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: 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

\dfrac B2=\sqrt{A^2+B^2+2AB\cos \theta} ....(i) \therefore \tan 90^o =\dfrac{B\sin \theta }{A+B\cos \theta }\Rightarrow A+B\cos \theta =0 \therefore \cos \theta =-\dfrac AB Hence, from (i) \dfrac{B^2}{4}=A^2+B^2-2A^2\Rightarrow A=\sqrt 3 \dfrac B2 \Rightarrow \cos \theta =-\dfrac AB=-\dfrac{\sqrt 3}{2}\therefore \theta =150^o

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: 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 heta 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 riangle OCD,

O C^{2} = O D^{2} + C D^{2}

Now \vec CD = \vec AC \sin heta = \vec Q \sin heta AD= \vec AC \cos heta=\vec Q \cos heta \vec OD=\vec OA+\vec AD=\vec P+\vec Q \cos heta Putting these values and representing resultant vector OC by \vec R , magnitude of the resultant is given by

R^{2} = (P + Q cos θ)^{2} + (Q sin θ)^{2} = P^{2} + Q^{2} + 2 P Q cos θ

In riangle OCD,

tan α = \frac{C D}{O D} = \frac{Q sin θ}{P + Q cos θ}

Resultant acts in the direction making an angle \alpha= an^{-1}(\dfrac{\vec Q \sin heta}{\vec P+\vec Q \cos heta}) with direction of vector P .

Question 3

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

Accepted Answer

Given  :       A.B = A	imes B              Taking mod on both sides,       |A.B| = |A	imes B|  OR       |A||B| \cos	heta = |A||B| \sin	heta  OR      \sin	heta = \cos	heta OR      	an	heta = 1                     \implies 	heta = 45^o

Question 4

State the law of parallelogram of vector addition and find the magnitude and direction of the resultant of vectors P and Q inclined at angle  	heta  with each other. What happens when  	heta=0^0  and  	heta=90^0 .

Accepted Answer

Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude & direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure. Let  	heta  be the angle between P and Q and R be the resultant vectors. According to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.So,  R=P+Q From triangle OCB OB^2=OC^2+BC^2 OB^2=(OA+AC)^2+BC^2  ......... (1) In  \Delta ABC \cos	heta =\dfrac{AC}{AB} AC=AB\cos	heta AC=OD\cos	heta =Q\cos	heta as  [AB=OD=Q] Also \cos	heta =\dfrac{BC}{AB} BC=AB\sin	heta BC=OD\sin	heta =Q\sin	heta [as  AB=OD=Q] Magnitude of Resultant substituting AC & BC in equation  (1)  we get R^2=(P+Q\cos	heta)^2+(Q\sin	heta)^2 	herefore R=\sqrt{P^2+2PQ\cos	heta +Q^2}   ightarrow  MagnitudeFrom  \Delta  ABC 	an\phi =\dfrac{BC}{OC}=\dfrac{BC}{OA+AC} 	an\phi =\dfrac{Q\sin	heta}{P+Q\cos	heta} 	herefore \phi =	an^{-1}\left(\dfrac{Q\sin 	heta}{P+Q\cos	heta}ight)  ....... (2) When  	heta =0^o \phi =	an^{-1}\left(\dfrac{Q\sin 0}{P+\cos 0}ight) \phi =	an^{-1}\left(\dfrac{0}{1}ight) \phi =	an^{-1}0 \phi =0 When  	heta =90^o \phi =	an^{-1}\left(\dfrac{Q\sin 90^o}{P+Q\cos 90^o}ight) \phi =	an^{-1}\left(\dfrac{Q}{P+0}ight) \phi =	an^{-1}\left(\dfrac{Q}{P}ight) .

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

\dfrac B2=\sqrt{A^2+B^2+2AB\cos 	heta}  ....(i) 	herefore 	an 90^o =\dfrac{B\sin 	heta }{A+B\cos 	heta }\Rightarrow A+B\cos 	heta =0 	herefore \cos 	heta =-\dfrac AB Hence, from (i)  \dfrac{B^2}{4}=A^2+B^2-2A^2\Rightarrow A=\sqrt 3 \dfrac B2 \Rightarrow \cos 	heta =-\dfrac AB=-\dfrac{\sqrt 3}{2}	herefore 	heta =150^o

Question 6

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 	herefore    \sqrt{(0.5)^2+(0.8)^2+c^2} = 1 Or   c^2 = 0.11 \implies  \ c = \sqrt{0.11}

Question 7

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	imes \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	imes \vec b|=19\sqrt{2} The unit vector perpendicular to both the vectors  \vec a,\vec b  is  \dfrac{\vec a	imes \vec b}{|\vec a	imes \vec b|}=\dfrac{19(\hat j+\hat k)}{19\sqrt{2}}=\dfrac{\hat j+\hat k}{\sqrt{2}}

Question 8

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

extbf{Step 1: Resolve the forces along x and y direction [Refer Fig. 1 & 2]}  Angle  	heta  can be calculated as:  	heta = 180^o - 60^o - 30^o - 30^o = 60^{\circ} 	extbf{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 9

Give the derivation for resultant of two vectors and for the angle between resultant and a vector.

Accepted Answer

Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.So, we have                  R = P + QNow, expand A to C and draw BC perpendicular to OC.From triangle OCB, OB^2 = C^2 + BC^2 or     OB^2 = ( OA + AC )^2 + BC^2 . . . . . . ( i )Intriangle ABC, cos 	heta = \frac{AC}{AB} or ,      AC =  AB cos 	heta or ,   AC = OD cos 	heta               = Q cos 	heta         [ AB =  OD = Q ]Also,  cos	heta  = \frac{BC}{AB} or ,    BC = AB sin 	heta or ,   BC = OD sin 	heta               = Q sin 	heta                   [ AB = OD = Q }Magnitude of resultant:Substituting value of AC and BC in ( i ), we get R^2 = ( P + Q cos 	heta )^2 + ( Q sin	heta )^2 or ,    R^2 = P^2 + 2P Q cos 	heta + Q^2 cos^2 	heta + Q^2 sin^2 	heta or ,    R^2 = P^2 + 2PQ cos 	heta + Q^2 R =  \sqrt{P^2 + 2PQ cos 	heta + Q^2} Which is the magnitude of resultant.Direction of resultant :Let  \phi  be the angle made by resultant R with P . Then,From triangle OBC, tan\phi =  \dfrac{BC}{OC} = \dfrac{BC}{OA + AC} or ,    tan \phi = \dfrac{Q sin 	heta}{P + Q cos 	heta} \phi = tan^{-1} ( \dfrac{Q sin 	heta}{P + Q cos 	heta}) which is the direction of resultant.

Question 10

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

Accepted Answer

\left(\vec{B}	imes\vec{A}ight)\cdot\vec{A}=-BA\sin{	heta}\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

Important Questions

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

State the law of parallelogram of vector addition and find the magnitude and direction of the resultant of vectors P and Q inclined at angle $θ$ with each other. What happens when $θ = 0^{0}$ and $θ = 9 0^{0}$ .

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

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

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

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

Give the derivation for resultant of two vectors and for the angle between resultant and a vector.

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

Important Questions

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

State the law of parallelogram of vector addition and find the magnitude and direction of the resultant of vectors P and Q inclined at angle θ with each other. What happens when θ=00 and θ=900.

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

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

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

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

Give the derivation for resultant of two vectors and for the angle between resultant and a vector.

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

State the law of parallelogram of vector addition and find the magnitude and direction of the resultant of vectors P and Q inclined at angle $θ$ with each other. What happens when $θ = 0^{0}$ and $θ = 9 0^{0}$ .

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

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

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

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