NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

# NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

Edited By Ramraj Saini | Updated on Sep 19, 2023 09:16 AM IST | #CBSE Class 12th

## NCERT Vector Algebra Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 10 provided here. The name of this chapter is Vector Algebra. NCERT Solutions for Class 12 Maths Chapter 10 are explained in a detailed manner to help students prepare for their board exams and competitive exams. Important topics that are going to be discussed in Vector Algebra class 12 are vector quantities, operations on vectors, geometric properties, and algebraic properties like addition, multiplication, etc. Questions related to all these topics are covered in the Vector Algebra Class 12 NCERT solutions. Check all NCERT solutions from classes 6 to 12 in a single place, which will help in better understanding of concepts in a much easier way. Also, check NCERT solutions for class 12 also.

Vector Algebra not only helps to solve problems in Mathematics. It is also helps in solving problems of Class 11 and 12 Physics also. Students may be familiar with some of the concepts discussed in Vectors Class 12 as Class 11 Physics also discuss the concepts of vectors. The concepts studied in the Class 12 Maths ch 10 are also used in the upcoming chapter Three Dimensional Geometry.

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## NCERT Vector Algebra Class 12 Questions And Answers - Important Formulae

For vector a represented as a = xi + yj + zk The magnitude (length) of the vector is given by:

|a| = √(x² + y² + z²)

Vector Addition:

A + B = B + A (Commutative Law)

A + (B + C) = (A + B) + C (Associative Law)

Dot Product (Scalar Product) of Vectors:

A • B = |A| |B| cos θ (Where θ is the angle between vectors A and B)

Cross Product (Vector Product) of Vectors:

A × B = |A| |B| sin θ (Where θ is the angle between vectors A and B)

Scalar Multiplication:

k(A + B) = kA + kB

Additive Identity: A + 0 = 0 + A

Free download NCERT Vector Algebra Class 12 Solutions for CBSE Exam.

## NCERT Class 12 Maths Chapter 10 Question Answer (Intext Questions and Exercise)

Class 12 Maths Chapter 10 NCERT Solutions Vector Algebra - Exercise: 10.1

Answer:

Represent graphically a displacement of 40 km, $30 \degree$ east of north.

N,S,E,W are all 4 direction north,south,east,west respectively.

$\underset{OP}{\rightarrow}$ is displacement vector which $\left | \underset{OP}{\rightarrow} \right |$

= 40 km.

$\underset{OP}{\rightarrow}$ makes an angle of 30 degrees east of north as shown in the figure.

Question:2 (1) Classify the following measures as scalars and vectors.

Answer:

10kg is a scalar quantity as it has only magnitude.

Answer:

This is a vector quantity as it has both magnitude and direction.

Answer:

This is a scalar quantity as it has only magnitude.

Answer:

This is a scalar quantity as it has only magnitude.

Answer:

This is a scalar quantity as it has only magnitude.

Answer:

This is a Vector quantity as it has magnitude as well as direction.by looking at the unit, we conclude that measure is acceleration which is a vector.

Answer:

This is a scalar quantity as it has only magnitude.

Answer:

Distance is a scalar quantity as it has only magnitude.

Answer:

Force is a vector quantity as it has both magnitude as well as direction.

Answer:

Velocity is a vector quantity as it has both magnitude and direction.

Answer:

work done is a scalar quantity, as it is the product of two vectors.

Answer:

Since vector $\vec{a}$ and vector $\vec{d}$ are starting from the same point, they are coinitial.

Answer:

Since Vector $\vec{b}$ and Vector $\vec{d}$ both have the same magnitude and same direction, they are equal.

Answer:

Since vector $\vec{a}$ and vector $\vec{c}$ have the same magnitude but different direction, they are colinear and not equal.

Answer:

True, $\vec a$ and $-\vec a$ are collinear. they both are parallel to one line hence they are colinear.

Answer:

False, because colinear means they are parallel to the same line but their magnitude can be anything and hence this is a false statement.

Question:5 Answer the following as true or false.

Answer:

False, because any two non-colinear vectors can have the same magnitude.

Question:5 Answer the following as true or false.

Answer:

False, because two colinear vectors with the same magnitude can have opposite direction

Class 12 Maths Chapter 10 NCERT Solutions Vector Algebra-Exercise: 10.2

Answer:

Here

$\vec a = \hat i + \hat j + \hat k$

Magnitude of $\vec a$

$\vec a=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Answer:

Here,

$\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Magnitude of $\vec b$

$\left | \vec b \right |=\sqrt{2^2+(-7)^2+(-3)^2}=\sqrt{62}$

Answer:

Here,

$\vec c = \frac{1}{\sqrt 3 }\hat i + \frac{1}{\sqrt 3 }\hat j -\frac{1}{\sqrt 3 }\hat k$

Magnitude of $\vec c$

$\left |\vec c \right |=\sqrt{\left ( \frac{1}{\sqrt{3}} \right )^2+\left ( \frac{1}{\sqrt{3}} \right )^2+\left ( \frac{1}{\sqrt{3}} \right )^2}=1$

Answer:

Two different Vectors having the same magnitude are

$\vec a= 3\hat i+6\hat j+9\hat k$

$\vec b= 9\hat i+6\hat j+3\hat k$

The magnitude of both vector

$\left | \vec a \right |=\left | \vec b \right | = \sqrt{9^2+6^2+3^2}=\sqrt{126}$

Answer:

Two different vectors having the same direction are:

$\vec a=\hat i+2\hat j+3\hat k$

$\vec b=2\hat i+4\hat j+6\hat k$

Answer:

$2 \hat i + 3 \hat j$ will be equal to $x \hat i + y \hat j$ when their corresponding components are equal.

Hence when,

$x=2$ and

$y=3$

Answer:

Let point P = (2, 1) and Q = (– 5, 7).

Now,

$\vec {PQ}=(-5-2)\hat i+(7-1)\hat j=-7\hat i +6\hat j$

Hence scalar components are (-7,6) and the vector is $-7\hat i +6\hat j$

Answer:

Given,

$\\ \vec a = \hat i - 2 \hat j + \hat k ,\\ \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \\\vec c = \hat i - 6 \hat j - 7 \hat k$

Now, The sum of the vectors:

$\vec a +\vec b+\vec c = \hat i - 2 \hat j + \hat k + -2 \hat i + 4 \hat j + 5 \hat k + \hat i - 6 \hat j - 7 \hat k$

$\vec a +\vec b+\vec c = (1-2+1)\hat i +(-2+4-6) \hat j + (1+5-7)\hat k$

$\vec a +\vec b+\vec c =-4\hat j-\hat k$

Answer:

Given

$\vec a = \hat i + \hat j + 2 \hat k$

Magnitude of $\vec a$

$\left |\vec a \right |=\sqrt{1^2+1^2+2^2}=\sqrt{6}$

A unit vector in the direction of $\vec a$

$\vec u = \frac{\hat i}{\left | a \right |} + \frac{\hat j}{\left | a \right |} +\frac{2\hat k}{\left | a \right |} =\frac{\hat i}{\sqrt{6}}+\frac{\hat j}{\sqrt{6}}+\frac{2\hat k}{\sqrt{6}}$

Answer:

Given P = (1, 2, 3) and Q = (4, 5, 6)

A vector in direction of PQ

$\vec {PQ}=(4-1)\hat i+(5-2)\hat j +(6-3)\hat k$

$\vec {PQ}=3\hat i+3\hat j +3\hat k$

Magnitude of PQ

$\left | \vec {PQ} \right |=\sqrt{3^2+3^2+3^2}=3\sqrt{3}$

Now, unit vector in direction of PQ

$\hat u=\frac{\vec {PQ}}{\left | \vec {PQ} \right |}=\frac{3\hat i+3\hat j+3\hat k}{3\sqrt{3}}$

$\hat u=\frac{\hat i}{\sqrt{3}}+\frac{\hat j}{\sqrt{3}}+\frac{\hat k}{\sqrt{3}}$

Answer:

Given

$\vec a = 2 \hat i - \hat j + 2 \hat k$

$\vec b = - \hat i + \hat j - \hat k$

Now,

$\vec a + \vec b=(2-1)\hat i+(-1+1)\hat j+ (2-1)\hat k$

$\vec a + \vec b=\hat i+\hat k$

Now a unit vector in the direction of $\vec a + \vec b$

$\vec u= \frac{\vec a + \vec b}{\left |\vec a + \vec b \right |}=\frac{\hat i+\hat j}{\sqrt{1^2+1^2}}$

$\vec u= \frac{\hat i}{\sqrt{2}}+\frac{\hat j}{\sqrt{2}}$

Answer:

Given a vector

$\vec a=5 \hat i - \hat j + 2 \hat k$

the unit vector in the direction of $5 \hat i - \hat j + 2 \hat k$

$\vec u=\frac{5\hat i - \hat j + 2 \hat k}{\sqrt{5^2+(-1)^2+2^2}}=\frac{5\hat i}{\sqrt{30}}-\frac{\hat j}{\sqrt{30}}+\frac{2\hat k}{\sqrt{30}}$

A vector in direction of $5 \hat i - \hat j + 2 \hat k$ and whose magnitude is 8 =

$8\vec u=\frac{40\hat i}{\sqrt{30}}-\frac{8\hat j}{\sqrt{30}}+\frac{16\hat k}{\sqrt{30}}$

Answer:

Let

$\vec a =2 \hat i -3 \hat j + 4 \hat k$

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k$

It can be seen that

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k=-2(2 \hat i -3 \hat j + 4 \hat k)=-2\vec a$

Hence here $\vec b=-2\vec a$

As we know

Whenever we have $\vec b=\lambda \vec a$ , the vector $\vec a$ and $\vec b$ will be colinear.

Here $\lambda =-2$

Hence vectors $2 \hat i -3 \hat j + 4 \hat k$ and $- 4 \hat i + 6 \hat j - 8 \hat k$ are collinear.

Answer:

Let

$\vec a=\hat i + 2 \hat j + 3 \hat k$

$\left |\vec a \right |=\sqrt{1^2+2^2+3^2}=\sqrt{14}$

Hence direction cosine of $\vec a$ are

$\left ( \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}} ,\frac{3}{\sqrt{14}}\right )$

Answer:

Given

point A=(1, 2, –3)

point B=(–1, –2, 1)

Vector joining A and B Directed from A to B

$\vec {AB}=(-1-1)\hat i +(-2-2)\hat j+(1-(-3))\hat k$

$\vec {AB}=-2\hat i +-4\hat j+4\hat k$

$\left | \vec {AB} \right |=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6$

Hence Direction cosines of vector AB are

$\left ( \frac{-2}{6},\frac{-4}{6},\frac{4}{6} \right )=\left ( \frac{-1}{3},\frac{-2}{3},\frac{2}{3} \right )$

Answer:

Let

$\vec a=\hat i + \hat j + \hat k$

$\left | \vec a \right |=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Hence direction cosines of this vectors is

$\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )$

Let $\alpha$ , $\beta$ and $\gamma$ be the angle made by x-axis, y-axis and z- axis respectively

Now as we know,

$cos\alpha=\frac{1}{\sqrt{3}}$ , $cos\beta=\frac{1}{\sqrt{3}}$ $and\:cos\gamma=\frac{1}{\sqrt{3}}$

Hence Given vector is equally inclined to axis OX,OY and OZ.

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n internally:

$\vec r=\frac{m\vec b+n\vec a}{m+n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

$\vec r = \frac{2(-\hat i+\hat j +\hat k)+1(\hat i+2\hat j-\hat k)}{2+1}=\frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}$

$\vec r = \frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}=\frac{-\hat i+4\hat j+\hat k}{3}$

$\vec r = \frac{-\hat i}{3}+\frac{4\hat j}{3}+\frac{\hat k}{3}$

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n externally:

$\vec r=\frac{m\vec b-n\vec a}{m-n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

$\vec r = \frac{2(-\hat i+\hat j +\hat k)-1(\hat i+2\hat j-\hat k)}{2-1}=\frac{-2\hat i+2\hat j +2\hat k-\hat i-2\hat j+\hat k}{1}$

$\vec r = -3\hat i +3\hat k$

Answer:

Given

The position vector of point P = $2\hat i+3\hat j +4\hat k$

Position Vector of point Q = $4\hat i+\hat j -2\hat k$

The position vector of R which divides PQ in half is given by:

$\vec r =\frac{2\hat i+3\hat j +4\hat k+4\hat i+\hat j -2\hat k}{2}$

$\vec r =\frac{6\hat i+4\hat j +2\hat k}{2}=3\hat i+2\hat j +\hat k$

Answer:

Given

the position vector of A, B, and C are

$\\\vec a = 3 \hat i - 4 \hat j - 4 \hat k ,\\\vec b = 2 \hat i - \hat j + \hat k \: \: and \\\vec c = \hat i - 3 \hat j - 5 \hat k$

Now,

$\vec {AB}=\vec b-\vec a=-\hat i+3\hat j+5\hat k$

$\vec {BC}=\vec c-\vec b=-\hat i-2\hat j-6\hat k$

$\vec {CA}=\vec a-\vec c=2\hat i-\hat j+\hat k$

$\left | \vec {AB} \right |=\sqrt{(-1)^2+3^2+5^2}=\sqrt{35}$

$\left | \vec {BC} \right |=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}$

$\left | \vec {CA} \right |=\sqrt{(2)^2+(-1)^2+(1)^2}=\sqrt{6}$

AS we can see

$\left | \vec {BC} \right |^2=\left | \vec {CA} \right |^2+\left | \vec {AB} \right |^2$

Hence ABC is a right angle triangle.

Answer:

From triangles law of addition we have,

$\vec {AB}+\vec {BC}=\vec {AC}$

From here

$\vec {AB}+\vec {BC}-\vec {AC}=0$

also

$\vec {AB}+\vec {BC}+\vec {CA}=0$

Also

$\vec {AB}-\vec {CB}+\vec {CA}=0$

Hence options A,B and D are true SO,

Option C is False.

Answer:

If two vectors are collinear then, they have same direction or are parallel or anti-parallel.
Therefore,
They can be expressed in the form $\vec{b}= \lambda \vec{a}$ where a and b are vectors and $\lambda$ is some scalar quantity.

Therefore, (a) is true.
Now,
(b) $\lambda$ is a scalar quantity so its value may be equal to $\pm 1$

Therefore,
(b) is also true.

C) The vectors and are proportional,
Therefore, (c) is not true.

D) The vectors and can have different magnitude as well as different directions.

Therefore, (d) is not true.

Therefore, the correct options are (C) and (D).

Class 12 Maths Chapter 10 NCERT Solutions Vector Algebra - Exercise: 10.3

Answer:

Given

$\left | \vec a \right |=\sqrt{3}$

$\left | \vec b \right |=2$

$\vec a . \vec b = \sqrt 6$

As we know

$\vec a . \vec b = \left | \vec a \right |\left | \vec b \right |cos\theta$

where $\theta$ is the angle between two vectors

So,

$cos\theta =\frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}=\frac{\sqrt{6}}{\sqrt{3}*2}=\frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{4}$

Hence the angle between the vectors is $\frac{\pi}{4}$ .

Answer:

Given two vectors

$\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

Now As we know,

The angle between two vectors $\vec a$ and $\vec b$ is given by

$\theta=cos^{-1}\left ( \frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}\right )$

Hence the angle between $\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

$\theta=cos^{-1}\left ( \frac{(\hat i-2\hat j+3\hat k).(3\hat i-2\hat j+\hat k)}{\left | \hat i-2\hat j+3\hat k \right |\left |3\hat i-2\hat j+\hat k \right |}\right )$

$\theta=cos^{-1}\left ( \frac{3+4+3}{\sqrt{1^2+(-2)^2+3^3}\sqrt{3^2+(-2)^2+1^2}} \right )$

$\theta=cos^{-1}\frac{10}{14}$

$\theta=cos^{-1}\frac{5}{7}$

Answer:

Let

$\vec a=\hat i - \hat j$

$\vec b=\hat i + \hat j$

Projection of vector $\vec a$ on $\vec b$

$\frac{\vec a.\vec b}{\left | \vec b \right |}=\frac{(\hat i-\hat j)(\hat i+\hat j)}{\left |\hat i+\hat j \right |}=\frac{1-1}{\sqrt{2}}=0$

Hence, Projection of vector $\vec a$ on $\vec b$ is 0.

Answer:

Let

$\vec a =\hat i + 3 \hat j + 7 \hat k$

$\vec b=7\hat i - \hat j + 8 \hat k$

The projection of $\vec a$ on $\vec b$ is

$\frac{\vec a.\vec b}{\left | \vec b \right |}=\frac{(\hat i+3\hat j+7\hat k)(7\hat i-\hat j+8\hat k)}{\left | 7\hat i-\hat j+8\hat k \right |}=\frac{7-3+56}{\sqrt{7^2+(-1)^2+8^2}}=\frac{60}{\sqrt{114}}$

Hence, projection of vector $\vec a$ on $\vec b$ is

$\frac{60}{\sqrt{114}}$

Answer:

Given

$\\\vec a=\frac{1}{7}( 2 \hat i + 3 \hat j + 6 \hat k ), \\\ \vec b =\frac{1}{7}( 3 \hat i- 6 \hat j + 2 \hat k ),\\\vec c = \frac{1}{7}( 6\hat i + 2 \hat j -3\hat k )$

Now magnitude of $\vec a,\vec b \:and\: \vec c$

$\left | \vec a \right |=\frac{1}{7} \sqrt{2^2+3^2+6^2}=\frac{\sqrt{49}}{7}=1$

$\left | \vec b \right |=\frac{1}{7} \sqrt{3^2+(-6)^2+2^2}=\frac{\sqrt{49}}{7}=1$

$\left | \vec c \right |=\frac{1}{7} \sqrt{6^2+2^2+(-3)^2}=\frac{\sqrt{49}}{7}=1$

Hence, they all are unit vectors.

Now,

$\vec a.\vec b=\frac{1}{7}(2\hat i+3\hat j+6\hat k)\frac{1}{7}(3\hat i-6\hat j+2\hat k)=\frac{1}{49}(6-18+12)=0$

$\vec b.\vec c=\frac{1}{7}(3\hat i-6\hat j+2\hat k)\frac{1}{7}(6\hat i+2\hat j-3\hat k)=\frac{1}{49}(18-12-6)=0$

$\vec c.\vec a=\frac{1}{7}(6\hat i+2\hat j-3\hat k)\frac{1}{7}(2\hat i+3\hat j-6\hat k)=\frac{1}{49}(12+6-18)=0$

Hence all three are mutually perpendicular to each other.

Answer:

Given in the question

$( \vec a + \vec b ). ( \vec a - \vec b )=8$

$\left | \vec a \right |^2-\left | \vec b \right |^2=8$

Since $|\vec a |\: \:= 8 \: \:|\vec b |$

$\left | \vec {8b} \right |^2-\left | \vec b \right |^2=8$

$\left | \vec {63b} \right |^2=8$

$\left | \vec {b} \right |^2=\frac{8}{63}$

$\left | \vec {b} \right |=\sqrt{\frac{8}{63}}$

So, answer of the question is

$\left | \vec {a} \right |=8\left | \vec {b} \right |=8\sqrt{\frac{8}{63}}$

Answer:

To evaluate the product $( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b )$

$( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b )=6\vec a.\vec a+21\vec a.\vec b-10\vec b.\vec a-35\vec b.\vec b$

$=6\vec a.^2+11\vec a.\vec b-35\vec b^2$

$=6\left | \vec a \right |^2+11\vec a.\vec b-35\left | \vec b \right |^2$

Answer:

Given two vectors $\vec a \: \: and \: \: \vec b$

$\left | \vec a \right |=\left | \vec b\right |$

$\vec a.\vec b=\frac{1}{2}$

Now Angle between $\vec a \: \: and \: \: \vec b$

$\theta=60^0$

Now As we know that

$\vec a.\vec b=\left | \vec a \right |\left | \vec b \right |cos\theta$

$\frac{1}{2}=\left | \vec a \right |\left | \vec a \right |cos60^0$

$\left | a \right |^2=1$

Hence, the magnitude of two vectors $\vec a \: \: and \: \: \vec b$

$\left | a \right |=\left | b \right |=1$

Answer:

Given in the question that

$( \vec x -\vec a ) . ( \vec x + \vec a ) = 12$

And we need to find $\left | \vec x \right |$

$\left | \vec x \right |^2-\left | \vec a \right |^2 = 12$

$\left | \vec x \right |^2-1 = 12$

$\left | \vec x \right |^2 = 13$

$\left | \vec x \right | = \sqrt{13}$

So the value of $\left | \vec x \right |$ is $\dpi{100} \sqrt{13}$

Answer:

Given in the question is

$\vec a = 2 \hat i + 2 \hat j + 3 \hat k , \vec b = - \hat i + 2 \hat j + \hat k \: \: and \: \: \vec c = 3 \hat i + \hat j$

and $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

and we need to find the value of $\lambda$ ,

so the value of $\vec a + \lambda \vec b$ -

$\vec a + \lambda \vec b=2\hat i +2\hat j +3\hat k+\lambda (-\hat i+2\hat j+\hat k)$

$\vec a + \lambda \vec b=(2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k$

As $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

$(\vec a + \lambda \vec b).\vec c=0$

$((2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k)(3\hat i+\hat j)=0$

$3(2-\lambda)+2+2\lambda=0$

$6-3\lambda+2+2\lambda=0$

$\lambda=8$

the value of $\lambda=8$ ,

Answer:

Given in the question that -

$\vec a \: \: \: and \: \: \vec b$ are two non-zero vectors

According to the question

$\left ( |\vec a | \vec b + |\vec b | \vec a\right )\left (|\vec a | \vec b - |\vec b | \vec a \right )$

$=|\vec a |^2 |\vec b|^2 - |\vec b |^2 |\vec a|^2+|\vec b||\vec a|\vec a.\vec b-|\vec a||\vec b|\vec b.\vec a=0$

Hence $|\vec a | \vec b + |\vec b | \vec a$ is perpendicular to $|\vec a | \vec b - |\vec b | \vec a$ .

Answer:

Given in the question

$\\\vec a . \vec a = 0 \\|\vec a|^2=0$

$\\|\vec a|=0$

Therefore $\vec a$ is a zero vector. Hence any vector $\vec b$ will satisfy $\vec a . \vec b = 0$

Answer:

Given in the question

$\vec a , \vec b , \vec c$ are unit vectors $\Rightarrow |\vec a|=|\vec b|=|\vec c|=1$

and $\vec a + \vec b + \vec c = \vec 0$

and we need to find the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$

$(\vec a + \vec b + \vec c)^2 = \vec 0$

$\vec a^2 + \vec b^2 + \vec c ^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$|\vec a|^2 + |\vec b|^2 + |\vec c |^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$1+1+1+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$\vec a . \vec b + \vec b. \vec c + \vec c . \vec a=\frac{-3}{2}$

Answer- the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$ is $\dpi{80} \frac{-3}{2}$

Answer:

Let

$\vec a=\hat i-2\hat j +3\hat k$

$\vec b=5\hat i+4\hat j +1\hat k$

we see that

$\vec a.\vec b=(\hat i-2\hat j +3\hat k)(5\hat i+4\hat j +1\hat k)=5-8+3=0$

we now observe that

$|\vec a|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}$

$|\vec b|=\sqrt{5^2+4^2+1^2}=\sqrt{42}$

Hence here converse of the given statement is not true.

Answer:

Given points,

A=(1, 2, 3),

B=(–1, 0, 0),

C=(0, 1, 2),

As need to find Angle between $\overline{BA}\: \: and\: \: \overline{BC} ]$

$\vec {BA}=(1-(-1))\hat i+(2-0)\hat j+(3-0)\hat k=2\hat i+2\hat j+3\hat k$

$\vec {BC}=(0-(-1))\hat i+(1-0)\hat j+(2-0)\hat k=\hat i+\hat j+2\hat k$

Hence angle between them ;

$\theta=cos^{-1}(\frac{\vec {BA}.\vec {BC}}{\left | \vec {BA} \right |\left | \vec {BC} \right |})$

$\theta=cos^{-1}\frac{2+2+6}{\sqrt{17}\sqrt{6}}$

$\theta=cos^{-1}\frac{10}{\sqrt{102}}$

Answer - Angle between the vectors $\overline{BA}\: \: and\: \: \overline{BC}$ is $\dpi{80} \theta=cos^{-1}\frac{10}{\sqrt{102}}$

Answer:

Given in the question

A=(1, 2, 7), B=(2, 6, 3) and C(3, 10, –1)

To show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear

$\vec {AB}=(2-1)\hat i+(6-2)\hat j+(3-7)\hat k$

$\vec {AB}=\hat i+4\hat j-4\hat k$

$\vec {BC}=(3-2)\hat i+(10-6)\hat j+(-1-3)\hat k$

$\vec {BC}=\hat i+4\hat j-4\hat k$

$\vec {AC}=(3-1)\hat i+(10-2)\hat j+(-1-7)\hat k$

$\vec {AC}=2\hat i+8\hat j-8\hat k$

$|\vec {AB}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {BC}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {AC}|=\sqrt{2^2+8^2+(-8)^2}=2\sqrt{33}$

As we see that

$|\vec {AC}|=|\vec {AB}|+|\vec {BC}|$

Hence point A, B , and C are colinear.

Answer:

Given the position vector of A, B , and C are

$2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$

To show that the vectors $2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$ form the vertices of a right angled triangle

$\vec {AB}=(1-2)\hat i + (-3-(-1))\hat j+(-5-1)\hat k=-1\hat i -2\hat j-6\hat k$

$\vec {BC}=(3-1)\hat i + (-4-(-3))\hat j+(-4-(-5))\hat k=-2\hat i -\hat j+\hat k$

$\vec {AC}=(3-2)\hat i + (-4-(-1))\hat j+(-4-(1))\hat k=\hat i -3\hat j-5\hat k$

$|\vec {AB}|=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}$

$|\vec {BC}|=\sqrt{(-2)^2+(-1)^2+(1)^2}=\sqrt{6}$

$|\vec {AC}|=\sqrt{(1)^2+(-3)^2+(-5)^2}=\sqrt{35}$

Here we see that

$|\vec {AC}|^2+|\vec {BC}|^2=|\vec {AB}|^2$

Hence A,B, and C are the vertices of a right angle triangle.

Answer:

Given $\vec a$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar

$\lambda \vec a$ is a unit vector when

$|\lambda \vec a|=1$

$|\lambda|| \vec a|=1$

$| \vec a|=\frac{1}{|\lambda|}$

Hence the correct option is D.

Class 12 vector algebra NCERT solutions - Exercise: 10.4

Answer:

Given in the question,

$\\ \vec a = \hat i - 7 \hat j + 7 \hat k \: \: and \: \: \\\vec b = 3 \hat i - 2 \hat j + 2 \hat k$

and we need to find $\dpi{100} |\vec a \times \vec b |$

Now,

$|\vec a \times \vec b | =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &-7 &7 \\ 3& -2 &2 \end{vmatrix}$

$|\vec a \times \vec b | =\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)$

$|\vec a \times \vec b | =19\hat j+19\hat k$

So the value of $\dpi{100} |\vec a \times \vec b |$ is $19\hat j+19\hat k$

Answer:

Given in the question

$\vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k$

$\vec a + \vec b =3\hat i +2\hat j+2\hat k+\hat i +2\hat j-2\hat k=4\hat i +4\hat j$

$\vec a - \vec b =3\hat i +2\hat j+2\hat k-\hat i -2\hat j+2\hat k=2\hat i +4\hat k$

Now , A vector which perpendicular to both $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is $(\vec a + \vec b) \times (\vec a - \vec b)$

$(\vec a + \vec b) \times (\vec a - \vec b)=\begin{vmatrix} \hat i &\hat j &\hat k \\ 4&4 &0 \\ 2& 0& 4 \end{vmatrix}$

$(\vec a + \vec b) \times (\vec a - \vec b)= \hat i (16-0)-\hat j(16-0)+\hat k(0-8)$

$(\vec a + \vec b) \times (\vec a - \vec b)= 16\hat i -16\hat j-8\hat k$

And a unit vector in this direction :

$\vec u =\frac{16\hat i-16\hat j-8\hat k}{|16\hat i-16\hat j-8\hat k|}=\frac{16\hat i-16\hat j-8\hat k}{\sqrt{16^2+(-16)^2+(-8)^2}}$

$\vec u =\frac{16\hat i-16\hat j-8\hat k}{24}=\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$

Hence Unit vector perpendicular to each of the vector $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is $\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$ .

Answer:

Given in the question,

angle between $\vec a$ and $\hat i$ :

$\alpha =\frac{\pi}{3}$

angle between $\vec a$ and $\hat j$

$\beta =\frac{\pi}{4}$

angle with $\vec a$ and $\hat k$ :

$\gamma =\theta$

Now, As we know,

$cos^2\alpha+cos^2\beta+cos^2\gamma=1$

$cos^2\frac{\pi}{3}+cos^2\frac{\pi }{4}+cos^2\theta=1$

$\left ( \frac{1}{2} \right )^2+\left ( \frac{1}{\sqrt{2}} \right )^2+cos^2\theta=1$

$cos^2\theta=\frac{1}{4}$

$cos\theta=\frac{1}{2}$

$\theta=\frac{\pi}{3}$

Now components of $\vec a$ are:

$\left ( cos\frac{\pi}{3},cos\frac{\pi}{2},cos\frac{\pi}{3} \right )=\left ( \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2} \right )$

Answer:

To show that $( \vec a - \vec b ) \times (\vec a + \vec b ) = 2 ( \vec a \times \vec b )$

LHS=

$\\( \vec a - \vec b ) \times (\vec a + \vec b )=( \vec a - \vec b ) \times (\vec a)+( \vec a - \vec b ) \times (\vec b)$

$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times \vec a-\vec b \times\vec a+\vec a \times \vec b-\vec b \times \vec b$

As product of a vector with itself is always Zero,

$( \vec a - \vec b ) \times (\vec a + \vec b )= 0-\vec b \times\vec a+\vec a \times \vec b-\0$

As cross product of a and b is equal to negative of cross product of b and a.

$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times\vec b+\vec a \times \vec b$

$( \vec a - \vec b ) \times (\vec a + \vec b )= 2(\vec a \times\vec b)$ = RHS

LHS is equal to RHS, Hence Proved.

Answer:

Given in the question

$( 2 \hat i + 6 \hat j + 27 \hat k ) \times ( \hat i + \lambda j + \mu \hat k ) = \vec 0$

and we need to find values of $\lambda$ and $\mu$

$\begin{vmatrix} \hat i &\hat j & \hat k\\ 2& 6&27 \\ 1& \lambda &\mu \end{vmatrix}=0$

$\hat i (6\mu-27\lambda)-\hat j(2\mu-27)+\hat k(2\lambda-6)=0$

From Here we get,

$6\mu-27\lambda=0$

$2\mu-27=0$

$2\lambda -6=0$

From here, the value of $\lambda$ and $\mu$ is

$\lambda = 3 , \: and \: \mu=\frac{27}{2}$

Answer:

Given in the question

$\vec a . \vec b = 0$ and $\vec a \times \vec b = 0$

When $\vec a . \vec b = 0$ , either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$ are perpendicular to each other

When $\vec a \times \vec b = 0$ either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$ are parallel to each other

Since two vectors can never be both parallel and perpendicular at same time,we conclude that

$|\vec a| =0\:or\: |\vec b|=0$

Answer:

Given in the question

$\\\vec a=\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k , \\\vec b=\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k , \\\vec c=\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k$

We need to show that $\vec a \times ( \vec b + \vec c ) = \vec a \times \vec b + \vec a \times \vec c$

Now,

$\vec a \times ( \vec b + \vec c ) =(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times(\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k +\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k)$

$=(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times((\vec b_ 1+\vec c_1) \hat i + (\vec b_ 2+\vec c_2) \hat j +( \vec b_3 +\vec c_3)\hat k)$

$=\begin{vmatrix} \hat i &\hat j &\hat k \\ a_1&a_2 &a_3 \\ (b_1+c_1)&(b_2+c_2) &(b_3+c_3) \end{vmatrix}$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)- a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

Now

$\vec a \times \vec b + \vec a \times \vec c=\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{vmatrix}+\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ c_1&c_2 &c_3 \end{vmatrix}$

$\vec a \times \vec b + \vec a \times \vec c=\hat i(a_2b_3-a_3b_2)-\hat j (a_1b_3-a_3b_1)+\hat k(a_1b_2-b_1a_2)+\hat i(a_2c_3-a_3c_2)-\hat j (a_1c_3-a_3c_1)+\hat k(a_1c_2-c_1a_2)$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

Hence they are equal.

Answer:

No, the converse of the statement is not true, as there can be two non zero vectors, the cross product of whose are zero. they are colinear vectors.

Consider an example

$\vec a=\hat i +\hat j + \hat k$

$\vec b =2\hat i +2\hat j + 2\hat k$

Here $|\vec a| =\sqrt{1^2+1^2+1^2}=\sqrt{3}$

$|\vec b| =\sqrt{2^2+2^2+2^2}=2\sqrt{3}$

$\vec a \times \vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 1&1 &1 \\ 2&2 &2 \end{vmatrix}=\hat i(2-2)-\hat j(2-2)+\hat k(2-2)=0$

Hence converse of the given statement is not true.

Answer:

Given in the question

vertices A=(1, 1, 2), B=(2, 3, 5) and C=(1, 5, 5). and we need to find the area of the triangle

$AB=(2-1)\hat i+(3-1)\hat j+(5-2)\hat k=\hat i+2\hat j+3\hat k$

$BC=(1-2)\hat i+(5-3)\hat j+(5-5)\hat k=-\hat i+2\hat j$

Now as we know

Area of triangle

$A=\frac{1}{2}|\vec {AB}\times\vec {BC}|=\frac{1}{2}|(\hat i+2\hat j +3\hat k)\times(-\hat i+2\hat j)|$

$\\A=\frac{1}{2}\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ -1 &2 &0 \end{vmatrix}=\frac{1}{2}|\hat i(0-6)-\hat j(0-(-3))+\hat k(2-(-2))| \\A=\frac{1}{2}|-6\hat i-3\hat j+4\hat k|$

$A=\frac{1}{2}*\sqrt{(-6)^2+(-3)^2+(4)^2}=\frac{\sqrt{61}}{2}$

The area of the triangle is $\dpi{100} \frac{\sqrt{61}}{2}$ square units

Answer:

Given in the question

$\vec a = \hat i - \hat j + 3 \hat k$

$\vec b = 2\hat i -7 \hat j + \hat k$

Area of parallelogram with adjescent side $\vec a$ and $\vec b$ ,

$A=|\vec a\times\vec b|=|(\vec i-\vec j+3\vec k)\times (2\hat i-7\hat j+\hat k)|$

$A=\begin{vmatrix} \hat i& \hat j & \hat k\\ 1&-1 &3 \\ 2&-7 &1 \end{vmatrix}=|\hat i(-1+21)-\hat j (1-6)+\hat k (-7+2)|$

$A=|\hat i(20)-\hat j (-5)+\hat k (-5)|=\sqrt{20^2+5^2+(-5)^2}$

$A=\sqrt{450}=15\sqrt{2}$

The area of the parallelogram whose adjacent sides are determined by the vectors $\vec a = \hat i - \hat j + 3 \hat k$ and $\vec b = 2\hat i -7 \hat j + \hat k$ is $A=\sqrt{450}=15\sqrt{2}$

Answer:

Given in the question,

$|\vec a| = 3 \: \: and\: \: |\vec b | = \frac{\sqrt 2 }{3}$

As given $\vec a \times \vec b$ is a unit vector, which means,

$|\vec a \times \vec b|=1$

$|\vec a| | \vec b|sin\theta=1$

$3*\frac{\sqrt{2}}{3}sin\theta=1$

$sin\theta=\frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{4}$

Hence the angle between two vectors is $\frac{\pi}{4}$ . Correct option is B.

Answer:

Given 4 vertices of rectangle are

$\\\vec a=- \hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec b=\hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec c= \hat i - \frac{1}{2}\hat j + 4 \hat k \: \: and \: \: \\\vec d= - \hat i - \frac{1}{2} \hat j + 4 \hat k$

$\vec {AB}=\vec b-\vec a=(1+1)\hat i+(\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=2\hat i$

$\vec {BC}=\vec c-\vec b=(1-1)\hat i+(-\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=-\hat j$

Now,

Area of the Rectangle

$A=|\vec {AB}\times\vec {BC}|=|2\hat i \times (-\hat j)|=2$

Hence option C is correct.

Class 12 vector algebra NCERT Solutions - Miscellaneous Exercise

Answer:

As we know

a unit vector in XY-Plane making an angle $\theta$ with x-axis :

$\vec r=cos\theta \hat i+sin\theta \hat j$

Hence for $\theta = 30^0$

$\vec r=cos(30^0) \hat i+sin(30^0) \hat j$

$\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j$

Answer- the unit vector in XY-plane, making an angle of $30 \degree$ with the positive direction of x-axis is

$\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j$

Answer:

Given in the question

$P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).$

And we need to finrd the scalar components and magnitude of the vector joining the points P and Q

$\vec {PQ}=(x_2-x_1)\hat i +(y_2-y_1)\hat j+(z_2-z_1)\hat k$

Magnitiude of vector PQ

$|\vec {PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Scalar components are

$(x_2-x_1),(y_2-y_1),(z_2-z_1)$

Answer:

As the girl walks 4km towards west

Position vector = $-4\hat i$

Now as she moves 3km in direction 30 degree east of north.

$-4\hat i+3sin30^0\hat i+3cos30^0\hat j$

$-4\hat i+\frac{3}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

$\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

hence final position vector is;

$\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

Answer:

No, if $\vec a = \vec b + \vec c$ then we can not conclude that $|\vec a| =| \vec b |+| \vec c |$ .

the condition $\vec a = \vec b + \vec c$ satisfies in the triangle.

also, in a triangle, $|\vec a| <| \vec b |+| \vec c |$

Since, the condition $|\vec a| =| \vec b |+| \vec c |$ is contradicting with the triangle inequality, if $\vec a = \vec b + \vec c$ then we can not conclude that $|\vec a| =| \vec b |+| \vec c |$

Answer:

Given in the question,

a unit vector, $\vec u=x ( \hat i+ \hat j + \hat k )$

We need to find the value of x

$|\vec u|=1$

$|x ( \hat i+ \hat j + \hat k )|=1$

$x\sqrt{1^2+1^2+1^2}=1$

$x\sqrt{3}=1$

$x=\frac{1}{\sqrt{3}}$

The value of x is $\dpi{80} \frac{1}{\sqrt{3}}$

Answer:

Given two vectors

$\vec a = 2 \hat i + 3 \hat j - \hat k \: \: and \: \: \vec b = \hat i - 2 \hat j + \hat k$

Resultant of $\vec a$ and $\vec b$ :

$\vec R = \vec a +\vec b$ $=2 \hat i + 3 \hat j - \hat k + \hat i - 2 \hat j + \hat k=3\hat i + \hat j$

Now, a unit vector in the direction of $\vec R$

$\vec u =\frac{3\hat i+\hat j}{\sqrt{3^2+1^2}}=\frac{3}{\sqrt{10}}\hat i+\frac{1}{\sqrt{10}}\hat j$

Now, a unit vector of magnitude in direction of $\vec R$

$\vec v=5\vec u =5*\frac{3}{\sqrt{10}}\hat i+5*\frac{1}{\sqrt{10}}\hat j=\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

Hence the required vector is $\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

Answer:

Given in the question,

$\vec a = \hat i + \hat j + \hat k , \vec b = 2 \hat i - \hat j + 3 \hat k \: \: and\: \: \vec c = \hat i - 2 \hat j + \hat k$

Now,

let vector $\vec V=2\vec a - \vec b + 3 \vec c$

$\vec V=2(\hat i +\hat j +\hat k) - (2\hat i-\hat j+3\hat k)+ 3 (\hat i-2\hat j+\hat k)$

$\vec V=3\hat i-3\hat j+2\hat k$

Now, a unit vector in direction of $\vec V$

$\vec u =\frac{3\hat i-3\hat j+2\hat k}{\sqrt{3^2+(-3)^2+2^2}}=\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k$

Now,

A unit vector parallel to $\vec V$

$\vec u =\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k$

OR

$-\vec u =-\frac{3}{\sqrt{22}}\hat i+\frac{3}{\sqrt{22}} \hat j-\frac{2}{\sqrt{22}}\hat k$

Answer:

Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

$\vec {AB }=(5-1)\hat i+(0-(-2))\hat j+(-2-(-8))\hat k=4\hat i+2\hat j+6\hat k$

$\vec {BC }=(11-5)\hat i+(3-0)\hat j+(7-(-2))\hat k=6\hat i+3\hat j+9\hat k$

$\vec {CA }=(11-1)\hat i+(3-(-2))\hat j+(7-(-8))\hat k=10\hat i+5\hat j+15\hat k$

now let's calculate the magnitude of the vectors

$|\vec {AB }|=\sqrt{4^2+2^2+6^2}=\sqrt{56}=2\sqrt{14}$

$|\vec {BC }|=\sqrt{6^2+3^2+9^2}=\sqrt{126}=3\sqrt{14}$

$|\vec {CA }|=\sqrt{10^2+5^2+15^2}=\sqrt{350}=5\sqrt{14}$

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio 2 : 3.

Answer:

Given, two vectors $\vec P=( 2 \vec a + \vec b ) \: \:and \: \:\vec Q= ( \vec a - 3 \vec b )$

the point R which divides line segment PQ in ratio 1:2 is given by

$=\frac{2(2\vec a +\vec b)-(\vec a-3\vec b)}{2-1}=4\vec a +2\vec b -\vec a+3\vec b=3\vec a+5\vec b$

Hence position vector of R is $3\vec a+5\vec b$ .

Now, Position vector of the midpoint of RQ

$=\frac{( 3\vec a + 5\vec b + \vec a - 3 \vec b )}{2}=2\vec a+\vec b$

which is the position vector of Point P . Hence, P is the mid-point of RQ

Answer:

Given, two adjacent sides of the parallelogram

$2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k$

The diagonal will be the resultant of these two vectors. so

resultant R:

$\vec R=2 \hat i - 4 \hat j + 5 \hat k \: +\: \hat i - 2 \hat j - 3 \hat k=3\hat i-6\hat j+2\hat k$

Now unit vector in direction of R

$\vec u=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{49}}=\frac{3\hat i-6\hat j+2\hat k}{7}$

Hence unit vector along the diagonal of the parallelogram

$\vec u={\frac{3}{7}\hat i-\frac{6}{7}\hat j+\frac{2}{7}\hat k}$

Now,

Area of parallelogram

$A=(2 \hat i - 4 \hat j + 5 \hat k )\: \times \: \: (\hat i - 2 \hat j - 3 \hat k)$

$A=\begin{vmatrix} \hat i &\hat j &\hat k \\ 2& -4 &5 \\ 1&-2 &-3 \end{vmatrix}=|\hat i(12+10)-\hat j(-6-5)+\hat k(-4+4)|=|22\hat i +11\hat j|$

$A=\sqrt{22^2+11^2}=11\sqrt{5}$

Hence the area of the parallelogram is $11\sqrt{5}$ .

Answer:

Let a vector $\vec a$ is equally inclined to axis OX, OY and OZ.

let direction cosines of this vector be

$cos\alpha,cos\alpha \:and \:cos\alpha$

Now

$cos^2\alpha+cos^2\alpha +cos^2\alpha=1$

$cos^2\alpha=\frac{1}{3}$

$cos\alpha=\frac{1}{\sqrt{3}}$

Hence direction cosines are:

$\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )$

Answer:

Given,

$\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k$

Let $\vec d=d_1\hat i+d_2\hat j +d_3\hat k$

now, since it is given that d is perpendicular to $\vec a$ and $\vec b$ , we got the condition,

$\vec b.\vec d=0$ and $\vec a.\vec d=0$

$(\hat i+4\hat j +2\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$ And $(3\hat i-2\hat j +7\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$

$d_1+4d_2+2d_3=0$ And $3d_1-2d_2+7d_3=0$

here we got 2 equation and 3 variable. one more equation will come from the condition:

$\vec c . \vec d = 15$

$(2\hat i-\hat j +4\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=15$

$2d_1-d_2+4d_3=15$

so now we have three equation and three variable,

$d_1+4d_2+2d_3=0$

$3d_1-2d_2+7d_3=0$

$2d_1-d_2+4d_3=15$

On solving this three equation we get,

$d_1=\frac{160}{3},d_2=-\frac{5}{3}\:and\:d_3=-\frac{70}{3}$ ,

Hence Required vector :

$\vec d=\frac{160}{3}\hat i-\frac{5}{3}\hat j-\frac{70}{3}\hat k$

Answer:

Let, the sum of vectors $2\hat i + 4 \hat j -5 \hat k$ and $\lambda \hat i + 2 \hat j +3 \hat k$ be

$\vec a=(\lambda +2)\hat i + 6 \hat j -2 \hat k$

unit vector along $\vec a$

$\vec u=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{(\lambda+2)^2+6^2+(-2)^2}}=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}$

Now, the scalar product of this with $\hat i + \hat j + \hat k$

$\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}.(\hat i+\hat j +\hat k)$

$\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\frac{(\lambda +6) }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\lambda =1$

Answer:

Given

$|\vec a|=|\vec b|=|\vec c|$ and

$\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=0$

Now, let vector $\vec a+\vec b +\vec c$ is inclined to $\vec a , \vec b \: \: and \: \: \vec c$ at $\theta_1,\theta_2\:and\:\theta_3$ respectively.

Now,

$cos\theta_1=\frac{(\vec a+\vec b+\vec c).\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a +\vec a.\vec b +\vec c.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{|\vec a|}{|\vec a+\vec b+\vec c|}$

$cos\theta_2=\frac{(\vec a+\vec b+\vec c).\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec a.\vec b +\vec b.\vec b +\vec c.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec b.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{|\vec b|}{|\vec a+\vec b+\vec c|}$

$cos\theta_3=\frac{(\vec a+\vec b+\vec c).\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec a.\vec c +\vec b.\vec c +\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{|\vec c|}{|\vec a+\vec b+\vec c|}$

Now, Since, $|\vec a|=|\vec b|=|\vec c|$

$cos\theta_1=cos\theta_2=cos\theta_3$

$\theta_1=\theta_2=\theta_3$

Hence vector $\vec a+\vec b +\vec c$ is equally inclined to $\vec a , \vec b \: \: and \: \: \vec c$ .

Answer:

Given in the question,

are perpendicular and we need to prove that

LHS=

if are perpendicular,

= RHS

LHS ie equal to RHS

Hence proved.

Answer:

Given in the question

$\theta$ is the angle between two vectors $\vec a \: \: and \: \: \vec b$

$\vec a \cdot \vec b \geq 0$

$|\vec a| | \vec b |cos\theta\geq 0$

this will satisfy when

$cos\theta\geq 0$

$0\leq\theta\leq \frac{\pi}{2}$

Hence option B is the correct answer.

Answer:

Gicen in the question

$\vec a \: \: and \: \: \vec b$ be two unit vectors and $\theta$ is the angle between them

$|\vec a|=1,\:and\:\:|\vec b|=1$

also

$|\vec a + \vec b|=1$

$|\vec a + \vec b|^2=1$

$|\vec a|^2 + |\vec b|^2+2\vec a.\vec b=1$

$1 + 1+2\vec a.\vec b=1$

$\vec a.\vec b=-\frac{1}{2}$

$|\vec a||\vec b|cos\theta =-\frac{1}{2}$

$cos\theta =-\frac{1}{2}$

$\theta =\frac{2\pi}{3}$

Then $\vec a + \vec b$ is a unit vector if $\theta =\frac{2\pi}{3}$

Hence option D is correct.

(A) 0

(B) –1

(C) 1

(D) 3

Answer:

To find the value of $\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j )$

$\\\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j ) \\=\hat i.\hat i+\hat j(-\hat j)+\hat k.\hat k\\=1-1+1\\=1$

Hence option C is correct.

Answer:

Given in the question

$\theta$ is the angle between any two vectors $\vec a \: \:and \: \: \vec b$ and $|\vec a \cdot \vec b |=|\vec a \times \vec b |$

To find the value of $\theta$

Hence option D is correct.

## Vector Algebra Class 12 - Topics

The main topics that will cover while practising Vector Algebra Class 12 NCERT solutions are:

• Concepts of Vectors
• Position Vector, Magnitude and Direction of Vectors and Unit Vector
• Direction Cosines and Ratios
• Addition and Subtraction of Vectors
• Collinear Vectors
• Dot Product and Cross Product of Vectors
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

### NCERT Vector Algebra Class 12 Solutions

A total of 5 exercises are covered in Class 12 Maths chapter 10 Vector Algebra. That are four exercises and one miscellaeous exercise. The exercises are listed below. Students can doenload the solutions for Individual exercises also. There are a total of 73 questions from these 5 exercises.

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## NCERT solutions for class 12 maths - Chapter wise

 Chapter 1 NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Chapter 2 NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions Chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices Chapter 4 NCERT solutions for class 12 maths chapter 4 Determinants Chapter 5 NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Chapter 6 NCERT solutions for class 12 maths chapter 6 Application of Derivatives Chapter 7 NCERT solutions for class 12 maths chapter 7 Integrals Chapter 8 NCERT solutions for class 12 maths chapter 8 Application of Integrals Chapter 9 NCERT solutions for class 12 maths chapter 9 Differential Equations Chapter 10 NCERT solutions for class 12 maths chapter 10 Vector Algebra Chapter 11 NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry Chapter 12 NCERT solutions for class 12 maths chapter 12 Linear Programming Chapter 13 NCERT solutions for class 12 maths chapter 13 Probability

## Key Features of NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra

NCERT Solutions for ch 10 maths class 12 - Vector Algebra is a comprehensive guide that helps students understand the concepts of vector algebra. Some of the key features of these solutions are:

1. Comprehensive coverage: The vectors class 12 ncert solutions cover all the topics included in the Class 12 Maths syllabus, ensuring that students are well-prepared for their exams.

2. Simplified language: The vectors maths class 12 solutions are written in simple language, making it easy for students to understand the concepts of vector algebra.

3. Detailed explanations: The solutions provide detailed explanations of concepts, which help students to understand the fundamental principles of vector algebra.

4. Step-by-step approach: The algebra chapters of class 12 solutions follow a step-by-step approach, which helps students to understand the solution process in a structured way.

5. Illustrated solutions: The solutions are accompanied by diagrams and illustrations, which help students to visualize the solution process and understand the concepts better.

6. Conceptual clarity: The solutions aim to develop the conceptual clarity of students, rather than just providing them with the final answers. This helps students to build a strong foundation in the subject.

7. Previous years' question papers: The solutions also include the solutions to previous years' question papers, and miscellaneous exercise chapter 10 class 12, which help students to get a better understanding of the exam pattern and types of questions asked in the exams.

## What are Vectors and Scalars

Vector Quantity- Quantity which involves both the value magnitude and direction. Vector quantities like weight, velocity, acceleration, displacement, force, momentum, etc.

Scalar Quantity- Quantity which involves only one value (magnitude) which is a real number. Scalar quantities like distance, length, time, mass, speed, area, temperature, work, money, volume, voltage, density, resistance, etc.

### Benefits of NCERT vector algebra class 12 solutions

• NCERT solutions for maths chapter 10 class 12 are explained in a step-by-step manner, so it will be very easy for you to understand the concepts.

• NCERT Solutions for maths chapter 10 class 12 vector algebra will give you some new way to solve the problem.

• Performance in the 12th board exam plays a very important role in deciding the future, so you can get admission to a good college. Scoring good marks in the exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 10 vector algebra.

• To develop a grip on the concept, you should solve the miscellaneous exercise also. In NCERT Class 12 Maths solutions chapter 10 vector algebra article, you will get a solution of miscellaneous exercise also.

## NCERT Books and NCERT Syllabus

### Frequently Asked Question (FAQs)

1. What is the weightage of the chapter vector algebra for CBSE board exam?

Two chapters combined vector algebra & three-dimensional geometry has 17% weightage in 12th board maths final exam. So the vector algebra class 12 NCERT solutions can be practised to get a good score in the CBSE board exam. Students can solve problems related to the NCERT syllabus of vector algebra using NCERT solutions and NCERT Exemplar solutions

2. What are the important topics in chapter vector algebra?

Some basic concepts of vectors, addition and multiplication of a vector by a scalar, vector joining two points, scalar (or dot) product of two vectors, projection of a vector on a line and vector (or cross) product of two vectors are the important topics of this chapter.

3. Where can I find solutions to the exercises for NCERT Solutions for class 12 math chapter 10?

Choosing appropriate study material is essential for chapter 10 class 12 Maths, as it aids in the efficient resolution of textbook problems. It requires considerable patience to choose the appropriate reference guide from the various options available in the market. Careers360 provides solutions for both chapter-wise and exercise-wise problems in PDF format. Students can use this resource to clear their doubts instantly while solving problems. For ease students can study vector algebra class 12 pdf both online and offline.

4. Where can I find the complete solutions of NCERT for class 12 maths?

A Here you will get the detailed NCERT solutions for class 12 maths  by clicking on the link.

5. Which is the best book for CBSE class 12 Maths?

NCERT textbook is the best book for CBSE class 12 maths. You don't need to buy any supplementary book. All you need to do is rigorous practice of all the questions given in the NCERT textbook.

6. Which are the most difficult chapters of NCERT Class 12 Maths syllabus?

Most of the students consider integration and applications of integrations as the most difficult chapters in CBSE class 12 maths but with the regular practice of NCERT questions you will be able to have a strong grip on them also.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

• Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

• Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

• Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

• Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.

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In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

• Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

• Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

• Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

• Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

• Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

• JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
• NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

• Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
• NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
• Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

• High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

• Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

• Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

• Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

• Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9
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