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

NCERT solutions for exercise 10.3 Class 12 Maths chapter 10 explains the questions related to dot products of vectors. As far as the unit vector algebra is concerned dot product is an important topic. NCERT syllabus Exercise 10.3 Class 12 Math familiarise the students with the concept of dot products. NCERT solutions for Class 12 Maths chapter 10 exercise 10.3 are framed by maths expert and is in accordance with the CBSE syllabus. Students can use Class 12 Maths chapter 10 exercise 10.3 for the preparations of CBSE Class 12 Board Exams. Along with NCERT practice CBSE Previous Year Class 12 Maths Questions to get a good score in board exam. Class 12th Maths chapter 10 exercise 10.3 helps in preparation for exams like JEE Main also.

Also practice-

  • Vector Algebra Exercise 10.1

  • Vector Algebra Exercise 10.2

  • Vector Algebra Exercise 10.4

  • Vector Algebra Miscellaneous Exercise

Vector Algebra Class 12 Chapter 10 Exercise: 10.3

Question:1 Find the angle between two vectors \vec a \: \:and \: \: \vec b with magnitudes \sqrt 3 \: \:and \: \: 2 , respectively having . \vec a . \vec b = \sqrt 6

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} .

Question:2 Find the angle between the vectors \hat i - 2 \hat j + 3 \hat k \: \:and \: \: 3 \hat i - 2 \hat j + \hat k

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}

Question:3 Find the projection of the vector \hat i - \hat j on the vector \hat i + \hat j

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.


Question:4 Find the projection of the vector \hat i + 3 \hat j + 7 \hat k on the vector 7\hat i - \hat j + 8 \hat k

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}}

Question:5 Show that each of the given three vectors is a unit vector: \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}{7}( 6\hat i + 2 \hat j -3\hat k ) Also, show that they are mutually perpendicular to each other.

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.

Question:6 Find |\vec a| \: \: and\: \:| \vec b | , if ( \vec a + \vec b ). ( \vec a - \vec b )=8 \: \:and \: \: |\vec a |\: \:= 8 \: \:|\vec b | .

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}}

Question:7 Evaluate the product ( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b ) .

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

Question:8 Find the magnitude of two vectors \vec a \: \: and \: \: \vec b , having the same magnitude and such that the angle between them is 60 \degree and their scalar product is 1/2

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

Question:9 Find |\vec x | , if for a unit vector \vec a , ( \vec x -\vec a ) . ( \vec x + \vec a ) = 12

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 \sqrt{13}

Question:10 If \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 are such that \vec a + \lambda \vec b is perpendicular to \vec c , then find the value of \lambda

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 ,

Question:11 Show that |\vec a | \vec b + |\vec b | \vec a is perpendicular to |\vec a | \vec b - |\vec b | \vec a , for any two nonzero vectors \vec a \: \: \: and \: \: \vec b .

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 .

Question:12 If \vec a . \vec a = 0 \: \: and \: \: \vec a . \vec b = 0 , then what can be concluded about the vector \vec b ?

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

Question:13 If \vec a , \vec b , \vec c are unit vectors such that \vec a + \vec b + \vec c = \vec 0 , find the value of \vec a . \vec b + \vec b. \vec c + \vec c . \vec a

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 \frac{-3}{2}

Question:14 If either vector \vec a = 0 \: \: or \: \: \vec b = 0 \: \: then \: \: \vec a . \vec b = 0 . But the converse need not be true. Justify your answer with an example

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.

Question:15 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find \angle ABC , [\angle ABC is the angle between the vectors \overline{BA}\: \: and\: \: \overline{BC} ] .

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

Question:16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

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.

Question:17 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.

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.

Question:18 If \vec a is a nonzero vector of magnitude ‘a’ and \lambda a nonzero scalar, then \lambda \vec a is unit vector if

\\A ) \lambda = 1 \\\\ B ) \lambda = -1 \\\\ C ) a = |\lambda | \\\\ D ) a = 1 / |\lambda |

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.

More About NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3

There are a total of 18 questions from exercise 10.3 Class 12 Maths. These questions mainly are based on the concepts of dot products. Dot products are used to identify orthogonal vectors and the projections of a vector. Question number 4 of Class 12 Maths chapter 10 exercise 10.3 gives an example of the projection of a vector on another.

Read Also| Vector Algebra Class 12 Chapter 10 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3

  • Similar questions as in Class 12th Maths chapter 10 exercise 10.3 can be expected for the CBSE Class 12 Maths exam. The questions may not be exactly the same.

  • Problems based on dot products are also asked in various competitive exams like JEE Main, KEAM etc.

  • Problems discussed in exercise 10.3 Class 12 Maths are useful not only for Class 12 Maths but also for Class 11 and 12 Physics also.

Also see-

  • NCERT Exemplar Solutions Class 12 Maths Chapter 10

  • NCERT Solutions for Class 12 Maths Chapter 10

NCERT Solutions Subject Wise

  • NCERT Solutions Class 12 Chemistry

  • NCERT Solutions for Class 12 Physics

  • NCERT Solutions for Class 12 Biology

  • NCERT Solutions for Class 12 Mathematics

Subject Wise NCERT Exemplar Solutions

  • NCERT Exemplar Class 12 Maths

  • NCERT Exemplar Class 12 Physics

  • NCERT Exemplar Class 12 Chemistry

  • NCERT Exemplar Class 12 Biology