1981
JEE Advanced
MCQ
IIT-JEE 1981
The scalar $\overrightarrow A .\left( {\overrightarrow B + \overrightarrow C } \right) \times \left( {\overrightarrow A + \overrightarrow B + \overrightarrow C } \right)$ equals :
A.
$0$
B.
$\left[ {\overrightarrow A \,\overrightarrow B \,\overrightarrow C } \right] + \left[ {\overrightarrow B \,\overrightarrow C \,\overrightarrow A } \right]$
C.
$\left[ {\overrightarrow A \,\overrightarrow B \,\overrightarrow C } \right]$
D.
None of these
2025
JEE Advanced
Numerical
JEE Advanced 2025 Paper 2 Online
Consider the vectors
$ \vec{x}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}, \quad \vec{y}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}, \quad \text { and } \quad \vec{z}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} $
For two distinct positive real numbers $\alpha$ and $\beta$, define
$ \vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}, \quad \vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}, \quad \text { and } \quad \vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y} . $
If the vectors $\vec{X}, \vec{Y}$, and $\vec{Z}$ lie in a plane, then the value of $\alpha+\beta-3$ is ____________.
Correct Answer: -2
Explanation:
$ \begin{aligned} & {[\vec{x} \vec{y} \vec{z}]=0} \\ & \Rightarrow\left|\begin{array}{ccc} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{array}\right| \underbrace{\left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right|}_{\neq 0}=0 \\ & \Rightarrow\left(\alpha^3+\beta^3-1\right)-(-\alpha \beta-\alpha \beta-\alpha \beta)=0 \\ & \Rightarrow \alpha^3+\beta^3+3 \alpha \beta=1 \\ & \Rightarrow \alpha^3+\beta^3+(-1)^3=3(\alpha)(\beta)(-1) \\ & \Rightarrow \alpha+\beta-1=0 \end{aligned} $
So, $\alpha+\beta-3=-2$
2025
JEE Advanced
Numerical
JEE Advanced 2025 Paper 1 Online
For any two points $M$ and $N$ in the $XY$-plane, let $\overrightarrow{MN}$ denote the vector from $M$ to $N$, and $\vec{0}$ denote the zero vector. Let $P, Q$ and $R$ be three distinct points in the $XY$-plane. Let $S$ be a point inside the triangle $\triangle PQR$ such that
$\overrightarrow{SP} + 5\; \overrightarrow{SQ} + 6\; \overrightarrow{SR} = \vec{0}.$
Let $E$ and $F$ be the mid-points of the sides $PR$ and $QR$, respectively. Then the value of
$\frac{\text { length of the line segment } E F}{\text { length of the line segment } E S}$
is ________________.
Correct Answer: 1.15TO1.25
2024
JEE Advanced
Numerical
JEE Advanced 2024 Paper 2 Online
Let $\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$. If for some real numbers $\alpha, \beta$, and $\gamma$, we have
$ 15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} \times \vec{q}), $
then the value of $\gamma$ is ________.
Correct Answer: 2
Explanation:
$15 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=\alpha(2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})+\beta(\overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}})+\gamma(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$
taking dot with $(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$
$ \begin{aligned} & \left|\begin{array}{ccc} 15 & 10 & 6 \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{array}\right|=0+0+{\gamma}\left(\mathrm{p}^2 \mathrm{q}^2-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2\right) \quad\left[\because(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})^2+(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2=\mathrm{p}^2 \mathrm{q}^2\right] \\ & \Rightarrow 52=26 \boldsymbol{\gamma} \\ & \therefore \gamma=2 \end{aligned}$
2024
JEE Advanced
Numerical
JEE Advanced 2024 Paper 1 Online
Let $\overrightarrow{O P}=\frac{\alpha-1}{\alpha} \hat{i}+\hat{j}+\hat{k}, \overrightarrow{O Q}=\hat{i}+\frac{\beta-1}{\beta} \hat{j}+\hat{k}$ and $\overrightarrow{O R}=\hat{i}+\hat{j}+\frac{1}{2} \hat{k}$ be three vectors, where $\alpha, \beta \in \mathbb{R}-\{0\}$ and $O$ denotes the origin. If $(\overrightarrow{O P} \times \overrightarrow{O Q}) \cdot \overrightarrow{O R}=0$ and the point $(\alpha, \beta, 2)$ lies on the plane $3 x+3 y-z+l=0$, then the value of $l$ is ____________.
Correct Answer: 5
Explanation:
$\begin{aligned} & (\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OQ}}) \cdot \overrightarrow{\mathrm{OR}}=0 \\ & \left|\begin{array}{ccc} \frac{\alpha-1}{\alpha} & 1 & 1 \\ 1 & \frac{\beta-1}{\beta} & 1 \\ 1 & 1 & \frac{1}{2} \end{array}\right|=0 \end{aligned}$
$\begin{array}{ll} \qquad\alpha+\beta+1=0 \quad \text{... (i)}\\ \text { Also } \quad (\alpha, \beta, 2) \text { lies on } 3 \mathrm{x}+3 \mathrm{y}-\mathrm{z}+l=0 \\ \Rightarrow \quad 3 \alpha+3 \beta-2+l=0 \quad \Rightarrow \quad l=2-3(\alpha+\beta) \\ \text { use (1) in it } \Rightarrow l=5 \end{array}$
2023
JEE Advanced
Numerical
JEE Advanced 2023 Paper 1 Online
Let $P$ be the plane $\sqrt{3} x+2 y+3 z=16$ and let
$S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1\right.$ and the distance of $(\alpha, \beta, \gamma)$ from the plane $P$ is $\left.\frac{7}{2}\right\}$.
Let $\vec{u}, \vec{v}$ and $\vec{w}$ be three distinct vectors in $S$ such that $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\vec{u}, \vec{v}$ and $\vec{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ is :
Correct Answer: 45
Explanation:
$\begin{aligned} & P: \sqrt{3} x+2 y+3 z=16 \\\\ & \qquad S=\left\{\alpha \hat{\imath}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1,\right. \left.\quad d_p=\frac{7}{2}\right\} \\\\ & \because|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}| .............. (1)\end{aligned}$
$\vec{u}, \vec{v}, \vec{w}$ are elements of set $S$ and in set $S$ magnitude of vector is 1
$\therefore \vec{u}, \vec{v}, \vec{w}$ are unit vectors and by equation (1) we can system $\vec{u}, \vec{v}, \vec{w}$ are equally inclined and vertices of equilateral triangle also lying on a circle which is intersection of sphere $|\vec{r}|=1$
Distance from origin to $P$,
$ d=\frac{|-16|}{\sqrt{3+4+9}}=\frac{16}{4}=4 $
$\therefore$ Plane containing $\hat{u}, \hat{v}, \hat{w}$ are at a distance
$4-\frac{7}{2}=\frac{1}{2}$ from origin and Parallel to $\sqrt{3 x}+2 y+3 z$ $=16$.
$\therefore$ Equation of the plane is
$ \begin{array}{rlrl} & \sqrt{3 x}+2 y+3 z & =\gamma \\\\ & \therefore \frac{1}{2} =\left|\frac{\gamma}{4}\right| \\\\ & \Rightarrow \gamma = \pm 2 \\\\ & \sqrt{3 x}+2 y+3 z =2 \end{array} $
Equation of sphere $x^2+y^2+z^2=1$
$\therefore$ Radius or circle
$\begin{aligned} r & =\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} \\\\ \text { then } \quad \frac{a}{2} & =\frac{\sqrt{3}}{2} \cos 30^{\circ} \\\\ a & =\sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}\end{aligned}$
$\therefore$ Area or triangle
$ =\frac{\sqrt{3}}{2} a^2=\frac{\sqrt{3}}{2} \times \frac{9}{4}=\frac{9 \sqrt{3}}{16} $
$\therefore$ Volume of Parallelepiped
$ \begin{aligned} & =2 \times \frac{1}{2} \times \frac{9 \sqrt{3}}{16} \\\\ V & =\frac{9 \sqrt{3}}{16} \end{aligned} $
$\therefore$ Volume of Parallelepiped
$ \begin{aligned} = & 2 \times \frac{1}{2} \times \frac{9 \sqrt{3}}{16} \\\\ V & =\frac{9 \sqrt{3}}{16} \end{aligned} $
$\therefore \frac{80 \mathrm{~V}}{\sqrt{3}}=\frac{80}{\sqrt{3}} \times \frac{9 \sqrt{3}}{16}=45$
$\vec{u}, \vec{v}, \vec{w}$ are elements of set $S$ and in set $S$ magnitude of vector is 1
$\therefore \vec{u}, \vec{v}, \vec{w}$ are unit vectors and by equation (1) we can system $\vec{u}, \vec{v}, \vec{w}$ are equally inclined and vertices of equilateral triangle also lying on a circle which is intersection of sphere $|\vec{r}|=1$
Distance from origin to $P$,
$ d=\frac{|-16|}{\sqrt{3+4+9}}=\frac{16}{4}=4 $
$\therefore$ Plane containing $\hat{u}, \hat{v}, \hat{w}$ are at a distance
$4-\frac{7}{2}=\frac{1}{2}$ from origin and Parallel to $\sqrt{3 x}+2 y+3 z$ $=16$.
$\therefore$ Equation of the plane is
$ \begin{array}{rlrl} & \sqrt{3 x}+2 y+3 z & =\gamma \\\\ & \therefore \frac{1}{2} =\left|\frac{\gamma}{4}\right| \\\\ & \Rightarrow \gamma = \pm 2 \\\\ & \sqrt{3 x}+2 y+3 z =2 \end{array} $
Equation of sphere $x^2+y^2+z^2=1$
$\therefore$ Radius or circle
$\begin{aligned} r & =\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} \\\\ \text { then } \quad \frac{a}{2} & =\frac{\sqrt{3}}{2} \cos 30^{\circ} \\\\ a & =\sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}\end{aligned}$
$\therefore$ Area or triangle
$ =\frac{\sqrt{3}}{2} a^2=\frac{\sqrt{3}}{2} \times \frac{9}{4}=\frac{9 \sqrt{3}}{16} $
$\therefore$ Volume of Parallelepiped
$ \begin{aligned} & =2 \times \frac{1}{2} \times \frac{9 \sqrt{3}}{16} \\\\ V & =\frac{9 \sqrt{3}}{16} \end{aligned} $
$\therefore$ Volume of Parallelepiped
$ \begin{aligned} = & 2 \times \frac{1}{2} \times \frac{9 \sqrt{3}}{16} \\\\ V & =\frac{9 \sqrt{3}}{16} \end{aligned} $
$\therefore \frac{80 \mathrm{~V}}{\sqrt{3}}=\frac{80}{\sqrt{3}} \times \frac{9 \sqrt{3}}{16}=45$
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
Let $\overrightarrow u $, $\overrightarrow v $ and $\overrightarrow w $ be vectors in three-dimensional space, where $\overrightarrow u $ and $\overrightarrow v $ are unit vectors which are not perpendicular to each other and $\overrightarrow u $ . $\overrightarrow w $ = 1, $\overrightarrow v $ . $\overrightarrow w $ = 1, $\overrightarrow w $ . $\overrightarrow w $ = 4
If the volume of the paralleopiped, whose adjacent sides are represented by the vectors, $\overrightarrow u $, $\overrightarrow v $ and $\overrightarrow w $, is $\sqrt 2 $, then the value of $\left| {3\overrightarrow u + 5\overrightarrow v } \right|$ is ___________.
If the volume of the paralleopiped, whose adjacent sides are represented by the vectors, $\overrightarrow u $, $\overrightarrow v $ and $\overrightarrow w $, is $\sqrt 2 $, then the value of $\left| {3\overrightarrow u + 5\overrightarrow v } \right|$ is ___________.
Correct Answer: 7
Explanation:
Given: $|\vec{u}|=1,|\vec{v}|=1$
$\vec{u}$ is not perpendicular to $\vec{v}$ $\vec{u} \cdot \vec{v} \neq 0$
And $\vec{u} \cdot \vec{w}=1 \quad \vec{v} \cdot \vec{w}=1 \quad \vec{w} \cdot \vec{w}=4$
$ \begin{aligned} & \Rightarrow \vec{w} \cdot \vec{w}=|\vec{w}|^2=4 \\\\ & \Rightarrow|\vec{w}|=2 \end{aligned} $
Volume of parallelepiped with $\vec{u}, \vec{v}$ and $\vec{w}$ as its sides,
$ \begin{aligned} & =[\vec{u} \vec{v} \vec{w}]=\sqrt{2} \\\\ & \text { Now, }[\vec{u} \vec{v} \vec{w}]^2=\left|\begin{array}{ccc} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} & \vec{v} \cdot \vec{w} \\ \vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & \vec{w} \cdot \vec{w} \end{array}\right|=2 \\\\ & \Rightarrow\left|\begin{array}{ccc} 1 & \vec{u} \cdot \vec{v} & 1 \\ \vec{v} \cdot \vec{u} & 1 & 1 \\ 1 & 1 & 4 \end{array}\right|=2 \end{aligned} $
$ \begin{aligned} & \Rightarrow 1(4-1)-\vec{u} \cdot \vec{v}(4 \vec{u} \cdot \vec{v}-1)+1(\vec{u} \cdot \vec{v}-1)=2 \\\\ & \Rightarrow 3-4(\vec{u} \vec{v})^2+\vec{u} \vec{v}+\vec{u} \vec{v}-1=2 \\\\ & \Rightarrow-4(\vec{u} \vec{v})^2+2 \vec{u} \vec{v}+2=2 \\\\ & \Rightarrow-4(\vec{u} \vec{v})^2+2 \vec{u} \vec{v}=0 \\\\ & \Rightarrow 2 \vec{u} \vec{v}(-2 \vec{u} \vec{v}+1)=0 \\\\ & \Rightarrow \vec{u} \cdot \vec{v}=\frac{1}{2}, \vec{u} \cdot \vec{v} \neq 0 \\\\ & \text { Now, }|3 \vec{u}+5 \vec{v}|=\sqrt{9+25+30\left(\frac{1}{2}\right)} \\\\ & \Rightarrow|3 \vec{u}+5 \vec{v}|=\sqrt{49} \\\\ & \Rightarrow|3 \vec{u}+5 \vec{v}|=7 \end{aligned} $
$\vec{u}$ is not perpendicular to $\vec{v}$ $\vec{u} \cdot \vec{v} \neq 0$
And $\vec{u} \cdot \vec{w}=1 \quad \vec{v} \cdot \vec{w}=1 \quad \vec{w} \cdot \vec{w}=4$
$ \begin{aligned} & \Rightarrow \vec{w} \cdot \vec{w}=|\vec{w}|^2=4 \\\\ & \Rightarrow|\vec{w}|=2 \end{aligned} $
Volume of parallelepiped with $\vec{u}, \vec{v}$ and $\vec{w}$ as its sides,
$ \begin{aligned} & =[\vec{u} \vec{v} \vec{w}]=\sqrt{2} \\\\ & \text { Now, }[\vec{u} \vec{v} \vec{w}]^2=\left|\begin{array}{ccc} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} & \vec{v} \cdot \vec{w} \\ \vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & \vec{w} \cdot \vec{w} \end{array}\right|=2 \\\\ & \Rightarrow\left|\begin{array}{ccc} 1 & \vec{u} \cdot \vec{v} & 1 \\ \vec{v} \cdot \vec{u} & 1 & 1 \\ 1 & 1 & 4 \end{array}\right|=2 \end{aligned} $
$ \begin{aligned} & \Rightarrow 1(4-1)-\vec{u} \cdot \vec{v}(4 \vec{u} \cdot \vec{v}-1)+1(\vec{u} \cdot \vec{v}-1)=2 \\\\ & \Rightarrow 3-4(\vec{u} \vec{v})^2+\vec{u} \vec{v}+\vec{u} \vec{v}-1=2 \\\\ & \Rightarrow-4(\vec{u} \vec{v})^2+2 \vec{u} \vec{v}+2=2 \\\\ & \Rightarrow-4(\vec{u} \vec{v})^2+2 \vec{u} \vec{v}=0 \\\\ & \Rightarrow 2 \vec{u} \vec{v}(-2 \vec{u} \vec{v}+1)=0 \\\\ & \Rightarrow \vec{u} \cdot \vec{v}=\frac{1}{2}, \vec{u} \cdot \vec{v} \neq 0 \\\\ & \text { Now, }|3 \vec{u}+5 \vec{v}|=\sqrt{9+25+30\left(\frac{1}{2}\right)} \\\\ & \Rightarrow|3 \vec{u}+5 \vec{v}|=\sqrt{49} \\\\ & \Rightarrow|3 \vec{u}+5 \vec{v}|=7 \end{aligned} $
2019
JEE Advanced
Numerical
JEE Advanced 2019 Paper 2 Offline
Let $\overrightarrow a = 2\widehat i + \widehat j - \widehat k$ and $\overrightarrow b = \widehat i + 2\widehat j + \widehat k$ be two vectors. Consider a vector c = $\alpha $$\overrightarrow a$ + $\beta $$\overrightarrow b$, $\alpha $, $\beta $ $ \in $ R. If the projection of $\overrightarrow c$ on the vector ($\overrightarrow a$ + $\overrightarrow b$) is $3\sqrt 2 $, then the
minimum value of ($\overrightarrow c$ $-$($\overrightarrow a$ $ \times $ $\overrightarrow b$)).$\overrightarrow c$ equals ................
minimum value of ($\overrightarrow c$ $-$($\overrightarrow a$ $ \times $ $\overrightarrow b$)).$\overrightarrow c$ equals ................
Correct Answer: 18
Explanation:
Given vectors $\overrightarrow a $$ = 2\widehat i + \widehat j - \widehat k$
and $\overrightarrow b = \widehat i + 2\widehat j + \widehat k$
So, $\overrightarrow a + \overrightarrow b = 3\widehat i + 3\widehat j \Rightarrow |\overrightarrow a + \overrightarrow b| = 3\sqrt 2 $
Since, it is given that projection of $\overrightarrow c $ = $\alpha $a + $\beta $b on the vector ($\overrightarrow a $ + $\overrightarrow b $) is $3\sqrt 2 $, then
${{(\overrightarrow a + \overrightarrow b ).\overrightarrow c } \over {|\overrightarrow a + \overrightarrow b|}} = 3\sqrt 2 $
$ \Rightarrow (\overrightarrow a + \overrightarrow b).(\alpha \overrightarrow a + \beta \overrightarrow b) = 18$
$ \Rightarrow \alpha (\overrightarrow a.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) + \alpha (\overrightarrow b.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) = 18$
$ \Rightarrow 6\alpha + 3\beta + 3\alpha + 6\beta = 18$
$ \Rightarrow 9\alpha + 9\beta = 18 \Rightarrow (\alpha + \beta ) = 2$ .....(i)
Now, for minimum value of
($\overrightarrow c$ $-$($\overrightarrow a$ $ \times $ $\overrightarrow b$)).$\overrightarrow c$
$ = (\alpha \overrightarrow a + \beta \overrightarrow b - (\overrightarrow a \times \overrightarrow b)).(\alpha \overrightarrow a + \beta \overrightarrow b)$
$ = {\alpha ^2}(\overrightarrow a.\overrightarrow a) + \alpha \beta (\overrightarrow a.\overrightarrow b) + \alpha \beta (\overrightarrow a.\overrightarrow b) + {\beta ^2}(\overrightarrow b.\overrightarrow b)$
[$ \because $ ($\overrightarrow a$ $ \times $ $\overrightarrow b$) . $\overrightarrow a$ = 0 = ($\overrightarrow a $$ \times $ $\overrightarrow b$) . $\overrightarrow b$]
$6{\alpha ^2} + 6\alpha \beta + 6{\beta ^2} = 6({\alpha ^2} + {\beta ^2} + \alpha \beta )$
$ = 6\,[{(\alpha + \beta )^2} - \alpha \beta ] = 6\,[4 - \alpha \beta ]$
$ = 6\,[4 - \alpha (2 - \alpha )]$
$ = 6\,[4 - 2\alpha + {\alpha ^2}]$
Let f ( $\alpha $) = $4 - 2\alpha + {\alpha ^2}$
f′( $\alpha $) = –2 + 2$\alpha $
At maximum and minimum f′( $\alpha $) = 0
$ \Rightarrow $ –2 + 2$\alpha $ $ \Rightarrow $ $\alpha $ = 1
f′'( $\alpha $) = 2 (+ve)
Therefore, minimum value of $4 - 2\alpha + {\alpha ^2}$ is (4 – 2 + 1) = 3.
$ \therefore $ The minimum value of
$6(4 - 2\alpha + {\alpha^2}) = 6(3) = 18$
and $\overrightarrow b = \widehat i + 2\widehat j + \widehat k$
So, $\overrightarrow a + \overrightarrow b = 3\widehat i + 3\widehat j \Rightarrow |\overrightarrow a + \overrightarrow b| = 3\sqrt 2 $
Since, it is given that projection of $\overrightarrow c $ = $\alpha $a + $\beta $b on the vector ($\overrightarrow a $ + $\overrightarrow b $) is $3\sqrt 2 $, then
${{(\overrightarrow a + \overrightarrow b ).\overrightarrow c } \over {|\overrightarrow a + \overrightarrow b|}} = 3\sqrt 2 $
$ \Rightarrow (\overrightarrow a + \overrightarrow b).(\alpha \overrightarrow a + \beta \overrightarrow b) = 18$
$ \Rightarrow \alpha (\overrightarrow a.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) + \alpha (\overrightarrow b.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) = 18$
$ \Rightarrow 6\alpha + 3\beta + 3\alpha + 6\beta = 18$
$ \Rightarrow 9\alpha + 9\beta = 18 \Rightarrow (\alpha + \beta ) = 2$ .....(i)
Now, for minimum value of
($\overrightarrow c$ $-$($\overrightarrow a$ $ \times $ $\overrightarrow b$)).$\overrightarrow c$
$ = (\alpha \overrightarrow a + \beta \overrightarrow b - (\overrightarrow a \times \overrightarrow b)).(\alpha \overrightarrow a + \beta \overrightarrow b)$
$ = {\alpha ^2}(\overrightarrow a.\overrightarrow a) + \alpha \beta (\overrightarrow a.\overrightarrow b) + \alpha \beta (\overrightarrow a.\overrightarrow b) + {\beta ^2}(\overrightarrow b.\overrightarrow b)$
[$ \because $ ($\overrightarrow a$ $ \times $ $\overrightarrow b$) . $\overrightarrow a$ = 0 = ($\overrightarrow a $$ \times $ $\overrightarrow b$) . $\overrightarrow b$]
$6{\alpha ^2} + 6\alpha \beta + 6{\beta ^2} = 6({\alpha ^2} + {\beta ^2} + \alpha \beta )$
$ = 6\,[{(\alpha + \beta )^2} - \alpha \beta ] = 6\,[4 - \alpha \beta ]$
$ = 6\,[4 - \alpha (2 - \alpha )]$
$ = 6\,[4 - 2\alpha + {\alpha ^2}]$
Let f ( $\alpha $) = $4 - 2\alpha + {\alpha ^2}$
f′( $\alpha $) = –2 + 2$\alpha $
At maximum and minimum f′( $\alpha $) = 0
$ \Rightarrow $ –2 + 2$\alpha $ $ \Rightarrow $ $\alpha $ = 1
f′'( $\alpha $) = 2 (+ve)
Therefore, minimum value of $4 - 2\alpha + {\alpha ^2}$ is (4 – 2 + 1) = 3.
$ \therefore $ The minimum value of
$6(4 - 2\alpha + {\alpha^2}) = 6(3) = 18$
2018
JEE Advanced
Numerical
JEE Advanced 2018 Paper 1 Offline
Let a and b be two unit vectors such that a . b = 0. For some x, y$ \in $R, let $\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b $. If | $\overrightarrow c $| = 2 and the vector c is inclined at the same angle $\alpha $ to both a and b, then the value of $8{\cos ^2}\alpha $ is ..............
Correct Answer: 3
Explanation:
We have,
$\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b $ and $\overrightarrow a .\overrightarrow b $ = 0
|$\overrightarrow a $| = |$\overrightarrow b $| = 1
and |$\overrightarrow c $| = 2
Also, given $\overrightarrow c $ is inclined on $\overrightarrow a $ and $\overrightarrow b $ with same angle $\alpha $.
$ \therefore $ $\overrightarrow a .\overrightarrow c = x|\overrightarrow a {|^2} + y(\overrightarrow a .\overrightarrow b ) + \overrightarrow a .(\overrightarrow a \times \overrightarrow b )$
$|\overrightarrow a ||\overrightarrow c |cos\alpha = x + 0 + 0$
x = 2cos$\alpha $
Similarly,
$|\overrightarrow b ||\overrightarrow c |cos\alpha = 0 + y + 0$
$ \Rightarrow $ y = 2cos$\alpha $
$|\overrightarrow c {|^2} = {x^2} + {y^2} + |\overrightarrow a \, \times \,\overrightarrow b {|^2}$
$4 = 8{\cos ^2}\alpha + |a{|^2}|b{|^2}{\sin ^2}90^\circ $
$4 = 8{\cos ^2}\alpha + 1$
$ \Rightarrow $ $8{\cos ^2}\alpha $ = 3
$\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b $ and $\overrightarrow a .\overrightarrow b $ = 0
|$\overrightarrow a $| = |$\overrightarrow b $| = 1
and |$\overrightarrow c $| = 2
Also, given $\overrightarrow c $ is inclined on $\overrightarrow a $ and $\overrightarrow b $ with same angle $\alpha $.
$ \therefore $ $\overrightarrow a .\overrightarrow c = x|\overrightarrow a {|^2} + y(\overrightarrow a .\overrightarrow b ) + \overrightarrow a .(\overrightarrow a \times \overrightarrow b )$
$|\overrightarrow a ||\overrightarrow c |cos\alpha = x + 0 + 0$
x = 2cos$\alpha $
Similarly,
$|\overrightarrow b ||\overrightarrow c |cos\alpha = 0 + y + 0$
$ \Rightarrow $ y = 2cos$\alpha $
$|\overrightarrow c {|^2} = {x^2} + {y^2} + |\overrightarrow a \, \times \,\overrightarrow b {|^2}$
$4 = 8{\cos ^2}\alpha + |a{|^2}|b{|^2}{\sin ^2}90^\circ $
$4 = 8{\cos ^2}\alpha + 1$
$ \Rightarrow $ $8{\cos ^2}\alpha $ = 3
2015
JEE Advanced
Numerical
JEE Advanced 2015 Paper 2 Offline
Suppose that $\overrightarrow p ,\overrightarrow q $ and $\overrightarrow r $ are three non-coplanar vectors in ${R^3}$. Let the components of a vector $\overrightarrow s $ along $\overrightarrow p ,$ $\overrightarrow q $ and $\overrightarrow r $ be $4, 3$ and $5,$ respectively. If the components of this vector $\overrightarrow s $ along $\left( { - \overrightarrow p + \overrightarrow q + \overrightarrow r } \right),\left( {\overrightarrow p - \overrightarrow q + \overrightarrow r } \right)$ and $\left( { - \overrightarrow p - \overrightarrow q + \overrightarrow r } \right)$ are $x, y$ and $z,$ respectively, then the value of $2x+y+z$ is
Correct Answer: 9
Explanation:
Here, $\overrightarrow s = 4\overrightarrow p + 3\overrightarrow q + 5\overrightarrow r $ ....... (i)
and $\overrightarrow s = ( - \overrightarrow p + \overrightarrow q + \overrightarrow r )x + (\overrightarrow p - \overrightarrow q + \overrightarrow r )y + ( - \overrightarrow p - \overrightarrow q + \overrightarrow r )z$ ...... (ii)
$\therefore$ $4\overrightarrow p + 3\overrightarrow q + 5\overrightarrow r = \overrightarrow p ( - x + y - z) + \overrightarrow q (x - y - z) + \overrightarrow r (x + y + z)$
On comparing both sides, we get
$ - x + y - z = 4$, $x - y - z = 3$ and $x + y + z = 5$
On solving above equations, we get
$x = 4$, $y = {9 \over 2}$, $z = {{ - 7} \over 2}$
$\therefore$ $2x + y + z = 8 + {9 \over 2} - {7 \over 2} = 9$
2014
JEE Advanced
Numerical
JEE Advanced 2014 Paper 1 Offline
Let $\overrightarrow a \,\,,\,\,\overrightarrow b $ and $\overrightarrow c $ be three non-coplanar unit vectors such that the angle between every pair of them is ${\pi \over 3}.$ If $\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = p\overrightarrow a + q\overrightarrow b + r\overrightarrow c ,$ where $p,q$ and $r$ are scalars, then the value of ${{{p^2} + 2{q^2} + {r^2}} \over {{q^2}}}$ is
Correct Answer: 4
Explanation:
Given $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}=p \vec{a}+q \vec{b}+r \vec{c}$ ..........(1)
Taking dot product with $\vec{a}$ :
Hence,
$ 0+\vec{a} \cdot \vec{b} \times \vec{c}=p(1 \cdot 1 \cdot \cos 0)+q\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right)+r\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right) $
$\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}=p+\frac{q}{2}+\frac{r}{2}$ ..........(2)
Taking the dot product of (1) with $\vec{b}$ :
$ 0+0=\frac{p}{2}+q+\frac{r}{2} $ ............(3)
Taking the dot product of (1) with $\vec{c}$ :
$ \vec{c} \cdot \vec{a} \times \vec{b}+0=\frac{p}{2}+\frac{q}{2}+r$ .............(4)
From (2) and (4), we get
$ \begin{aligned} p+\frac{q}{2}+\frac{r}{2} & =\frac{p}{2}+\frac{q}{2}+r \\\\ \frac{p}{2} & =\frac{r}{2} \Rightarrow p=r \end{aligned} $
Now, from Eq. (3), we get $0=\frac{r}{2}+q+\frac{r}{2} \Rightarrow q=-r$.
Now, $\frac{p^2+2 q^2+r^2}{q^2}=\frac{r^2+2(-r)^2+r^2}{(-r)^2}=\frac{4 r^2}{r^2}=4$.
Taking dot product with $\vec{a}$ :
Hence,
$ 0+\vec{a} \cdot \vec{b} \times \vec{c}=p(1 \cdot 1 \cdot \cos 0)+q\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right)+r\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right) $
$\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}=p+\frac{q}{2}+\frac{r}{2}$ ..........(2)
Taking the dot product of (1) with $\vec{b}$ :
$ 0+0=\frac{p}{2}+q+\frac{r}{2} $ ............(3)
Taking the dot product of (1) with $\vec{c}$ :
$ \vec{c} \cdot \vec{a} \times \vec{b}+0=\frac{p}{2}+\frac{q}{2}+r$ .............(4)
From (2) and (4), we get
$ \begin{aligned} p+\frac{q}{2}+\frac{r}{2} & =\frac{p}{2}+\frac{q}{2}+r \\\\ \frac{p}{2} & =\frac{r}{2} \Rightarrow p=r \end{aligned} $
Now, from Eq. (3), we get $0=\frac{r}{2}+q+\frac{r}{2} \Rightarrow q=-r$.
Now, $\frac{p^2+2 q^2+r^2}{q^2}=\frac{r^2+2(-r)^2+r^2}{(-r)^2}=\frac{4 r^2}{r^2}=4$.
2012
JEE Advanced
Numerical
IIT-JEE 2012 Paper 1 Offline
If $\overrightarrow a ,\overrightarrow b $ and $\overrightarrow c $ are unit vectors satisfying
${\left| {\overrightarrow a - \overrightarrow b } \right|^2} + {\left| {\overrightarrow b - \overrightarrow c } \right|^2} + {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = 9,$ then $\left| {2\overrightarrow a + 5\overrightarrow b + 5\overrightarrow c } \right|$ is
${\left| {\overrightarrow a - \overrightarrow b } \right|^2} + {\left| {\overrightarrow b - \overrightarrow c } \right|^2} + {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = 9,$ then $\left| {2\overrightarrow a + 5\overrightarrow b + 5\overrightarrow c } \right|$ is
Correct Answer: 3
Explanation:
$\begin{aligned} & \text { Given, }|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9 \\ & \Rightarrow|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b}+|\vec{b}|^2 \\ &|\vec{c}|^2-2 \vec{b} \cdot \vec{c}+|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=9 \\ & \Rightarrow 2\left(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\right)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\ & \Rightarrow 2(1+1+1)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\ & \Rightarrow(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2} \quad \text{... (i)} \\ & \operatorname{Now},|\vec{a}+\vec{b}+\vec{c}|^2 \geq 0 \\ & \Rightarrow|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \geq 0 \\ & \Rightarrow 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \geq 0 \\ & \Rightarrow(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \geq \frac{-3}{2} \quad \text{... (ii)} \end{aligned}$
From (i) and (ii), we get
$\begin{array}{lr} \Rightarrow & (\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2} \\ \Rightarrow & |\vec{a}+\vec{b}+\vec{c}|=0 \\ \Rightarrow & \vec{a}+\vec{b}+\vec{c}=0 \\ \Rightarrow & \vec{b}+\vec{c}=-\vec{a} \quad \text{.... (iii)} \end{array}$
$\begin{aligned} \therefore \quad|2 \vec{a}+5 \vec{b}+5 \vec{c}| & =|2 \vec{a}+5(\vec{b}+\vec{c})| \\ & =|2 \vec{a}+5(-\vec{a})| \quad[\text { [from (iii)] } \\ & =|-3 \vec{a}| \\ \begin{aligned} & =|3 \vec{a}| \\ & =3 \quad[\because \vec{a} \text { is unit vector }] \end{aligned} \end{aligned}$
2011
JEE Advanced
Numerical
IIT-JEE 2011 Paper 2 Offline
Let $\overrightarrow a = - \widehat i - \widehat k,\overrightarrow b = - \widehat i + \widehat j$ and $\overrightarrow c = \widehat i + 2\widehat j + 3\widehat k$ be three given vectors. If $\overrightarrow r $ is a vector such that $\overrightarrow r \times \overrightarrow b = \overrightarrow c \times \overrightarrow b $ and $\overrightarrow r .\overrightarrow a = 0,$ then the value of $\overrightarrow r .\overrightarrow b $ is
Correct Answer: 9
Explanation:
Since it is given that $\overrightarrow r \times \overrightarrow b = \overrightarrow c \times \overrightarrow b $, taking cross product with
$\overrightarrow a \times (\overrightarrow r \times \overrightarrow b ) = \overrightarrow a \times (\overrightarrow c \times \overrightarrow b )$
$(\overrightarrow a \,.\,\overrightarrow b )\overrightarrow r - (\overrightarrow a \,.\,\overrightarrow r )\overrightarrow b = \overrightarrow a \times (\overrightarrow c \times \overrightarrow b )$
$ \Rightarrow \overrightarrow r = - 3\widehat i + 6\widehat j + 3\widehat k$
$\overrightarrow r \,.\,\overrightarrow b = 3 + 6 = 9$
2010
JEE Advanced
Numerical
IIT-JEE 2010 Paper 1 Offline
If $\overrightarrow a $ and $\overrightarrow b $ are vectors in space given by $\overrightarrow a = {{\widehat i - 2\widehat j} \over {\sqrt 5 }}$ and $\overrightarrow b = {{2\widehat i + \widehat j + 3\widehat k} \over {\sqrt {14} }},$ then find the value of $\,\left( {2\overrightarrow a + \overrightarrow b } \right).\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a - 2\overrightarrow b } \right)} \right].$
Correct Answer: 5
Explanation:
$ \begin{aligned} & (2 \vec{a}+\vec{b})[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})] \\\\ & =4(\vec{a} \cdot \vec{a})+\vec{b} \cdot \vec{b}+4 \vec{a} \cdot \vec{b} \text { where } \\\\ & \vec{a} \cdot \vec{b}=\frac{2-2}{\sqrt{70}}=0 \quad|\vec{a}|=1 \text { and }|\vec{b}|=1 \\\\ & =5 \end{aligned} $
2022
JEE Advanced
MSQ
JEE Advanced 2022 Paper 2 Online
Let $\hat{\imath}, \hat{\jmath}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let
$ \begin{aligned} & \vec{a}=3 \hat{\imath}+\hat{\jmath}-\hat{k} \text {, } \\ & \vec{b}=\hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}, \quad b_{2}, b_{3} \in \mathbb{R} \text {, } \\ & \vec{c}=c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}, \quad c_{1}, c_{2}, c_{3} \in \mathbb{R} \end{aligned} $
be three vectors such that $b_{2} b_{3}>0, \vec{a} \cdot \vec{b}=0$ and
$ \left(\begin{array}{ccc} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ b_{2} \\ b_{3} \end{array}\right)=\left(\begin{array}{r} 3-c_{1} \\ 1-c_{2} \\ -1-c_{3} \end{array}\right) . $
Then, which of the following is/are TRUE?
$ \begin{aligned} & \vec{a}=3 \hat{\imath}+\hat{\jmath}-\hat{k} \text {, } \\ & \vec{b}=\hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}, \quad b_{2}, b_{3} \in \mathbb{R} \text {, } \\ & \vec{c}=c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}, \quad c_{1}, c_{2}, c_{3} \in \mathbb{R} \end{aligned} $
be three vectors such that $b_{2} b_{3}>0, \vec{a} \cdot \vec{b}=0$ and
$ \left(\begin{array}{ccc} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ b_{2} \\ b_{3} \end{array}\right)=\left(\begin{array}{r} 3-c_{1} \\ 1-c_{2} \\ -1-c_{3} \end{array}\right) . $
Then, which of the following is/are TRUE?
A.
$\vec{a} \cdot \vec{c}=0$
B.
$\vec{b} \cdot \vec{c}=0$
C.
$|\vec{b}|>\sqrt{10}$
D.
$|\vec{c}| \leq \sqrt{11}$
2021
JEE Advanced
MSQ
JEE Advanced 2021 Paper 2 Online
Let O be the origin and $\overrightarrow {OA} = 2\widehat i + 2\widehat j + \widehat k$ and $\overrightarrow {OB} = \widehat i - 2\widehat j + 2\widehat k$ and $\overrightarrow {OC} = {1 \over 2}\left( {\overrightarrow {OB} - \lambda \overrightarrow {OA} } \right)$ for some $\lambda$ > 0. If $\left| {\overrightarrow {OB} \times \overrightarrow {OC} } \right| = {9 \over 2}$, then which of the following statements is (are) TRUE?
A.
Projection of $\overrightarrow {OC} $ on $\overrightarrow {OA} $ is $ - {3 \over 2}$
B.
Area of the triangle OAB is ${9 \over 2}$
C.
Area of the triangle ABC is ${9 \over 2}$
D.
The acute angle between the diagonals of the parallelogram with adjacent sides ${\overrightarrow {OA} }$ and ${\overrightarrow {OC} }$ is ${\pi \over 3}$
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 2 Offline
Let a and b be positive real numbers. Suppose $PQ = a\widehat i + b\widehat j$ and $PS = a\widehat i - b\widehat j$ are adjacent sides of a parallelogram PQRS. Let u and v be the projection vectors of $w = \widehat i + \widehat j$ along PQ and PS, respectively. If |u| + |v| = |w| and if the area of the parallelogram PQRS is 8, then which of the following statements is/are TRUE?
A.
a + b = 4
B.
a $-$ b = 2
C.
The length of the diagonal PR of the parallelogram PQRS is 4
D.
w is an angle bisector of the vectors PQ and PS
2016
JEE Advanced
MSQ
JEE Advanced 2016 Paper 2 Offline
Let $\widehat u = {u_1} \widehat i + {u_2}\widehat j + {u_3}\widehat k$ be a unit vector in ${{R^3}}$ and
$\widehat w = {1 \over {\sqrt 6 }}\left( {\widehat i + \widehat j + 2\widehat k} \right).$ Given that there exists a vector ${\overrightarrow v }$ in ${{R^3}}$ such that $\left| {\widehat u \times \overrightarrow v } \right| = 1$ and $\widehat w.\left( {\widehat u \times \overrightarrow v } \right) = 1.$ Which of the following statement(s) is (are) correct?
$\widehat w = {1 \over {\sqrt 6 }}\left( {\widehat i + \widehat j + 2\widehat k} \right).$ Given that there exists a vector ${\overrightarrow v }$ in ${{R^3}}$ such that $\left| {\widehat u \times \overrightarrow v } \right| = 1$ and $\widehat w.\left( {\widehat u \times \overrightarrow v } \right) = 1.$ Which of the following statement(s) is (are) correct?
A.
There is exactly one choice for such ${\overrightarrow v }$
B.
There are infinitely many choices for such ${\overrightarrow v }$
C.
If $\widehat u$ lies in the $xy$-plane then $\left| {{u_1}} \right| = \left| {{u_2}} \right|$
D.
If $\widehat u$ lies in the $xz$-plane then $2\left| {{u_1}} \right| = \left| {{u_3}} \right|$
2015
JEE Advanced
MSQ
JEE Advanced 2015 Paper 1 Offline
Let $\Delta PQR$ be a triangle. Let $\vec a = \overrightarrow {QR} ,\vec b = \overrightarrow {RP} $ and $\overrightarrow c = \overrightarrow {PQ} .$ If $\left| {\overrightarrow a } \right| = 12,\,\,\left| {\overrightarrow b } \right| = 4\sqrt 3 ,\,\,\,\overrightarrow b .\overrightarrow c = 24,$ then which of the following is (are) true?
A.
${{{{\left| {\overrightarrow c } \right|}^2}} \over 2} - \left| {\overrightarrow a } \right| = 12$
B.
${{{{\left| {\overrightarrow c } \right|}^2}} \over 2} + \left| {\overrightarrow a } \right| = 30$
C.
$\left| {\overrightarrow a \times \overrightarrow b + \overrightarrow c \times \overrightarrow a } \right| = 48\sqrt 3 $
D.
$\overrightarrow a .\overrightarrow b = - 72$
2014
JEE Advanced
MSQ
JEE Advanced 2014 Paper 1 Offline
Let $\overrightarrow x ,\overrightarrow y $ and $\overrightarrow z $ be three vectors each of magnitude $\sqrt 2 $ and the angle between each pair of them is ${\pi \over 3}$. If $\overrightarrow a $ is a non-zero vector perpendicular to $\overrightarrow x $ and $\overrightarrow y \times \overrightarrow z $ and $\overrightarrow b $ is a non-zero vector perpendicular to $\overrightarrow y $ and $\overrightarrow z \times \overrightarrow x ,$ then
A.
$\overrightarrow b = \left( {\overrightarrow b \,.\,\overrightarrow z } \right)\left( {\overrightarrow z - \overrightarrow x } \right)$
B.
$\overrightarrow a = \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow y - \overrightarrow z } \right)$
C.
$\overrightarrow a \,.\,\overrightarrow b = - \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow b \,.\,\overrightarrow z } \right)$
D.
$\overrightarrow a = \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow z - \overrightarrow y } \right)$
2011
JEE Advanced
MSQ
IIT-JEE 2011 Paper 1 Offline
The vector (s) which is/are coplanar with vectors ${\widehat i + \widehat j + 2\widehat k}$ and ${\widehat i + 2\widehat j + \widehat k,}$ and perpendicular to the vector ${\widehat i + \widehat j + \widehat k}$ is/are
A.
$\widehat j - \widehat k$
B.
$-\widehat i + \widehat j$
C.
$\widehat i - \widehat j$
D.
$-\widehat j + \widehat k$
1999
JEE Advanced
MSQ
IIT-JEE 1999
Let $a$ and $b$ two non-collinear unit vectors. If $u = a - \left( {a\,.\,b} \right)\,b$ and $v = a \times b,$ then $\left| v \right|$ is
A.
$\left| u \right|$
B.
$\,\left| u \right| + \left| {u\,.\,a} \right|$
C.
$\,\left| u \right| + \left| {u\,.\,b} \right|$
D.
$\left| u \right| + u.\left( {a + b} \right)$
1998
JEE Advanced
MSQ
IIT-JEE 1998
Which of the following expressions are meaningful?
A.
$u\left( {v \times w} \right)$
B.
$\left( {u \bullet v} \right) \bullet w$
C.
$\left( {u \bullet v} \right)w$
D.
$\,u\, \times \left( {v \bullet w} \right)$
1994
JEE Advanced
MSQ
IIT-JEE 1994
The vector $\,{1 \over 3}\left( {2\widehat i - 2\widehat j + \widehat k} \right)$ is
A.
a unit vector
B.
makes an angle ${\pi \over 3}$ with the vector $\left( {2\widehat i - 4\widehat j + 3\widehat k} \right)$
C.
parallel to the vector $\left( { - \widehat i + \widehat j - {1 \over 2}\widehat k} \right)$
D.
perpendicular to the vector ${3\widehat i + 2\widehat j - 2\widehat k}$
1993
JEE Advanced
MSQ
IIT-JEE 1993
Let $\vec a = 2\hat i - \hat j + \hat k,\vec b = \hat i + 2\hat j - \hat k$ and $\overrightarrow c = \widehat i + \widehat j - 2\widehat k - 2\widehat k$ be three vectors. A vector in the plane of ${\overrightarrow b }$ and ${\overrightarrow c }$, whose projection on ${\overrightarrow a }$ is of magnitude $\sqrt {2/3,} $ is :
A.
$2\widehat i + 3\widehat j - 3\widehat k$
B.
$2\widehat i + 3\widehat j + 3\widehat k$
C.
$-2\widehat i - \widehat j + 5\widehat k$
D.
$2\widehat i + \widehat j + 5\widehat k$
2005
JEE Advanced
Numerical
IIT-JEE 2005
If the incident ray on a surface is along the unit vector $\widehat v\,\,,$ the reflected ray is along the unit vector $\widehat w\,\,$ and the normal is along unit vector $\widehat a\,\,$ outwards. Express $\widehat w\,\,$ in terms of $\widehat a\,\,$ and $\widehat v\,\,.$
Correct Answer: $$\widehat w = \widehat v - 2\left( {\widehat a\,.\,\widehat v} \right)\widehat a$$
2004
JEE Advanced
Numerical
IIT-JEE 2004
If $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ and $\overrightarrow d $ are distinct vectors such that
$\,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $ and $\overrightarrow a \times \overrightarrow b = \overrightarrow c \times \overrightarrow d \,.$ Prove that
$\left( {\overrightarrow a - \overrightarrow d } \right).\left( {\overrightarrow b - \overrightarrow c } \right) \ne 0\,\,i.e.\,\,\,\overrightarrow a .\overrightarrow b + \overrightarrow d .\overrightarrow c \ne \overrightarrow d .\overrightarrow b + \overrightarrow a .\overrightarrow c $
$\,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $ and $\overrightarrow a \times \overrightarrow b = \overrightarrow c \times \overrightarrow d \,.$ Prove that
$\left( {\overrightarrow a - \overrightarrow d } \right).\left( {\overrightarrow b - \overrightarrow c } \right) \ne 0\,\,i.e.\,\,\,\overrightarrow a .\overrightarrow b + \overrightarrow d .\overrightarrow c \ne \overrightarrow d .\overrightarrow b + \overrightarrow a .\overrightarrow c $
Correct Answer: Solve it.
2003
JEE Advanced
Numerical
IIT-JEE 2003
If $\overrightarrow u ,\overrightarrow v ,\overrightarrow w ,$ are three non-coplanar unit vectors and $\alpha ,\beta ,\gamma $ are the angles between $\overrightarrow u $ and $\overrightarrow v $ and $\overrightarrow w ,$ $\overrightarrow w $ and $\overrightarrow u $ respectively and $\overrightarrow x ,\overrightarrow y ,\overrightarrow z ,$ are unit vectors along the bisectors of the angles $\alpha ,\,\,\beta ,\,\,\gamma $ respectively. Prove that $\,\left[ {\overrightarrow x \times \overrightarrow y \,\,\overrightarrow y \times \overrightarrow z \,\,\overrightarrow z \times \overrightarrow x } \right] = {1 \over {16}}{\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w } \right]^2}\,{\sec ^2}{\alpha \over 2}{\sec ^2}{\beta \over 2}{\sec ^2}{\gamma \over 2}.$
Correct Answer: Solve it.
2002
JEE Advanced
Numerical
IIT-JEE 2002
Let $V$ be the volume of the parallelopiped formed by the vectors $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k,$ $\,\,\,\,\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k,$ $\,\,\,\,\,\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k.$ where $r=1, 2, 3,$ are non-negative real numbers and $\sum\limits_{r = 1}^3 {\left( {{a_r} + {b_r} + {c_r}} \right) = 3L,} $ show that $V \le {L^3}\,\,.$
Correct Answer: Solve it.
2001
JEE Advanced
Numerical
IIT-JEE 2001
Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.
Correct Answer: Solve it.
2001
JEE Advanced
Numerical
IIT-JEE 2001
Find $3-$dimensional vectors ${\overrightarrow v _1},{\overrightarrow v _2},{\overrightarrow v _3}$ satisfying
$\,{\overrightarrow v _1}.{\overrightarrow v _1} = 4,\,{\overrightarrow v _1}.{\overrightarrow v _2} = - 2,\,{\overrightarrow v _1}.{\overrightarrow v _3} = 6,\,\,{\overrightarrow v _2}.{\overrightarrow v _2}$
$ = 2,\,{\overrightarrow v _2}.{\overrightarrow v _3} = - 5,\,{\overrightarrow v _3}.{\overrightarrow v _3} = 29$
$\,{\overrightarrow v _1}.{\overrightarrow v _1} = 4,\,{\overrightarrow v _1}.{\overrightarrow v _2} = - 2,\,{\overrightarrow v _1}.{\overrightarrow v _3} = 6,\,\,{\overrightarrow v _2}.{\overrightarrow v _2}$
$ = 2,\,{\overrightarrow v _2}.{\overrightarrow v _3} = - 5,\,{\overrightarrow v _3}.{\overrightarrow v _3} = 29$
Correct Answer: $${\overrightarrow V _1} = 2\widehat i\,;\,\,\,{\overrightarrow V _2} = - \widehat i \pm \widehat j\,;\,\,\,{\overrightarrow V _3} = 3\widehat i \pm 2\widehat j \pm 4\widehat k$$
2001
JEE Advanced
Numerical
IIT-JEE 2001
Let $\overrightarrow A \left( t \right) = {f_1}\left( t \right)\widehat i + {f_2}\left( t \right)\widehat j$ and
$$\overrightarrow B \left( t \right) = {g_1}\left( t \right)\overrightarrow i + {g_2}\left( t \right)\widehat j,t \in \left[ {0,1} \right],$$
where ${f_1},{f_2},{g_1},{g_2}$ are continuous functions. If $\overrightarrow A \left( t \right)$ and $\overrightarrow B \left( t \right)$ are nonzero vectors for all $t$ and $\overrightarrow A \left( 0 \right) = 2\widehat i + 3\widehat j,$ $\,\overrightarrow A \left( 1 \right) = 6\widehat i + 2\widehat j,$ $\,\overrightarrow B \left( 0 \right) = 3\widehat i + 2\widehat j$ and $\,\overrightarrow B \left( 1 \right) = 2\widehat i + 6\widehat j.$ Then show that $\,\overrightarrow A \left( t \right)$ and $\,\overrightarrow B \left( t \right)$ are parallel for some $t.$
where ${f_1},{f_2},{g_1},{g_2}$ are continuous functions. If $\overrightarrow A \left( t \right)$ and $\overrightarrow B \left( t \right)$ are nonzero vectors for all $t$ and $\overrightarrow A \left( 0 \right) = 2\widehat i + 3\widehat j,$ $\,\overrightarrow A \left( 1 \right) = 6\widehat i + 2\widehat j,$ $\,\overrightarrow B \left( 0 \right) = 3\widehat i + 2\widehat j$ and $\,\overrightarrow B \left( 1 \right) = 2\widehat i + 6\widehat j.$ Then show that $\,\overrightarrow A \left( t \right)$ and $\,\overrightarrow B \left( t \right)$ are parallel for some $t.$
Correct Answer: Solve it.
1999
JEE Advanced
Numerical
IIT-JEE 1999
Let $u$ and $v$ be units vectors. If $w$ is a vector such that $w + \left( {w \times u} \right) = v,$ then prove that $\left| {\left( {u \times v} \right) \cdot w} \right| \le 1/2$ and that the equality holds if and only if $u$ is perpendicular to $v .$
Correct Answer: Solve it.
1998
JEE Advanced
Numerical
IIT-JEE 1998
For any two vectors $u$ and $v,$ prove that
(a) ${\left( {u\,.\,v} \right)^2} + {\left| {u \times v} \right|^2} = {\left| u \right|^2}{\left| v \right|^2}$ and
(b) $\left( {1 + {{\left| u \right|}^2}} \right)\left( {1 + {{\left| v \right|}^2}} \right) = {\left( {1 - u.v} \right)^2} + {\left| {u + v + \left( {u \times v} \right)} \right|^2}.$
(a) ${\left( {u\,.\,v} \right)^2} + {\left| {u \times v} \right|^2} = {\left| u \right|^2}{\left| v \right|^2}$ and
(b) $\left( {1 + {{\left| u \right|}^2}} \right)\left( {1 + {{\left| v \right|}^2}} \right) = {\left( {1 - u.v} \right)^2} + {\left| {u + v + \left( {u \times v} \right)} \right|^2}.$
Correct Answer: Solve it.
1998
JEE Advanced
Numerical
IIT-JEE 1998
Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)
Correct Answer: Solve it.
1997
JEE Advanced
Numerical
IIT-JEE 1997
If $A,B$ and $C$ are vectors such that $\left| B \right| = \left| C \right|.$ Prove that
$\left[ {\left( {A + B} \right) \times \left( {A + C} \right)} \right] \times \left( {B \times C} \right)\left( {B + C} \right) = 0\,\,.$
$\left[ {\left( {A + B} \right) \times \left( {A + C} \right)} \right] \times \left( {B \times C} \right)\left( {B + C} \right) = 0\,\,.$
Correct Answer: Solve it.
1994
JEE Advanced
Numerical
IIT-JEE 1994
If the vectors $\overrightarrow b ,\overrightarrow c ,\overrightarrow d ,$ are not coplanar, then prove that the vector
$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) + \left( {\overrightarrow a \times \overrightarrow c } \right) \times \left( {\overrightarrow d \times \overrightarrow b } \right) + \left( {\overrightarrow a \times \overrightarrow d } \right) \times \left( {\overrightarrow b \times \overrightarrow c } \right)$ is parallel to $\overrightarrow a .$
$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) + \left( {\overrightarrow a \times \overrightarrow c } \right) \times \left( {\overrightarrow d \times \overrightarrow b } \right) + \left( {\overrightarrow a \times \overrightarrow d } \right) \times \left( {\overrightarrow b \times \overrightarrow c } \right)$ is parallel to $\overrightarrow a .$
Correct Answer: Solve it.
1993
JEE Advanced
Numerical
IIT-JEE 1993
In a triangle $ABC, D$ and $E$ are points on $BC$ and $AC$ respectively, such that $BD=2DC$ and $AE=3EC.$ Let $P$ be the point of intersection of $AD$ and $BE.$ Find $BP/PE$ using vector methods.
Correct Answer: $$8:3$$
1991
JEE Advanced
Numerical
IIT-JEE 1991
Determine the value of $'c'$ so that for all real $x,$ the vector
$cx\widehat i - 6\widehat j - 3\widehat k$ and $x\widehat i + 2\widehat j + 2cx\widehat k$ make an obtuse angle with each other.
$cx\widehat i - 6\widehat j - 3\widehat k$ and $x\widehat i + 2\widehat j + 2cx\widehat k$ make an obtuse angle with each other.
Correct Answer: $$ - {4 \over 3} < C < 0$$
1990
JEE Advanced
Numerical
IIT-JEE 1990
Let $\overrightarrow A = 2\overrightarrow i + \overrightarrow k ,\,\overrightarrow B = \overrightarrow i + \overrightarrow j + \overrightarrow k ,$ and $\overrightarrow C = 4\overrightarrow i - 3\overrightarrow j + 7\overrightarrow k .$ Determine a vector $\overrightarrow R .$ Satisfying $\overrightarrow R \times \overrightarrow B = \overrightarrow C \times \overrightarrow B $ and $\overrightarrow R \,.\,\overrightarrow A = 0$
Correct Answer: $$ - \widehat i - 8\widehat j + 2\widehat k$$
1989
JEE Advanced
Numerical
IIT-JEE 1989
If vectors $\overrightarrow A ,\overrightarrow B ,\overrightarrow C $ are coplanar, show that
$$\left| {\matrix{
{} & {\overrightarrow {a.} } & {} & {\overrightarrow {b.} } & {} & {\overrightarrow {c.} } \cr
{\overrightarrow {a.} } & {\overrightarrow {a.} } & {\overrightarrow {a.} } & {\overrightarrow {b.} } & {\overrightarrow {a.} } & {\overrightarrow {c.} } \cr
{\overrightarrow {b.} } & {\overrightarrow {a.} } & {\overrightarrow {b.} } & {\overrightarrow {b.} } & {\overrightarrow {b.} } & {\overrightarrow {c.} } \cr
} } \right| = \overrightarrow 0 $$
Correct Answer: Solve it.
1989
JEE Advanced
Numerical
IIT-JEE 1989
In a triangle $OAB,E$ is the midpoint of $BO$ and $D$ is a point on $AB$ such that $AD:DB=2:1.$ If $OD$ and $AE$ intersect at $P,$ determine the ratio $OP:PD$ using vector methods.
Correct Answer: $$3:2$$
1988
JEE Advanced
Numerical
IIT-JEE 1988
Let $OA$ $CB$ be a parallelogram with $O$ at the origin and $OC$ a diagonal. Let $D$ be the midpoint of $OA.$ Using vector methods prove that $BD$ and $CO$ intersect in the same ratio. Determine this ratio.
Correct Answer: $$2:1$$
1987
JEE Advanced
Numerical
IIT-JEE 1987
If $A, B, C, D$ are any four points in space, prove that -
$\left| {\overrightarrow {AB} \times \overrightarrow {CD} + \overrightarrow {BC} \times \overrightarrow {AD} + \overrightarrow {CA} \times \overrightarrow {BD} } \right| = 4$ (area of triangle $ABC$)
$\left| {\overrightarrow {AB} \times \overrightarrow {CD} + \overrightarrow {BC} \times \overrightarrow {AD} + \overrightarrow {CA} \times \overrightarrow {BD} } \right| = 4$ (area of triangle $ABC$)
Correct Answer: Solve it.
1986
JEE Advanced
Numerical
IIT-JEE 1986
The position vectors of the points $A, B, C$ and $D$ are $3\widehat i - 2\widehat j - \widehat k,\,2\widehat i + 3\widehat j - 4\widehat k,\, - \widehat i + \widehat j + 2\widehat k$ and $4\widehat i + 5\widehat j + \lambda \widehat k,$
respectively. If the points $A, B, C$ and $D$ lie on a plane, find the value of $\lambda .$
respectively. If the points $A, B, C$ and $D$ lie on a plane, find the value of $\lambda .$
Correct Answer: $${{146} \over {17}}$$
1982
JEE Advanced
Numerical
IIT-JEE 1982
Find all values of $\lambda $ such that $x, y, z,$$\, \ne $$(0,0,0)$ and
$\left( {\overrightarrow i + \overrightarrow j + 3\overrightarrow k } \right)x + \left( {3\overrightarrow i - 3\overrightarrow j + \overrightarrow k } \right)y + \left( { - 4\overrightarrow i + 5\overrightarrow j } \right)z$
$ = \lambda \left( {x\overrightarrow i \times \overrightarrow j \,\,y + \overrightarrow k \,z} \right)$ where $\overrightarrow i ,\,\,\overrightarrow j ,\,\,\overrightarrow k $ are unit vectors along the coordinate axes.
$\left( {\overrightarrow i + \overrightarrow j + 3\overrightarrow k } \right)x + \left( {3\overrightarrow i - 3\overrightarrow j + \overrightarrow k } \right)y + \left( { - 4\overrightarrow i + 5\overrightarrow j } \right)z$
$ = \lambda \left( {x\overrightarrow i \times \overrightarrow j \,\,y + \overrightarrow k \,z} \right)$ where $\overrightarrow i ,\,\,\overrightarrow j ,\,\,\overrightarrow k $ are unit vectors along the coordinate axes.
Correct Answer: $$\lambda = 0,\, - 1$$
1982
JEE Advanced
Numerical
IIT-JEE 1982
${A_1},{A_2},.................{A_n}$ are the vertices of a regular plane polygon with $n$ sides and $O$ is its centre. Show that
$\sum\limits_{i = 1}^{n - 1} {\left( {\overrightarrow {O{A_i}} \times {{\overrightarrow {OA} }_{i + 1}}} \right) = \left( {1 - n} \right)\left( {{{\overrightarrow {OA} }_2} \times {{\overrightarrow {OA} }_1}} \right)} $
$\sum\limits_{i = 1}^{n - 1} {\left( {\overrightarrow {O{A_i}} \times {{\overrightarrow {OA} }_{i + 1}}} \right) = \left( {1 - n} \right)\left( {{{\overrightarrow {OA} }_2} \times {{\overrightarrow {OA} }_1}} \right)} $
Correct Answer: Solve it
1997
JEE Advanced
Numerical
IIT-JEE 1997
Let $OA=a,$ $OB=10a+2b$ and $OC=b$ where $O,A$ and $C$ are non-collinear points. Let $p$ denote the area of the quadrilateral $OABC,$ and let $q$ denote the area of the parallelogram with $OA$ and $OC$ as adjacent sides. If $p=kq,$ then $k=$.........
Correct Answer: $$6$$
1996
JEE Advanced
Numerical
IIT-JEE 1996
If $\overrightarrow b \,$ and $\overrightarrow c \,$ are two non-collinear unit vectors and $\overrightarrow a \,$ is any vector, then $\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b + \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow c + {{\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)} \over {\left| {\overrightarrow b \times \overrightarrow c } \right|}}\left( {\overrightarrow b \times \overrightarrow c } \right) = $ ..............
Correct Answer: $${\overrightarrow a }$$
1996
JEE Advanced
Numerical
IIT-JEE 1996
A nonzero vector $\overrightarrow a $ is parallel to the line of intersection of the plane determined by the vectors $\widehat i,\widehat i + \widehat j$ and the plane determined by the vectors $\widehat i - \widehat j,\widehat i + \widehat k.$ The angle between $\overrightarrow a $ and the vector $\widehat i - 2\widehat j + 2\widehat k$ is ................
Correct Answer: $${\pi \over 4}$$ or $${3\pi \over 4}$$