ABC$,
then $\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$ is equal to:
>Let $A, B, C$ be three points in xy-plane, whose position vector are given by $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $a \hat{i}+(1-a) \hat{j}$ respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ is $\frac{9}{\sqrt{2}}$, then the sum of all the possible values of $a$ is :
Let $\overrightarrow{OA}=\vec a,\ \overrightarrow{OB}=12\vec a+4\vec b$ and $\overrightarrow{OC}=\vec b$, where $O$ is the origin.
If $S$ is the parallelogram with adjacent sides $OA$ and $OC$, then
$\dfrac{\text{area of the quadrilateral }OABC}{\text{area of }S}$ is equal to:
Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is
Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$, $\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$ and $\overrightarrow c = \widehat i - \widehat j + \widehat k$ be three given vectors. Let $\overrightarrow v $ be a vector in the plane of $\overrightarrow a $ and $\overrightarrow b $ whose projection on $\overrightarrow c $ is ${2 \over {\sqrt 3 }}$. If $\overrightarrow v \,.\,\widehat j = 7$, then $\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)$ is equal to :
Let $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$ and $\vec b=b_1\hat i+b_2\hat j+b_3\hat k$ be two vectors such that
$|\vec a|=1,\ \vec a\cdot\vec b=2$ and $|\vec b|=4$. If $\vec c=2(\vec a\times\vec b)-3\vec b$, then the angle between $\vec b$ and $\vec c$ 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:
Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$.
Let $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$, $|(\vec{a}\times\vec{b})\times\vec{c}|=3$ and the angle between $\vec{c}$ and $\vec{a}\times\vec{b}$ is $30^\circ$.
Then $\vec{a}\cdot\vec{c}$ is equal to :
Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by
$\hat{i} + \hat{j}, \; \hat{i} + \hat{k}$ and $\hat{i} - \hat{j}, \; \hat{j} - \hat{k}$.
If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$
and $\vec{a} \cdot \vec{b} = 6$, then the ordered pair $(\theta, |\vec{a} \times \vec{b}|)$ is equal to:
Let $(\alpha,\beta,\gamma)$ be the foot of the perpendicular from the point $(1,2,3)$
on the line
\[
\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}.
\]
Then $19(\alpha+\beta+\gamma)$ is equal to:
Let $\vec{\alpha}=4\hat{i}+3\hat{j}+5\hat{k}$ and $\vec{\beta}=\hat{i}+2\hat{j}-4\hat{k}$.
Let $\vec{\beta}_{1}$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_{2}$ be perpendicular to $\vec{\alpha}$.
If $\vec{\beta}=\vec{\beta}_{1}+\vec{\beta}_{2}$, then the value of $5\,\vec{\beta}_{2}\cdot(\hat{i}+\hat{j}+\hat{k})$ is:
Let $\vec a,\vec b,\vec c$ be three non-zero vectors such that $\vec b$ and $\vec c$ are non-collinear. If $\ \vec a+5\vec b\ $ is collinear with $\vec c$, and $\ \vec b+6\vec c\ $ is collinear with $\vec a$, and $\ \vec a+\alpha,\vec b+\beta,\vec c=\vec 0$, then $\alpha+\beta$ is equal to:
Let $\vec a=\hat i+2\hat j+\hat k$ and $\vec b=2\hat i+\hat j-\hat k$. Let $\vec c$ be a unit vector in the plane of the vectors $\vec a$ and $\vec b$ and be perpendicular to $\vec a$. Then such a vector $\vec c$ is:
Let $A(2,3,5)$ and $C(-3,4,-2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD}= \hat{i}+2\hat{j}+3\hat{k}$, then the area of the parallelogram is equal to:
Let $\overrightarrow a = \widehat i + \widehat j - \widehat k$ and $\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$. Then the number of vectors $\overrightarrow b $ such that $\overrightarrow b \times \overrightarrow c = \overrightarrow a $ and $|\overrightarrow b | \in $ {1, 2, ........, 10} is :
$L_1:;\vec r=(2+\lambda),\hat i+(1-3\lambda),\hat j+(3+4\lambda),\hat k,;\lambda\in\mathbb R$
$L_2:;\vec r=2(1+\mu),\hat i+3(1+\mu),\hat j+(5+\mu),\hat k,;\mu\in\mathbb R$
is $\dfrac{m}{\sqrt{n}}$, where $\gcd(m,n)=1$, then the value of $m+n$ equals
Let $O$ be the origin and the position vectors of $A$ and $B$ be
$\vec{A} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ respectively.
If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is:
The vector $\vec{a}=-\hat{i}+2\hat{j}+\hat{k}$ is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\vec{b}$.
Then the projection of $3\vec{a}+\sqrt{2}\,\vec{b}$ on $\vec{c}=5\hat{i}+4\hat{j}+3\hat{k}$ is:
Let $\vec{a}=4\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=11\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b})\times\vec{c}=\vec{c}\times(-2\vec{a}+3\vec{b})$. If $(2\vec{a}+3\vec{b})\cdot\vec{c}=1670$, then $|\vec{c}|^{2}$ is equal to:
The distance of the point $P(4,6,-2)$ from the line passing through the point $(-3,2,3)$ and parallel to a line with direction ratios $3,3,-1$ is equal to:
Consider the lines $L_{1}$ and $L_{2}$ given by
$L_{1}:\ \dfrac{x-1}{2}=\dfrac{y-3}{1}=\dfrac{z-2}{2}$
$L_{2}:\ \dfrac{x-2}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$
A line $L_{3}$ having direction ratios $1,-1,-2$ intersects $L_{1}$ and $L_{2}$ at the points $P$ and $Q$ respectively.
Then the length of line segment $PQ$ is:
Let $\vec a=3\hat{i}+2\hat{j}+x\hat{k}$ and $\vec b=\hat{i}-\hat{j}+\hat{k}$, for some real $x$. Then $\left|\vec a\times\vec b\right|=r$ is possible if:
Let $L_1:\ \vec r=(\hat i-\hat j+2\hat k)+\lambda(\hat i-\hat j+2\hat k),\ \lambda\in\mathbb R,$
$L_2:\ \vec r=(\hat j-\hat k)+\mu(3\hat i+\hat j+p\hat k),\ \mu\in\mathbb R,$ and
$L_3:\ \vec r=\delta(\ell\hat i+m\hat j+n\hat k),\ \delta\in\mathbb R,$
be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$.
Then, the point which lies on $L_3$ is:
Let $\overrightarrow a $ and $\overrightarrow b $ be the vectors along the diagonals of a parallelogram having area $2\sqrt 2 $. Let the angle between $\overrightarrow a $ and $\overrightarrow b $ be acute, $|\overrightarrow a | = 1$, and $|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$. If $\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow b $, then an angle between $\overrightarrow b $ and $\overrightarrow c $ is :
Let $\vec a=\hat i+\alpha\hat j+\beta\hat k,\ \alpha,\beta\in\mathbb R$.
Let $\vec b$ be such that the angle between $\vec a$ and $\vec b$ is $\dfrac{\pi}{4}$ and $|\vec b|^{2}=6$.
If $\vec a\cdot\vec b=3\sqrt{2}$, then the value of $(\alpha^{2}+\beta^{2})\,|\vec a\times\vec b|^{2}$ is:
Let a unit vector $\hat{\mathbf{u}}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ make angles $\dfrac{\pi}{2},\ \dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$ with the vectors
$\dfrac{1}{\sqrt{2}}\mathbf{i}+\dfrac{1}{\sqrt{2}}\mathbf{k},\ \dfrac{1}{\sqrt{2}}\mathbf{j}+\dfrac{1}{\sqrt{2}}\mathbf{k},\ \dfrac{1}{\sqrt{2}}\mathbf{i}+\dfrac{1}{\sqrt{2}}\mathbf{j}$ respectively.
If $\vec{\mathbf{v}}=\dfrac{1}{\sqrt{2}}\mathbf{i}+\dfrac{1}{\sqrt{2}}\mathbf{j}+\dfrac{1}{\sqrt{2}}\mathbf{k}$, then $|\hat{\mathbf{u}}-\vec{\mathbf{v}}|^{2}$ is equal to:
Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+3\hat{j}-5\hat{k}$ and $\vec{c}=3\hat{i}-\hat{j}+\lambda\hat{k}$ be three vectors.
Let $\vec{r}$ be a unit vector along $\vec{b}+\vec{c}$.
If $\vec{r}\cdot\vec{a}=3$, then $3\lambda$ is equal to:
The foot of the perpendicular from the point $(2,0,5)$ on the line
$\dfrac{x+1}{2}=\dfrac{y-1}{5}=\dfrac{z+1}{-1}$ is $(\alpha,\beta,\gamma)$.
Then, which of the following is NOT correct?
Let $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$, $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$.
If a vector $\vec{d}$ satisfies $\vec{d}\times\vec{b}=\vec{c}\times\vec{b}$ and $\vec{d}\cdot\vec{a}=24$,
then $|\vec{d}|^{2}$ is equal to:
Let O be the origin. Let $\overrightarrow{OP} = x\widehat i + y\widehat j - \widehat k$ and $\overrightarrow{OQ} = -\widehat i + 2\widehat j + 3x\widehat k$, $x, y \in R, x > 0$, be such that $|\overrightarrow{PQ}| = \sqrt{20}$ and the vector $\overrightarrow{OP}$ is perpendicular $\overrightarrow{OQ}$. If $\overrightarrow{OR} = 3\widehat i + z\widehat j - 7\widehat k$, $z \in R$, is coplanar with $\overrightarrow{OP}$ and $\overrightarrow{OQ}$, then the value of $x^2 + y^2 + z^2$ is equal to :
Let three vectors $\vec a=\alpha\hat i+4\hat j+2\hat k,;\vec b=5\hat i+3\hat j+4\hat k,;\vec c=x\hat i+y\hat j+z\hat k$ form a triangle such that $\vec c=\vec a-\vec b$ and the area of the triangle is $5\sqrt{6}$. If $\alpha$ is a positive real number, then $\lvert\vec c\rvert$ is equal to:
Let $\overrightarrow{OA}=\vec a,\ \overrightarrow{OB}=12\vec a+4\vec b$ and $\overrightarrow{OC}=\vec b$, where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $OA$ and $OC$, then $\dfrac{\text{area of quadrilateral }OABC}{\text{area of }S}$ is equal to:
Let $\vec{\alpha}=3\hat{i}+\hat{j}$ and $\vec{\beta}=2\hat{i}-\hat{j}+3\hat{k}$. If $\vec{\beta}=\vec{\beta}{1}-\vec{\beta}{2}$, where $\vec{\beta}{1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta}{2}$ is perpendicular to $\vec{\alpha}$, then $\vec{\beta}{1}\times\vec{\beta}{2}$ is equal to:
If the vector $\vec{b} = 3\vec{j} + 4\vec{k}$ is written as the sum of a vector $\vec{b_1}$ parallel to
$\vec{a} = \vec{i} + \vec{j}$ and a vector $\vec{b_2}$ perpendicular to $\vec{a}$,
then $\vec{b_1} \times \vec{b_2}$ is equal to:
A vector $\overrightarrow a $ has components 3p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, $\overrightarrow a $ has components p + 1 and $\sqrt {10} $, then the value of p is equal to :
Let $|\vec a|=2$, $|\vec b|=3$ and the angle between the vectors $\vec a$ and $\vec b$ be $\dfrac{\pi}{4}$. Then $|(\vec a+2\vec b)\times(2\vec a-3\vec b)|^2$ is equal to:
Let $\vec a=3\hat i+\hat j-2\hat k,\ \vec b=4\hat i+\hat j+7\hat k$ and $\vec c=\hat i-3\hat j+4\hat k$ be three vectors.
If a vector $\vec p$ satisfies $\vec p\times\vec b=\vec c\times\vec b$ and $\vec p\cdot\vec a=0$,
then $\vec p\cdot(\hat i-\hat j-\hat k)$ is equal to:
The distance of the point $Q(0,2,-2)$ from the line passing through the point
$P(5,-4,3)$ and perpendicular to the lines
$\ \vec r = (-3\hat i + 2\hat k) + \lambda(2\hat i + 3\hat j + 5\hat k),\ \lambda\in\mathbb R,$
and
$\ \vec r = (\hat i - 2\hat j + \hat k) + \mu(-\hat i + 3\hat j + 2\hat k),\ \mu\in\mathbb R,$
is:
Let $\overrightarrow a $ and $\overrightarrow b $ be two vectors such that $\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$ and the angle between $\overrightarrow a $ and $\overrightarrow b $ is 60$^\circ$. If ${1 \over 8}\overrightarrow a $ is a unit vector, then $\left| {\overrightarrow b } \right|$ is equal to :
Let $\overrightarrow{OA}=2\vec a,\ \overrightarrow{OB}=6\vec a+5\vec b,\ \overrightarrow{OC}=3\vec b$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units, then the area (in sq. units) of the quadrilateral $OABC$ is equal to:
Let $\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$, where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overrightarrow a $ and $\overrightarrow b $ is $\sqrt {15({\alpha ^2} + 4)} $, then the value of $2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a \,.\,\overrightarrow b } \right){\left| {\overrightarrow b } \right|^2}$ is equal to :
If $\vec{a}$ is a nonzero vector such that its projections on the vectors
$2\hat{i} - \hat{j} + 2\hat{k}$, $\hat{i} + 2\hat{j} - 2\hat{k}$ and $\hat{k}$ are equal,
then a unit vector along $\vec{a}$ is:
For a triangle $ABC$,
$\overrightarrow{AB}=-2\hat i+\hat j+3\hat k$
$\overrightarrow{CB}=\alpha\hat i+\beta\hat j+\gamma\hat k$
$\overrightarrow{CA}=4\hat i+3\hat j+\delta\hat k$
If $\delta>0$ and the area of the triangle $ABC$ is $5\sqrt{6}$, then $\overrightarrow{CB}\cdot\overrightarrow{CA}$ is equal to:
Let a, b c $ \in $ R be such that a2 + b2 + c2 = 1. If $a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$, where${\theta = {\pi \over 9}}$, then the angle between the vectors $a\widehat i + b\widehat j + c\widehat k$ and $b\widehat i + c\widehat j + a\widehat k$ is
Let $\overrightarrow a $ and $\overrightarrow b $ be two non-zero vectors perpendicular to each other and $|\overrightarrow a | = |\overrightarrow b |$. If $|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$, then the angle between the vectors $\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)$ and ${\overrightarrow a }$ is equal to :
If the vectors $\vec a=\lambda\,\hat i+\mu\,\hat j+4\,\hat k$, $\vec b=-2\,\hat i+4\,\hat j-2\,\hat k$ and
$\vec c=2\,\hat i+3\,\hat j+\hat k$ are coplanar and the projection of $\vec a$ on the vector $\vec b$ is
$\sqrt{54}$ units, then the sum of all possible values of $\lambda+\mu$ is equal to:
Let ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and
$\overrightarrow{(AB-BC)}+\overrightarrow{(AD-DC)}=k\,\overrightarrow{FE}$, then $k$ is equal to:
Let $(\alpha,\beta,\gamma)$ be the mirror image of the point $(2,3,5)$ in the line
\[
\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}.
\]
Then, $\,2\alpha+3\beta+4\gamma\,$ is equal to:
In a triangle ABC, if $|\overrightarrow {BC} | = 8,|\overrightarrow {CA} | = 7,|\overrightarrow {AB} | = 10$, then the projection of the vector $\overrightarrow {AB} $ on $\overrightarrow {AC} $ is equal to :
Let $\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$, $\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$ and $\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$ where $\alpha ,\,\beta \in R$, be three vectors. If the projection of $\overrightarrow a $ on $\overrightarrow c $ is ${{10} \over 3}$ and $\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$, then the value of $\alpha + \beta $ is equal to :
Between the following two statements:
Statement I:
Let $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$.
Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r} = \vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r} = 0$
is of magnitude $\sqrt{10}$.
Statement II:
In a triangle $ABC$, $\cos 2A + \cos 2B + \cos 2C \geq -\dfrac{3}{2}$.
If a unit vector $\vec{a}$ makes angles $\dfrac{\pi}{3}$ with $\hat{i}$, $\dfrac{\pi}{4}$ with $\hat{j}$ and $\theta\in(0,\pi)$ with $\hat{k}$, then a value of $\theta$ is:
Let $S$ be the set of all values of $\lambda$ for which the shortest distance between the lines
$\dfrac{x-\lambda}{0}=\dfrac{y-3}{4}=\dfrac{z+6}{1}$ and $\dfrac{x+\lambda}{3}=\dfrac{y}{-4}=\dfrac{z-6}{0}$ is $13$.
Then $8\Big|\displaystyle\sum_{\lambda\in S}\lambda\Big|$ is equal to:
Let \(\vec{a} = 4\hat{i} + 3\hat{j}\) and \(\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}\).
If \(\vec{c}\) is a vector such that
\[
\vec{c}\cdot(\vec{a}\times\vec{b}) + 25 = 0,\qquad
\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k}) = 4,
\]
and the projection of \(\vec{c}\) on \(\vec{a}\) is \(1\), then the projection of \(\vec{c}\) on \(\vec{b}\) equals:
The line $L_1$ is parallel to the vector $\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ and passes through the point $(7, 6, 2)$,
and the line $L_2$ is parallel to the vector $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$ and passes through the point $(5, 3, 4)$.
The shortest distance between the lines $L_1$ and $L_2$ is:
The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line $\dfrac{x+2}{3}=\dfrac{y-1}{0}=\dfrac{z}{4}$ such that $BC=5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1,-1,2)$, is:
Let $\vec a=2\hat i+\alpha\hat j+\hat k,\ \vec b=-\hat i+\hat k,\ \vec c=\beta\hat j-\hat k$, where $\alpha,\beta$ are integers and $\alpha\beta=-6$.
Let the values of the ordered pair $(\alpha,\beta)$ for which the area of the parallelogram whose diagonals are $\vec a+\vec b$ and $\vec b+\vec c$ is $\dfrac{\sqrt{21}}{2}$ be $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$.
Then $\alpha_1^{,2}+\beta_1^{,2}-\alpha_2\beta_2$ is equal to:
Let $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$, and a vector $\vec{c}$ be such that
$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$ and $\vec{a} \cdot \vec{c} = 3$.
If $\vec{b} \times \vec{c} = \vec{d}$, then $|\vec{a} \cdot \vec{d}|$ is equal to:
In a triangle $ABC$, right angled at the vertex $A$, if the position vectors of $A,B$ and $C$ are respectively
$3\hat{i} + \hat{j} - \hat{k}$, $-\hat{i} + 3\hat{j} + p\hat{k}$ and $5\hat{i} + q\hat{j} - 4\hat{k}$,
then the point $(p,q)$ lies on a line:
$ \text{Let A, B, C be three points whose position vectors respectively are } \vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}, ; \vec{b} = 2\hat{i} + \alpha \hat{j} + 4\hat{k}, ; \alpha \in \mathbb{R}, ; \vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}. ; \text{If } \alpha \text{ is the smallest positive integer for which } \vec{a}, \vec{b}, \vec{c} \text{ are non-collinear, then the length of the median in } \triangle ABC \text{ through A is :}$
The shortest distance between the lines
$\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1}$
and
$\dfrac{x + 2}{-1} = \dfrac{y - 4}{8} = \dfrac{z - 5}{4}$
lies in the interval:
Let a unit vector $\overrightarrow{OP}$ make angles $\alpha,\beta,\gamma$ with the positive directions of the coordinate axes $OX, OY, OZ$ respectively, where $\beta\in\left(0,\tfrac{\pi}{2}\right)$. If $\overrightarrow{OP}$ is perpendicular to the plane through points $(1,2,3)$, $(2,3,4)$ and $(1,5,7)$, then which one of the following is true?
Let $L_1:\ \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $L_2:\ \dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$ be two lines. Which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?
Let $A(3,0,-1),; B(2,10,6)$ and $C(1,2,1)$ be the vertices of a triangle and $M$ be the midpoint of $AC$. If $G$ divides $BM$ in the ratio $2:1$, then $\cos(\angle GOA)$ ($O$ being the origin) is equal to:
In a triangle ABC, if $\left| {\overrightarrow {BC} } \right| = 3$, $\left| {\overrightarrow {CA} } \right| = 5$ and $\left| {\overrightarrow {BA} } \right| = 7$, then the projection of the vector $\overrightarrow {BA} $ on $\overrightarrow {BC} $ is equal to :
Let $\vec a$ and $\vec b$ be two vectors. Let $|\vec a|=1$, $|\vec b|=4$ and $\vec a\cdot\vec b=2$. If $\vec c=(2\,\vec a\times\vec b)-3\vec b$, then the value of $\vec b\cdot\vec c$ is:
(A) $-48$
If the shortest distance between the lines
\[
\frac{x-\lambda}{2}=\frac{y-2}{1}=\frac{z-1}{1}
\quad\text{and}\quad
\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}
\]
is $1$, then the sum of all possible values of $\lambda$ is:
Consider a $\triangle ABC$ where $A(1,3,2)$, $B(-2,8,0)$ and $C(3,6,7)$.
If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:
Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive $y$- and $z$-axes, respectively, is half of the angle that this line makes with the positive $x$-axis. Then the sum of all possible values of the angle $\beta$ is
Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3\hat i+2\hat j+2\hat k$. If the distance of the point $Q$ from the point $R(1,3,3)$ is $5$, then the square of the area of $\triangle PQR$ is equal to:
Let $\triangle ABC$ be a triangle whose circumcentre is at $P$.
If the position vectors of $A, B, C$ and $P$ are $\vec a, \vec b, \vec c$ and $\dfrac{\vec a + \vec b + \vec c}{4}$ respectively,
then the position vector of the orthocentre of this triangle is:
Let $\vec a=\hat i+\hat j+\sqrt{2}\,\hat k$, $\vec b=b_1\hat i+b_2\hat j+\sqrt{2}\,\hat k$, $\vec c=5\hat i+\hat j+\sqrt{2}\,\hat k$ be three vectors such that the projection vector of $\vec b$ on $\vec a$ is $\vec a$. If $\vec a+\vec b$ is perpendicular to $\vec c$, then $|\vec b|$ is equal to:
Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda \vec{a}+2 \vec{b}$ and $3 \vec{a}-\lambda \vec{b}$ are perpendicular to each other, then the number of values of $\lambda$ in $[-1,3]$ is :
Let $\vec a=\hat{\imath}+2\hat{\jmath}+3\hat{k}$, $\vec b=\hat{\imath}-\hat{\jmath}+2\hat{k}$ and $\vec c=5\hat{\imath}-3\hat{\jmath}+3\hat{k}$ be three vectors. If $\vec r$ is a vector such that $\vec r\times\vec b=\vec c\times\vec b$ and $\vec r\cdot\vec a=0$, then $25\lvert\vec r\rvert^{2}$ is equal to:
If the mirror image of the point $P(3, 4, 9)$ in the line
$\dfrac{x-1}{3} = \dfrac{y+1}{2} = \dfrac{z-2}{1}$
is $(\alpha, \beta, \gamma)$, then $14(\alpha + \beta + \gamma)$ is:
Let $\vec a=2\hat i+\lambda_{1}\hat j+3\hat k$, $\vec b=4\hat i+(3-\lambda_{2})\hat j+6\hat k$, and $\vec c=3\hat i+6\hat j+(\lambda_{3}-1)\hat k$ be three vectors such that $\vec b=2\vec a$ and $\vec a$ is perpendicular to $\vec c$. Then a possible value of $(\lambda_{1},\lambda_{2},\lambda_{3})$ is:
Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\dfrac{x-3}{7}=\dfrac{y-2}{-1}=\dfrac{z+1}{-2}$. Then the area of the right-angled triangle $PQR$, where $R$ is the point $(3,-2,1)$, is:
The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $\sqrt {13.44} $, then the standard deviation of the second sample is :
Let a unit vector which makes an angle of $60^\circ$ with $\,2\hat i+2\hat j-\hat k\,$ and an angle of $45^\circ$ with $\,\hat i-\hat k\,$ be $\vec C$.
Then $\displaystyle \vec C+\Big(-\tfrac12\,\hat i+\tfrac{1}{3\sqrt2}\,\hat j-\tfrac{\sqrt2}{3}\,\hat k\Big)$ is:
Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be $\hat{i}+2\hat{j}+\hat{k}$, $\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median through $A$ of $\triangle ABC$ at the point $E$. If the length of $AD$ is $\dfrac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\dfrac{\sqrt{805}}{6\sqrt{2}}$, then the position vector of $E$ is:
Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$,
farther from the origin and at a distance of $9$ units from the point $P$, be $(\alpha,\beta,\gamma)$.
Then $\alpha^2+\beta^2+\gamma^2$ is equal to:
Consider two vectors $\vec{u}=3\hat{i}-\hat{j}$ and $\vec{v}=2\hat{i}+\hat{j}-\lambda\hat{k},\ \lambda>0$. The angle between them is given by $\cos^{-1}!\left(\dfrac{\sqrt{5}}{2\sqrt{7}}\right)$. Let $\vec{v}=\vec{v}_1+\vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$. Then the value $\left|\vec{v}_1\right|^{2}+\left|\vec{v}_2\right|^{2}$ is equal to
Let $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$ be two vectors.
If a vector perpendicular to both the vectors $\vec a+\vec b$ and $\vec a-\vec b$ has magnitude $12$, then one such vector is:
If $\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$ and $\left| {\overrightarrow a \times \overrightarrow b } \right|$ = 8, then $\left| {\overrightarrow a .\,\overrightarrow b } \right|$ is equal to :