Vector - Dot and Cross product (Lecture-02)
A vector has magnitude and direction and it is represented by a directed line segment. Its magnitude is denoted by $|\vec{a}|$. Unit vector along $\vec{a}$ is $\hat{a}=\dfrac{\vec{a}}{|\vec{a}|}$
Note : Zero vector has no definite direction
Addition of Vectors (Parallelogram Law)
If $\vec{a} \& \vec{b}$ are two adjacent sides of a parallelogram, then their sum $\vec{a}+\vec{b}$ represents the diagonal of the parallelogram through the common points
Note: Vector addition is commutative $\&$ associative
i.e. $\vec{a}+\vec{b}=\vec{a}+\vec{b}$ and $(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$
Angular bisectors
The internal bisector of angle between $\vec{a} \& \vec{b}$ is along $\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{b}}{|\vec{b}|}$ Hence the internal bisector is given by $\lambda(\hat{a}+\hat{b})$ and the external bisector of the angle is along $\dfrac{\vec{a}}{|\vec{a}|}-\dfrac{\vec{b}}{|\vec{b}|}$ Hence external bisector is given by $\lambda(\hat{a}-\hat{b})$. These bisectors are perpendicular to each other.
Orthonormal Vectors
Orthonormal Vectors are unit vectors along positive X, $\mathrm{Y} \& \mathrm{Z}$ axes.
Note :
Let $\vec{a}=\overrightarrow{\mathrm{OA}} \& \vec{b}=\overrightarrow{\mathrm{OB}} \cdot \& \overrightarrow{\mathrm{OC}}=p \overrightarrow{\mathrm{a}}+\mathrm{q} \vec{b}$ then
(i) $\mathrm{C}$ lies inside $\triangle$ OAB if $p, q>0$ \& $p+q<1$
(ii) $\mathrm{C}$ lies outside $\triangle \mathrm{OAB}$ but inside $\angle \mathrm{AOB}$ if $\mathrm{p}, \mathrm{q}>0$ \& $\mathrm{p}+\mathrm{q}>1$
(iii) $\mathrm{C}$ lies outside $\triangle \mathrm{OAB}$ but inside $\angle \mathrm{OAB}$ if $\mathrm{p}<0, \mathrm{q}>0$ and $\mathrm{p}+\mathrm{q}<1$
(iv) C lies outside $\triangle \mathrm{OAB}$ but inside $\angle \mathrm{OBA}$ if $\mathrm{p}>0, \mathrm{q}<0$ and $\mathrm{p}+\mathrm{q}<1$
Note : Let $\mathrm{P}$ be the point $\left(\mathrm{x} _{1}, \mathrm{y} _{1}, \mathrm{z} _{1}\right)$. If the coordinate system is rotated about
(i) $x$ axis through an angle $\theta$, then new coordinates of $P$ are
$\left(\mathrm{x} _{1}, \mathrm{y} _{1} \cos \theta+\mathrm{z} _{1} \sin \theta,-\mathrm{y} _{1} \sin \theta+\mathrm{z} _{1} \cos \theta\right)$
(ii) $\mathrm{y}$-axis through an angle $\theta$, then new coordinates of $\mathrm{Q}$ are $\left(-\mathrm{z} _{1} \sin \theta+\mathrm{x} _{1} \cos \theta, \mathrm{y} _{1}, \mathrm{z} _{1} \cos \theta+\mathrm{x} _{1} \sin \theta\right)$
(iii) $\mathrm{z}$ axis through an angle $\theta$, then the new coordinates of $\mathrm{P}$ are $\left(\mathrm{x} _{1} \cos \theta+\mathrm{y} _{1} \sin \theta,-\mathrm{x} _{1} \sin \theta+\mathrm{y} _{1} \cos \theta, \mathrm{z} _{1}\right)$
Scalar (dot) Product
For two non zero vectors $\vec{a} \& \vec{b}$, dot product $\vec{a} \cdot \vec{b}=a b \cos \theta$ where $a=|\vec{a}| ; b=|\vec{b}| ; \theta$ is the
angle between $\vec{a} \& \vec{b}$.
Note:
$\theta$ is acute $\Leftrightarrow \vec{a} \cdot \vec{b}>0$
$\theta$ is obtuse $\Leftrightarrow \vec{a} . \vec{b}<0$
Properties
(i) $\quad \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}=|\overrightarrow{\mathrm{a}}|^{2}$
(ii) $\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1$
$\hat{\mathrm{i}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{j}} \cdot \hat{\mathrm{k}}=\hat{\mathrm{k}} \cdot \hat{\mathrm{i}}=0$
(iv) $\vec{a} \cdot \vec{b}=a _{1} b _{1}+a _{2} b _{2}+a _{3} b _{3}$ where $\vec{a}=a _{1} \hat{i}+a _{2} \hat{j}+a _{3} \hat{k}$ and $\vec{b}=b _{1} \hat{i}+b _{2} \hat{j}+b _{3} \hat{k}$
(v) $\vec{a} \cdot(\vec{b} \pm \vec{c})=\vec{a} \cdot \vec{b} \pm \vec{a} \cdot \vec{c}$
(vi) $\vec{a} \cdot \vec{b}=0$ if and only if $\vec{a} \perp \vec{b}$
(vii) Length of projection of $\vec{a}$ on $\vec{b}=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ and orthogonal projection of $\vec{a}$ on $\vec{b}=\dfrac{(\vec{a} \cdot \vec{b}) \vec{b}}{|\vec{b}|^{2}}$
(viii) $\cos \theta=\dfrac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{\mathrm{ab}}=\dfrac{\mathrm{a} _{1} \mathrm{~b} _{1}+\mathrm{a} _{2} \mathrm{~b} _{2}+\mathrm{a} _{3} \mathrm{~b} _{3}}{\sqrt{\mathrm{a} _{1}{ }^{2}+\mathrm{a}^{2}{ } _{2}+\mathrm{a}^{2}{ } _{3}} \sqrt{\mathrm{b} _{1}{ }^{2}+\mathrm{b}^{2}{ } _{2}+\mathrm{b}^{2}{ } _{3}}}$
(ix) $\left(a _{1} b _{1}+a _{2} b _{2}+a _{3} b _{3}\right)^{2} \leq \sqrt{a _{1}{ }^{2}+a _{2}{ }^{2}+a _{3}{ }^{2}} \sqrt{b _{1}{ }^{2}+b _{2}{ }^{2}+b _{3}{ }^{2}}$
Here equality holds if $\dfrac{\mathrm{a} _{1}}{\mathrm{~b} _{1}}=\dfrac{\mathrm{a} _{2}}{\mathrm{~b} _{2}}=\dfrac{\mathrm{a} _{3}}{\mathrm{~b} _{3}}$
${\text { i.e. }(\vec{a} \cdot \vec{b}) \leq|\vec{a}||\vec{b}|}$
(x) $\overrightarrow{\mathrm{r}}=(\overrightarrow{\mathrm{r}} \cdot \hat{\mathrm{i}}) \hat{\mathrm{i}}+(\overrightarrow{\mathrm{r}} \cdot \hat{\mathrm{j}}) \hat{\mathrm{j}}+(\overrightarrow{\mathrm{r}} \hat{\mathrm{k}}) \hat{\mathrm{k}}$.
Note
- Angle between any two diagonals of a cube is $\cos ^{-1}(1 / 3)$
- Angle between diagonal of a cube and a diagonal of a face of the cube is $\cos ^{-1} \sqrt{\dfrac{2}{3}}$
Vector (cross) Product
$\vec{a} \times \vec{b}=a b \sin \theta$ n where $\hat{n}$ is a unit normal vector to the plane determined by $\vec{a} \& \vec{b}$ such that $\vec{a}, \vec{b}, \hat{n}$ form a right handed system.
Properties
(i) $\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}$
(ii) $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{0}$
(iii) $\vec{a} \times \vec{b}=\overrightarrow{0} \quad$ if and only if $\vec{a} | \vec{b}$
(iv) $\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overrightarrow{0}$
$\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i} ; \hat{k} \times \hat{i}=\hat{j}$
(v) $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a _{1} & a _{2} & a _{3} \\ b _{1} & b _{2} & b _{3}\end{array}\right|$ if $\vec{a}=a _{1} \hat{i}+a _{2} \hat{j}+a _{3} \hat{k}$ and $\vec{b}=b _{1} \hat{i}+b _{2} \hat{j}+b _{3} \hat{k}$
(vi) $\vec{a} \times(\vec{b} \pm \vec{c})=\vec{a} \times \vec{b} \pm \vec{a} \times \vec{c}$
(vii) $(\vec{a} \cdot \vec{b})+(\vec{a} \times \vec{b})=|\vec{a}|^{2}|\vec{b}|^{2}$
(viii) Geometrical significance:
-
Area of parallelogram = $ \left\{\begin{array}{l} |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\ \dfrac{1}{2}\left|\overrightarrow{\mathrm{d}}_1 \times \overrightarrow{\mathrm{d}}_2\right| \text { if } \overrightarrow{\mathrm{d}}_1=\overrightarrow{\mathrm{OC}} \& \overrightarrow{\mathrm{d}}_2=\overrightarrow{\mathrm{BA}} \text { are diagonals } \end{array}\right.$
-
$\text { Area of }{ }_{\Delta} \mathrm{OAB}=\dfrac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|$
(ix) Unit vector perpendicular to $\overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{b}}$ is $\hat{n}=\dfrac{ \pm(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}$
(x) If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of vertices $A, B, C$ of $\triangle A B C$. Then
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$
$\left\{\begin{array}{l}=2(\text { vector area of } \triangle \mathrm{ABC}) \\ =0 \Rightarrow \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} \text { are collinear } \\ \text { is a vector perpendicular to the plane } \mathrm{ABC}\end{array}\right.$
(xi) $(\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2}=2|\vec{a}|^{2}$
Scalar Triple (Dot, Mixed) Product
Let the angle between $\vec{a} \& \vec{b}$ be $\theta ; \vec{c} \& \vec{a} \times \vec{b}$ be $\phi$, then
$\vec{a} \times \vec{b} \cdot \vec{c}=[\vec{a} \vec{b} \vec{c}]=|\vec{a}||\vec{b}||\vec{c}| \sin \theta \cos \phi$
Properties
(i) $[\hat{i} \hat{j} \hat{k}]=1$
(ii) $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$
(iii) $[\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]$
But $[\bar{a} \bar{b} \bar{c}]=-[\vec{a} \vec{c} \vec{b}]$
(iv) $[\vec{a} \vec{b} \vec{c}]=|\vec{a}||\vec{b}||\vec{c}|$ if $\vec{a} \perp \vec{b} \perp \vec{c}$
(v) $[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{lll}a _{1} & a _{2} & a _{3} \\ b _{1} & b _{2} & b _{3} \\ c _{1} & c _{2} & c _{3}\end{array}\right|$ if $\vec{a}=a _{1} \hat{i} a _{2} \hat{j}+a _{3} \hat{k}, \vec{b}=b _{1} \hat{i}+b _{2} \hat{j}+b _{3} \hat{k}, \vec{c}=c _{1} \hat{i}+c _{2} \hat{j}+c _{3} \hat{k}$
(vi) $[\vec{a} \vec{b} \vec{c}]=0$ if any of two vectors are parallel or equal.
(vii) $[\lambda \vec{a} \vec{b} \vec{c}]=\lambda[\vec{a} \vec{b} \vec{c}]$
(viii) $[\vec{a}+\vec{d} \vec{b} \vec{c}]=[\vec{a} \vec{b} \vec{c}]+[\vec{d} \vec{b} \vec{c}]$
(ix) Geometrical meaning:
Volume of tetrahedron
- with $\vec{a}, \vec{b}, \vec{c}$ as coterminous edges is $\dfrac{1}{6}|[\vec{a} \vec{b} \vec{c}]|$
- with $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ as vertices $=\dfrac{1}{6}[[\overrightarrow{\mathrm{AB}} \overrightarrow{\mathrm{AC}} \overrightarrow{\mathrm{AD}}]$
- Centroid of tetrahedron divides the line joining any vertex to the centroid of its opposite face in the ratio $3: 1$.
(x) $[\vec{a} \vec{b} \vec{c}]\left[\begin{array}{lll}\vec{u} & \vec{v} & \vec{w}\end{array}\right]=\left|\begin{array}{lll}\vec{a} \cdot \vec{u} & \vec{a} \cdot \vec{v} & \vec{a} \cdot \vec{w} \ \vec{b} \cdot \vec{u} & \vec{b} \cdot \vec{v} & \vec{b} \cdot \vec{w} \ \vec{c} \cdot \vec{u} & \vec{c} \cdot \vec{v} & \vec{c} \cdot \vec{w}\end{array}\right|$
(xi) $\left[\begin{array}{lll}\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}\end{array}\right]=2\left[\begin{array}{ll}\vec{a} \vec{b} \vec{c}\end{array}\right]$
(xii) $\left[\begin{array}{lll}\vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a}\end{array}\right]=0$
(xiii) If $\vec{a}, \vec{b}$ and $\vec{c}$ are three coterminous edges of a parallelepiped, triangular prism and tetrahedron, $\mathrm{V} _{1}, \mathrm{~V} _{2} \& \mathrm{~V} _{3}$ be their volumes respectively, then
$V _{1}=[\vec{a} \vec{b} \vec{c}] ; \quad V _{2}=\dfrac{1}{2}[\vec{a} \vec{b} \vec{c}]$ and $V _{3}=\dfrac{1}{6}[\vec{a} \vec{b} \vec{c}]$
$\therefore \mathrm{V} _{1}: \mathrm{V} _{2}: \mathrm{V} _{3}=1: \dfrac{1}{2}: \dfrac{1}{6}=6: 3: 1$
Solved Examples
1. Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be three vectors. A vector $\overrightarrow{\mathrm{v}}$ is the plane of $\vec{a} \& \vec{b}$ whose projection on $\vec{c}$ is $\dfrac{1}{\sqrt{3}}$ is given by
(a) $\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
(b) $-3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
(c) $3 \hat{i}-\hat{j}+3 \hat{k}$
(d) $\hat{i}+3 \hat{j}-3 \hat{k}$
Show Answer
Solution:
Let $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$
$=(1+\lambda) \hat{\mathrm{i}}+(1-\lambda) \hat{\mathrm{j}}+(1+\lambda) \hat{\mathrm{k}}$
projection of $\overrightarrow{\mathrm{v}}$ on $\overrightarrow{\mathbf{c}}=\dfrac{1}{\sqrt{3}}$
$\begin{aligned} & \Rightarrow \dfrac{\overrightarrow{\mathrm{v}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}=\dfrac{1}{\sqrt{3}} \\ & \Rightarrow 1+\lambda-1+\lambda-1-\lambda=1 \Rightarrow \lambda=2 \end{aligned}$
$\therefore \overrightarrow{\mathrm{v}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
Answer (c)
2. Values of $\lambda$ such that $(\mathrm{x}, \mathrm{y}, \mathrm{z}) \neq(0,0,0)$ and $(\hat{i}+\hat{j}+3 \hat{k}) x+(3 \hat{i}-3 \hat{j}+\hat{k}) y+(-4 \hat{i}+5 \hat{j}) z=\lambda(x \hat{i}+y \hat{j}+z \hat{k})$
(a) 0
(b) -1
(c) 0
(d) None of these
Show Answer
Solution:
Comparing coefficents of $\hat{i}, \hat{j} \& \hat{k}$
$\mathrm{x}+2 \mathrm{y}-4 \mathrm{z}=\lambda \mathrm{x}$ $\quad \quad \quad$i.e. $(1-\lambda) x+3 y-4 z=0$ ……………….(1)
$\mathrm{x}-3 \mathrm{y}+5 \mathrm{z}=\lambda \mathrm{y}$ $\quad \quad \quad$ i.e. $x-(3+\lambda) y+5 z=0$. ……………….(2)
$3 \mathrm{x}+\mathrm{y}=\lambda \mathrm{z}$ $\quad \quad \quad$ i.e. $3 x+y-\lambda z=0$. …………………(3)
(1),(2) & (3) will have non trivial solutions if $\left|\begin{array}{ccc}1-\lambda & 3 & -4 \\ 1 & -(3+\lambda) & 5 \\ 3 & 1 & -\lambda\end{array}\right|=0 \quad(\because \mathrm{x}, \mathrm{y}, \mathrm{z}) \neq 0$ Expanding, we get $\lambda^{3}+2 \lambda^{2}+\lambda=0$
$\Rightarrow \lambda(\lambda+1)^{2}=0$
$\therefore \lambda=0,-1$
Answer (a), (b)
3. If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$, then possible values of $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is
(a) 0
(b) 3
(c) 4
(d) 8
Show Answer
Solution:
Let $\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
Then $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}} \Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{0}$
$\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} | \overrightarrow{\mathrm{c}}$
$\therefore \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{c}}$
$\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\lambda||\overrightarrow{\mathrm{c}}|$
$\Rightarrow \sqrt{29}=|\lambda| \sqrt{29} \quad \Rightarrow \lambda= \pm 1$
$\therefore \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}= \pm(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$
Now $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \pm(-14+6+12)= \pm 4$
Answer: (c)
4. In figure, $\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}, \overrightarrow{\mathrm{AC}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{DE}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}$ then the area of the shaded region in square units is
(a) 15
(b) 6
(c) 7
(d) 8
Show Answer
Solution:
We have $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AC}}$
$=(-3 \hat{i}+\hat{j})+(2 \hat{i}+3 \hat{j})=-\hat{i}+4 \hat{j}$
Vector Area of shaded region $=\dfrac{1}{2} \overrightarrow{\mathrm{ED}} \times \overrightarrow{\mathrm{EB}}+\dfrac{1}{2} \overrightarrow{\mathrm{EC}} \times \overrightarrow{\mathrm{ED}}$
$\begin{aligned} & =\dfrac{1}{2} \overrightarrow{\mathrm{ED}} \times(\overrightarrow{\mathrm{EB}}-\overrightarrow{\mathrm{EC}}) \\ & =\dfrac{1}{2} \overrightarrow{\mathrm{ED}} \times \overrightarrow{\mathrm{CB}} \\ & =\dfrac{1}{2}(16 \hat{\mathrm{k}}-2 \hat{\mathrm{k}})=7 \hat{\mathrm{k}} \end{aligned}$
$\therefore$ Area $=7$ sq.units
Answer: (c)
5. If ’ $\mathrm{a}$ ’ is a real constant and $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are variable angles and $\sqrt{\mathrm{a}^{2}-4} \tan \mathrm{A}+\mathrm{a} \tan \mathrm{B}+$ $\sqrt{\mathrm{a}^{2}+4} \tan \mathrm{C}=6 \mathrm{a}$, then the least value of $\tan ^{2} \mathrm{~A}+\tan ^{2} \mathrm{~B}+\tan ^{2} \mathrm{C}$ is
(a) 6
(b) 10
(c) 12
(d) 3
Show Answer
Solution:
The given relation can be written as
$\begin{aligned} & \left(\sqrt{a^{2}-4} \hat{i}+a \hat{j}+\sqrt{a^{2}+4} \hat{k}\right)(\tan A \hat{i}+\tan B \hat{j}+\tan C \hat{k})=6 a \\ & \Rightarrow \sqrt{a^{2}-4+a^{2}+a^{2}+4}\left(\tan ^{2} A+\tan ^{2} B+\tan ^{2} C\right) \cos \theta=6 a \\ & \Rightarrow \sqrt{3} a \sqrt{\tan ^{2} A+\tan ^{2} B+\tan ^{2} C}=6 a \sin \theta \\ & \Rightarrow \tan ^{2} A+\tan ^{2} B+\tan ^{2} C \geq 12 \sec ^{2} \theta \geq 12 \end{aligned}$
Answer: (c)
6. The vertex A of triangle $A B C$ is on the line $\vec{r}=\hat{i}+\hat{j}+\lambda \hat{k}$ and the vertices $B$ \& $C$ have respective position vectors $\hat{i}$ and $\hat{j}$. Let $\Delta$ be the area of the triangle and $\Delta \in\left[\dfrac{3}{2}, \dfrac{\sqrt{33}}{2}\right]$, then the range of values of $\lambda$ corresponding to $A$ is
(a) $[-8,-4] \cup[4,8]$
(b) $[-4,4]$
(c) $[-2,2]$
(d) $[-4,-2] \cup[2,4]$
Show Answer
Solution:
$\left.\Delta=\dfrac{1}{2}|(\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\lambda \hat{\mathrm{k}})|=\dfrac{1}{2} \right\rvert, \hat{\mathrm{k}}+\lambda \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}=\dfrac{1}{2} \sqrt{2 \lambda^{2}+1}$
$\dfrac{9}{4} \leq \dfrac{1}{4}\left(2 \lambda^{2}+1\right) \leq \dfrac{33}{4} \Rightarrow 4 \leq \lambda^{2} \leq 16$
$2 \leq|\lambda| \leq 4$
Answer: (d)
7. If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$, then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is
(a) $\sqrt{54}$
(b) 3
(c) 12
(d) None of these
Show Answer
Solution:
$\begin{aligned} & |\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9 \\ & \Rightarrow 6-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\ & \Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c} \cdot+\vec{c} \cdot \vec{a}=\dfrac{-3}{2} \\ & |\vec{a}+\vec{b}+\vec{c}|^{2}=\vec{a}^{2}+\vec{b}^{2}+\vec{c}^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c} \cdot+\vec{c} \cdot \vec{a})=0 \\ & \Rightarrow \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \\ & \Rightarrow \vec{a}=-(\vec{b}+\vec{c}) \\ & \therefore|(2 \vec{a}+5 \vec{b}+5 \vec{c})|=|2 \vec{a}-5 \vec{a}|=|-3 \vec{a}|=3 \end{aligned}$
Answer: 3
Exercise
1. Two adjacent sides of a parallelogram $\mathrm{ABCD}$ are given by $\overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}} \& \overrightarrow{\mathrm{AD}}=-$ $\hat{i}+2 \hat{j}+2 \hat{k}$. The side $A D$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $\mathrm{AD}$ becomes $\mathrm{AD}^{1}$. If $\mathrm{AD}^{1}$ makes a right angle with the side $\mathrm{AB}$, then the cosine of the angle $\alpha$ is given by
(a) $\dfrac{8}{9}$
(b) $\dfrac{\sqrt{17}}{9}$
(c) $\dfrac{1}{9}$
(d) $\dfrac{4 \sqrt{5}}{9}$
Show Answer
Answer: b2. Let two non collinear unit vectors $\hat{a}$ and $\hat{b}$ form an acute angle. A point $\mathrm{P}$ moves so that at any time the position vector $\overrightarrow{\mathrm{OP}}$ is given by $\hat{\mathrm{a}}$ cost $+\hat{\mathrm{b}}$ sint. When $\mathrm{P}$ is farthest from origin $\mathrm{O}$, let $\mathrm{M}$ be the length of $\overrightarrow{\mathrm{OP}}$ and $\hat{\mathrm{u}}$ be unit vector along $\overrightarrow{\mathrm{OP}}$. Then
(a) $\quad \hat{u}=\dfrac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$ and $M=(1+\hat{a} \cdot \hat{b})^{1 / 2}$
(b) $\quad \hat{u}=\dfrac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$ and $M=(1+\hat{a} \cdot \hat{b})^{1 / 2}$
(c) $\quad \hat{u}=\dfrac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$ and $M=(1+2 \hat{a} \cdot \hat{b})^{1 / 2}$
(d) $\quad \hat{u}=\dfrac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$ and $M=(1+2 \hat{a} \cdot \hat{b})^{1 / 2}$
Show Answer
Answer: a3. The values of $x$ for which the angle between the vectors $2 x^{2} \hat{i}+4 x \hat{j}+\hat{k}$ and $7 \hat{i}-2 \hat{j}+x \hat{k}$ are obtuse and the angle between the $z$-axis and $7 \hat{i}-2 \hat{j}+x \hat{k}$ is acute and less then $\dfrac{\pi}{6}$ is given by
(a) $0<x<\dfrac{1}{2}$
(b) $\mathrm{x}>\dfrac{1}{2}$ or $\mathrm{x}<0$
(c) $\dfrac{1}{2}<x<15$
(d) There is no such value for $x$.
Show Answer
Answer: d4. A unit tangent vector at $t=2$ on the serves $x=t^{2}+2, y=4 t-5, z=2 t^{2}-6 t$ is
(a) $\dfrac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
(b) $\dfrac{1}{3}(2 \hat{i}+2 \hat{j}+\hat{k})$
(c) $\dfrac{1}{\sqrt{6}}(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
(d) None of these
Show Answer
Answer: b5. When a right handed rectangular Cartesian system OXYZ is rotated about the z-axis through an angle $\dfrac{\pi}{4}$ in the counter clockwise direction it is found that a vector $\vec{a}$ has the components $2 \sqrt{2}, 3 \sqrt{2}$ and 4 . The components of $\vec{a}$ in the OXYZ coordinate system are
(a) $5,-1,4$
(b) $5,-1,4 \sqrt{2}$
(c) $-1,-5,4 \sqrt{2}$
(d) None of these
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Answer: d6. Let $V$ be the volume of the parallelopiped formed by the vectors $\vec{a}=a _{1} \hat{i}+a _{2} \hat{j}+a _{3} \hat{k}$, $\vec{b}=b _{1} \hat{i}+b _{2} \hat{j}+b _{3} \hat{k}$ and $\vec{c}=c _{1} \hat{i}+c _{2} \hat{j}+c _{3} \hat{k}$ If $a _{r}, b _{r}, c _{r}$ where $r=1,2,3$, are non negative real numbers and $\sum\limits _{\mathrm{r}=1}^{3}\left(\mathrm{a} _{\mathrm{r}}+\mathrm{b} _{\mathrm{r}}+\mathrm{c} _{\mathrm{r}}\right)=3 \mathrm{~L}$, then $\mathrm{V}$
(a) $\leq \mathrm{L}^{3}$
(b) $\geq \mathrm{L}^{3}$
(c) $\leq \mathrm{L}^{2}$
(d) $\geq \mathrm{L}^{2}$
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Answer: a7. Incident ray is along the unit vector $\hat{v}$ and the reflected ray is along the unit vector $\hat{w}$. The normal is along unit vector $\hat{a}$ outwards Then $\hat{v}-2(\hat{a} \cdot \hat{v}) \hat{a}=$
(a) $\hat{u}$
(b) $\hat{\mathbf{w}}$
(c) $\dfrac{\hat{\mathrm{w}}}{2}$
(d) None of these
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Answer: b8. If vector $\vec{a}, \vec{b}, \vec{c}$ are coplanar, then $\left|\begin{array}{ccc}\vec{a} & \vec{b} & \vec{c} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} . \vec{c} \\ \vec{b} . \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} . \vec{c}\end{array}\right|=$
(a) 0
(b) $\overrightarrow{0}$
(c) $\vec{a}+\vec{b}+\vec{c}$
(d) None of these
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Answer: b9. $\quad \mathrm{~A} _{1}, \mathrm{~A} _{2}, \mathrm{~A} _{3} \ldots \ldots . \mathrm{A} _{\mathrm{n}}$ are the vertices of a regular plane polygon with $\mathrm{n}$ sides and 0 is its centre, then $\sum\limits _{i=1}^{n-1}\left(\overrightarrow{\mathrm{OA}} _{\mathrm{i}} \times \overrightarrow{\mathrm{OA}} _{\mathrm{i}+1}\right)=$
(a) (n-1) $\left(\overrightarrow{\mathrm{OA}} _{2} \times \overrightarrow{\mathrm{OA}} _{1}\right)$
(b) $(1-\mathrm{n})\left(\overrightarrow{\mathrm{OA}} _{2} \times \overrightarrow{\mathrm{OA}} _{1}\right)$
(c) $(\mathrm{n}-1)\left(\overrightarrow{\mathrm{OA}} _{1} \times \overrightarrow{\mathrm{OA}} _{2}\right)$
(d) $(1-\mathrm{n})\left(\overrightarrow{\mathrm{OA}} _{1} \times \overrightarrow{\mathrm{OA}} _{2}\right)$
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Answer: b, c10. $\mathrm{O}$ is a point inside a tetrahedron $\mathrm{ABCD}$. If $\mathrm{AO}, \mathrm{BO}, \mathrm{CO}, \mathrm{DO}$ are produced to cut the opposite faces $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ respectively, then $\sum \dfrac{\mathrm{OP}}{\mathrm{AP}}=$
(a) 4
(b) 3
(c) 2
(d) 1
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Answer: d11. Read the paragraph and answer the following question
A tetrahedron is a triangular pyramid. If position vector of all the vertices of tetrahedron are $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$, then position vector of centroid is $\dfrac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$. If $\overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D}$ are adjacent sides of tetrahedron, then its volume is $\dfrac{1}{6}[\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}]$
(i) In a regular tetrahedron if the distance between centroid and mid-point of any edge of tetrahedron is equal to
(a) $\dfrac{1}{3}$ (edge of tetrahedron)
(b) $\dfrac{1}{2 \sqrt{2}}$ (edge of tetrahedron)
(c) $\dfrac{1}{2 \sqrt{3}}$ (edge of tetrahedron)
(d) $\dfrac{1}{3 \sqrt{2}}$ (edge of tetrahedron)
(ii) In regular tetrahedron angle between two opposite edges is
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{6}$
(c) $\dfrac{2 \pi}{3}$
(d) $\dfrac{\pi}{2}$
(iii) If vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are four vectors whose magnitudes are equal to area of the faces of a tetrahedron and directions perpendicular and outward directions to the faces respectively then the volume o of tetrahedron is
(a) $\dfrac{1}{6 \sqrt{2}} \mathrm{a}^{3}$
(b) $\dfrac{1}{3 \sqrt{2}} \mathrm{a}^{3}$
(c) $\dfrac{1}{4 \sqrt{2}} a^{3}$
(d) $\dfrac{1}{8 \sqrt{2}} \mathrm{a}^{3}$
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Answer: (i) b (ii) d (iii) a12. $\vec{a} \& \vec{b}$ form the consecutive sides of a regular hexagon $A B C D E F$
Column I | Column II | ||
---|---|---|---|
(a) | If $\overrightarrow{\mathrm{CD}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}$, then | (p) | $x=-2$ |
(b) | If $\overrightarrow{\mathrm{CE}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}$, then | (q) | $x=-2$ |
(c) | If $\overrightarrow{\mathrm{AC}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}$, then | (r) | $y=1$ |
(d) | If $\overrightarrow{\mathrm{AD}}=\mathrm{x} \overrightarrow{\mathrm{a}}$ then | (s) | $y=2$ |