Q: If \$A.B=A\times B\$, then angle between \$A\$ and \$B\$ is

Given : \$A.B = A\times B\$ Taking mod on both sides, \$|A.B| = |A\times B|\$ OR \$|A||B| \cos\theta = |A||B| \sin\theta\$ OR \$\sin\theta = \cos\theta\$OR \$\tan\theta = 1\$ \$\implies \theta = 45^o\$

Q: Find a unit vector perpendicular to both the vectors \$\vec{a}\$ and \$\vec{b}\$, where \$\vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k}\$ and \$\vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}\$.

Given \$\vec a=\hat i-7\hat j+7\hat k,\vec b=3\hat i-2\hat j+2\hat k\$\$\vec a\times \vec b=\begin {vmatrix}\hat i&\hat j&\hat k\\1&-7&7\\3&-2&2\end{vmatrix}=\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)=19(\hat j+\hat k)\implies |\vec a\times \vec b|=19\sqrt{2}\$The unit vector perpendicular to both the vectors \$\vec a,\vec b\$ is \$\dfrac{\vec a\times \vec b}{|\vec a\times \vec b|}=\dfrac{19(\hat j+\hat k)}{19\sqrt{2}}=\dfrac{\hat j+\hat k}{\sqrt{2}}\$

Q: The resultant of two vectors \$A\$ and \$B\$ is perpendicular to the vector \$A\$ and its magnitude is equal to half the magnitude of vector \$B\$. The angle between \$A\$ and \$B\$ is

\$\textbf{Hint: Dot product of two perpendicular vector is zero.}\$\$\textbf{Step 1: Calculation of resultant of}\$ \$\vec {A}\$ \$\bf{and}\$ \$\vec {B}\$ Since \$|\vec {A} + \vec {B}| = \dfrac {1}{2} |\vec {B}|\$ [Given] \$|\vec {R}| = |\vec{A} + \vec{B}| = \sqrt {A^{2} + B^{2} + 2AB\cos \theta}\$ \$\therefore \sqrt {A^{2} + B^{2} + 2AB \cos \theta} = \dfrac {1}{2} B\$ Squaring both sides \$A^{2} + B^{2} + 2AB \cos \theta = \dfrac {B^{2}}{4}\$ \$\Rightarrow\$ \$A^{2} + 2AB\cos \theta - \dfrac {3}{4} B^{2} = 0\$ \$.... (1)\$\$\textbf{Step 2: Take dot product between}\$ \$\vec {R}\$ \$\bf{and}\$ \$\vec {A}\$ \$\vec {R} \perp\$ to \$\vec A\$ \$\therefore\$ We can say that \$\vec {A} \cdot (\vec {A} + \vec {B}) = 0\$ \$\Rightarrow\$ \$\vec {A} \cdot \vec {A} + \vec {A} \cdot \vec {B} = 0\$ \$\Rightarrow\$ \$A^{2} + AB \cos \theta = 0\$ \$.... (2)\$\$\textbf{Step 3: Solving the equations}\$ From eqn \$(1)\$ and \$(2)\$ \$A^{2} - 2A^{2} + \dfrac {3}{4} B^{2} = 0\$ \$\Rightarrow\$ \$A^{2} = \dfrac {3}{4}B^{2}\$ \$\Rightarrow\$ \$A = \dfrac {\sqrt {3}}{2} B\$From step \$(2)\$Now, \$\cos \theta = -\dfrac {A}{B} = \dfrac {-\sqrt {3}}{2} \dfrac {B}{B}\$ \$\Rightarrow\$ \$\cos \theta = \dfrac {-\sqrt {3}}{2}\$ \$\Rightarrow\$ \$\theta = 150^{\circ}\$Hence option \$B\$ is correct.

Q: Three forces acting on a body are shown in fig. To have the resultant force only along the y-direction,the magnitude of the minimum additional force needed along OX is

\$\textbf{Step 1: Resolve the forces along x and y direction [Refer Fig. 1 & 2]}\$ Angle \$\theta\$ can be calculated as: \$\theta = 180^o - 60^o - 30^o - 30^o = 60^{\circ}\$\$\textbf{Step 2: Resultant force along x axis}\$OX is the additional force along x-directionIf we have the resultant force only along y direction then: \$\sum F_{x} = 0 \$ \$\Rightarrow\$ \$OX + 1\cos 60^o + 2\cos 60^o - 4\sin 30^o = 0\$ \$\Rightarrow\$ \$OX = 0.5\ N\$Hence additional force \$OX\$ is \$0.5\ N\$Option \$A\$ is correct.

Q: Find unit vector in the direction of vector \$a=2i+3j+k\$.

unit vector \$ \hat{a} = \frac{\vec{a}}{\left | \vec{a} \right |} \$Given \$ \vec{a} = 2\hat{i}+3\hat{j}+\hat{k} \$\$ \left | \vec{a} \right | = \sqrt{2^{2}+3^{2}+1} \$\$ = \sqrt{14} \$unit vector \$ \hat{a} = \frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}} \$\$ = \frac{2}{\sqrt{14}}\hat{i}+\frac{3}{\sqrt{14}}\hat{j}+\frac{1}{\sqrt{14}}\hat{k} \$zero vector \$ \vec{0} = 0\hat{i}+0\hat{j}+0\hat{k} \$

Q: If a unit vector is represented by \$0.5\overline{i}+0.8\overline{j}+c\overline{k}\$then the value of 'c' is

The unit vector is given as \$\hat{A} = 0.5 \hat{i}+0.8\hat{j} + c\hat{k}\$Magnitude of unit vector is always equal to \$1\$ i.e. \$|\hat{A}| = 1\$\$\therefore\$ \$\sqrt{(0.5)^2+(0.8)^2+c^2} = 1\$Or \$c^2 = 0.11\$\$\implies \ c = \sqrt{0.11}\$

Q: The sum of magnitudes of two forces acting at a point is \$16N\$. If the resultant force is \$8N\$ and its direction is perpendicular to smaller force, then the forces are :

\$\displaystyle A + B = 16\, ...(1)\$\$\displaystyle tan \alpha = \frac {B\,sin\,\theta} {A + B\,cos\,\theta} = tan\,90\$Thus, \$\displaystyle A + B\,cos\,\theta = 0 \Rightarrow cos\theta= - \frac {A} {B} ...(2)\$We also have: \$\displaystyle 8 = \sqrt {A^2 + B^2 + 2AB\,cos\,\theta} ...(3)\$Solving these we get, \$A=6N\$ and \$B=10N\$

Q: If the angle between the vectors \$\vec{A}\$ and \$\vec{B}\$ is \$\theta\$, then the value of the product \$(\vec{B} \times \vec{A})\$. \$\vec{A}\$ equals

\$\left(\vec{B}\times\vec{A}\right)\cdot\vec{A}=-BA\sin{\theta}\hat{n}\cdot\vec{A}=0\$ Vector \$\vec{A}\$ and \$\hat{n}\$ are perpendicular to each other as direction of \$\hat{n}\$ can be given by right hand screw rule.Dot product of two vectors is zero when they are perpendicular to each other.

Question 1

State and prove parallelogram law of vector addition.

Accepted Answer

- Parallelogram law of vector addition states thatif two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of the two vectors is given by the vector that is diagonal passing through the point of contact of two vectors.Proof:Let $\vec{A}$ and $\vec{B}$ are the two vectors be represented by two lines $\vec{OP}$ and $\vec{OQ}$ drawn from the same point. Let us complete the parallelogram and name it as OPTQ. Let the diagonal be $\vec{OT}$.Since $\vec{PT}$ is equal and parallel to $\vec{OQ}$, therefore, vector $\vec{B}$ can also be represented by $\vec{PT}$.Applying the triangle's law of vector to triangle OPT.$\vec{OT}=\vec{OP}+\vec{PT}$$\Rightarrow \vec{R}=\vec{A}+\vec{B}$.(proved).

Question 2

Derive an expression for the magnitude and direction of the resultant vector using parallelogram law

Accepted Answer

Parallelogram law states that if two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.In the figure $\vec{P}$ and $\vec{Q}$ are two vectors.with magnitudes equal to length OA and OB respectively and making angle $\theta$ between them. Complete the parallelogram, OACB, Join diagonal OC , that makes angle $\alpha$ with vector $\vec{P}$. According to parallelogram law of vectors the resultant is represented by the diagonal  passing through the point of contact of two vectors.To find the magnitude of resultant , produce a perpendicular CD to meet OA produced to D. From $\triangle$ OCD,\begin{equation}OC^2=OD^2+CD^2\end{equation}Now $\vec CD$=$\vec AC$ $\sin \theta$=$\vec Q \sin\theta$$AD= \vec AC \cos\theta=\vec Q \cos\theta$$\vec OD=\vec OA+\vec AD=\vec P+\vec Q \cos\theta$Putting these values and representing resultant vector OC by $\vec R$, magnitude of the resultant is given by \begin{equation}R^2=(\vec P+\vec Q \cos\theta)^2+(\vec Q \sin\theta)^2=\vec P^2+  \vec Q^2+2\vec P\vec Q \cos \theta\end{equation}In  $\triangle$ OCD,\begin{equation}\tan \alpha= \frac{CD}{OD}=\frac{\vec Q \sin\theta}{\vec P+\vec Q \cos\theta}\end{equation}Resultant acts in the direction making an angle $\alpha=\tan^{-1}(\dfrac{\vec Q \sin\theta}{\vec P+\vec Q \cos\theta})$ with direction of vector P .

Question 3

If $A.B=A\times B$, then angle between $A$ and $B$ is

Accepted Answer

Given  :      $A.B = A\times B$             Taking mod on both sides,      $|A.B| = |A\times B|$ OR      $|A||B| \cos\theta = |A||B| \sin\theta$ OR     $\sin\theta = \cos\theta$OR     $\tan\theta = 1$                   $\implies \theta = 45^o$

Question 4

Find a unit vector perpendicular to both the vectors $\vec{a}$ and $\vec{b}$, where $\vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k}$ and $\vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}$.

Accepted Answer

Given $\vec a=\hat i-7\hat j+7\hat k,\vec b=3\hat i-2\hat j+2\hat k$$\vec a\times \vec b=\begin {vmatrix}\hat i&\hat j&\hat k\1&-7&7\3&-2&2\end{vmatrix}=\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)=19(\hat j+\hat k)\implies |\vec a\times \vec b|=19\sqrt{2}$The unit vector perpendicular to both the vectors $\vec a,\vec b$ is $\dfrac{\vec a\times \vec b}{|\vec a\times \vec b|}=\dfrac{19(\hat j+\hat k)}{19\sqrt{2}}=\dfrac{\hat j+\hat k}{\sqrt{2}}$

Question 5

The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle between $A$ and $B$ is

Accepted Answer

$\textbf{Hint: Dot product of two perpendicular vector is zero.}$$\textbf{Step 1: Calculation of resultant of}$ $\vec {A}$ $\bf{and}$ $\vec {B}$     Since                $|\vec {A} + \vec {B}| = \dfrac {1}{2} |\vec {B}|$          [Given]                $|\vec {R}| = |\vec{A} + \vec{B}| = \sqrt {A^{2} + B^{2} + 2AB\cos \theta}$           $\therefore \sqrt {A^{2} + B^{2} + 2AB \cos \theta} = \dfrac {1}{2} B$     Squaring both sides                 $A^{2} + B^{2} + 2AB \cos \theta = \dfrac {B^{2}}{4}$          $\Rightarrow$       $A^{2} + 2AB\cos \theta - \dfrac {3}{4} B^{2} = 0$                $.... (1)$$\textbf{Step 2: Take dot product between}$ $\vec {R}$ $\bf{and}$ $\vec {A}$                   $\vec {R} \perp$ to $\vec A$            $\therefore$ We can say that                  $\vec {A} \cdot (\vec {A} + \vec {B}) = 0$      $\Rightarrow$            $\vec {A} \cdot \vec {A} + \vec {A} \cdot \vec {B} = 0$       $\Rightarrow$               $A^{2} + AB \cos \theta = 0$                           $.... (2)$$\textbf{Step 3: Solving the equations}$     From eqn $(1)$ and $(2)$                  $A^{2} - 2A^{2} + \dfrac {3}{4} B^{2} = 0$     $\Rightarrow$             $A^{2} = \dfrac {3}{4}B^{2}$     $\Rightarrow$             $A = \dfrac {\sqrt {3}}{2} B$From step $(2)$Now, $\cos \theta = -\dfrac {A}{B} = \dfrac {-\sqrt {3}}{2} \dfrac {B}{B}$     $\Rightarrow$     $\cos \theta = \dfrac {-\sqrt {3}}{2}$     $\Rightarrow$           $\theta = 150^{\circ}$Hence option $B$ is correct.

Question 6

Three forces acting on a body are shown in fig. To have the resultant force only along the y-direction,the magnitude of the minimum additional force needed along OX is

Accepted Answer

$\textbf{Step 1: Resolve the forces along x and y direction [Refer Fig. 1 & 2]}$ Angle $\theta$ can be calculated as: $\theta = 180^o - 60^o - 30^o - 30^o = 60^{\circ}$$\textbf{Step 2: Resultant force along x axis}$OX is the additional force along x-directionIf we have the resultant force only along y direction then:     $\sum F_{x} = 0 $ $\Rightarrow$   $OX + 1\cos 60^o + 2\cos 60^o - 4\sin 30^o = 0$ $\Rightarrow$    $OX = 0.5\ N$Hence additional force $OX$ is $0.5\ N$Option $A$ is correct.

Question 7

Find  unit vector in the direction of vector $a=2i+3j+k$.

Accepted Answer

unit vector $ \hat{a} = \frac{\vec{a}}{\left | \vec{a} \right |} $Given $ \vec{a} = 2\hat{i}+3\hat{j}+\hat{k} $$ \left | \vec{a} \right | = \sqrt{2^{2}+3^{2}+1} $$ = \sqrt{14} $unit vector $ \hat{a} = \frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}} $$ = \frac{2}{\sqrt{14}}\hat{i}+\frac{3}{\sqrt{14}}\hat{j}+\frac{1}{\sqrt{14}}\hat{k} $zero vector $ \vec{0} = 0\hat{i}+0\hat{j}+0\hat{k} $

Question 8

If a unit vector is represented by $0.5\overline{i}+0.8\overline{j}+c\overline{k}$then the value of 'c' is

Accepted Answer

The unit vector is given as  $\hat{A} = 0.5 \hat{i}+0.8\hat{j} + c\hat{k}$Magnitude of unit vector is always equal to $1$  i.e.  $|\hat{A}| = 1$$\therefore$  $\sqrt{(0.5)^2+(0.8)^2+c^2} = 1$Or  $c^2 = 0.11$$\implies  \ c = \sqrt{0.11}$

Question 9

The sum of magnitudes of two forces acting at a point is $16N$. If the resultant force is $8N$ and its direction is perpendicular to smaller force, then the forces are :

Accepted Answer

$\displaystyle A + B = 16\,         ...(1)$$\displaystyle tan \alpha = \frac {B\,sin\,\theta} {A + B\,cos\,\theta} = tan\,90$Thus, $\displaystyle A + B\,cos\,\theta = 0 \Rightarrow cos\theta= - \frac {A} {B}         ...(2)$We also have: $\displaystyle 8 = \sqrt {A^2 + B^2 + 2AB\,cos\,\theta}         ...(3)$Solving these we get, $A=6N$ and $B=10N$

Question 10

If the angle between the vectors $\vec{A}$ and $\vec{B}$ is $\theta$, then the value of the product $(\vec{B} \times \vec{A})$. $\vec{A}$ equals

Accepted Answer

$\left(\vec{B}\times\vec{A}\right)\cdot\vec{A}=-BA\sin{\theta}\hat{n}\cdot\vec{A}=0$ Vector $\vec{A}$ and $\hat{n}$ are perpendicular to each other as direction of $\hat{n}$ can be given by right hand screw rule.Dot product of two vectors is zero when they are perpendicular to each other.

Revise with Concepts

Addition and Subtraction of Vectors

Law of Sines and Law of Cosines and Use in Vector Addition

Resolution of Vectors

Unit Vector

Multiplication of a Vector by a Scalar

Vector Addition by Component Method

Learn with Videos

Quick Summary With Stories

Vector Addition in Rectangular Coordinate System

Properties related to Vector addition

Triangle Law of Vector Addition

Parallelogram Law Of Vector Addition

Polygon Law of Vector addition

Representation of Vectors in Rectangular Coordinate System

Resolution of Vectors

Unit vector

Product of Scalar with Vector

Multiplication Of A Vector By A Scalar

State and prove parallelogram law of vector addition.

Derive an expression for the magnitude and direction of the resultant vector using parallelogram law

If $A . B = A \times B$ , then angle between $A$ and $B$ is

Find a unit vector perpendicular to both the vectors $a$ and $b$ , where $a = \hat{i} - 7 \hat{j} + 7 \hat{k}$ and $b = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}$ .

The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$ . The angle between $A$ and $B$ is

Three forces acting on a body are shown in fig. To have the resultant force only along the y-direction,the magnitude of the minimum additional force needed along OX is

Find unit vector in the direction of vector $a = 2 i + 3 j + k$ .

If a unit vector is represented by
$0.5 \overline{i} + 0.8 \overline{j} + c \overline{k}$
then the value of 'c' is

The sum of magnitudes of two forces acting at a point is $16 N$ . If the resultant force is $8 N$ and its direction is perpendicular to smaller force, then the forces are :

If the angle between the vectors $A$ and $B$ is $θ$ , then the value of the product $(B \times A)$ . $A$ equals

Revise with Concepts

Addition and Subtraction of Vectors

Law of Sines and Law of Cosines and Use in Vector Addition

Resolution of Vectors

Unit Vector

Multiplication of a Vector by a Scalar

Vector Addition by Component Method

Learn with Videos

Quick Summary With Stories

Vector Addition in Rectangular Coordinate System

Properties related to Vector addition

Triangle Law of Vector Addition

Parallelogram Law Of Vector Addition

Polygon Law of Vector addition

Representation of Vectors in Rectangular Coordinate System

Resolution of Vectors

Unit vector

Product of Scalar with Vector

Multiplication Of A Vector By A Scalar

State and prove parallelogram law of vector addition.

Derive an expression for the magnitude and direction of the resultant vector using parallelogram law

If A.B=A×B, then angle between A and B is

Find a unit vector perpendicular to both the vectors a and b, where a=i^−7j^​+7k^ and b=3i^−2j^​+2k^.

The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is

Three forces acting on a body are shown in fig. To have the resultant force only along the y-direction,the magnitude of the minimum additional force needed along OX is

Find unit vector in the direction of vector a=2i+3j+k.

If a unit vector is represented by 0.5i+0.8j​+ck then the value of 'c' is

The sum of magnitudes of two forces acting at a point is 16N. If the resultant force is 8N and its direction is perpendicular to smaller force, then the forces are :

If the angle between the vectors A and B is θ, then the value of the product (B×A). A equals

If $A . B = A \times B$ , then angle between $A$ and $B$ is

Find a unit vector perpendicular to both the vectors $a$ and $b$ , where $a = \hat{i} - 7 \hat{j} + 7 \hat{k}$ and $b = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}$ .

The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$ . The angle between $A$ and $B$ is

Find unit vector in the direction of vector $a = 2 i + 3 j + k$ .

If a unit vector is represented by
$0.5 \overline{i} + 0.8 \overline{j} + c \overline{k}$
then the value of 'c' is

The sum of magnitudes of two forces acting at a point is $16 N$ . If the resultant force is $8 N$ and its direction is perpendicular to smaller force, then the forces are :

If the angle between the vectors $A$ and $B$ is $θ$ , then the value of the product $(B \times A)$ . $A$ equals