NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 is the solution of Exercise 10.4. This page is dedicated to the PDF solution and information on the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4. These solutions are made by collaborating all the important resources and data researched on the topic of Vector Algebra. Students have been provided with the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 on this website to make their preparation for the Class 12 board examinations an effortless journey.

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 is in downloadable PDF format, and all the exercises in the chapter, are solved by Extramarks experts. Extramarks’ PDFs are developed in line with the guidelines provided by CBSE and the standard of answers a CBSE examiner would expect. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 with all the problems and solutions are helpful in the revision of the complete syllabus of the chapter, which enhances the scoring abilities of the students’ answers.

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4

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Q.1 $Find | \vec{a} \times \vec{b} |, if \vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} .$

Ans

$\begin{array}{l} Given vectors are: \vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} \\ Then, \\ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 7 & 7 \\ 3 & - 2 & 2 \end{matrix} | \\ = (- 14 + 14) \hat{i} + (2 - 21) \hat{j} + (- 2 + 21) \hat{k} \\ = - 19 \hat{j} + 19 \hat{k} \\ ∴ | \vec{a} \times \vec{b} | = | - 19 \hat{j} + 19 \hat{k} | \\ = \sqrt{{(- 19)}^{2} + {(19)}^{2}} \\ = 2 \sqrt{19} \end{array}$

Q.2

$\begin{array}{l} Find a unit vector perpendicular to each of the vector \vec{a} + \vec{b} \\ and \vec{a} - \vec{b}, where \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} . \end{array}$

Ans

$\begin{array}{l} The given vectors are \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} . \\ Then, \vec{a} + \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} + \hat{i} + 2 \hat{j} - 2 \hat{k} \\ = 4 \hat{i} + 4 \hat{j} \\ and \vec{a} - \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} - \hat{i} - 2 \hat{j} + 2 \hat{k} \\ = 2 \hat{i} + 4 \hat{k} \\ (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{matrix} | \\ = (16 - 0) \hat{i} - (16 - 0) \hat{j} + (0 - 8) \hat{k} \\ = 16 \hat{i} - 16 \hat{j} - 8 \hat{k} \\ | (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) | \\ = | 16 \hat{i} - 16 \hat{j} - 8 \hat{k} | \\ = \sqrt{{(16)}^{2} + {(- 16)}^{2} + {(- 8)}^{2}} \\ = 24 \\ Hence, the unit vector perpendicular to each of the vectors \\ (\vec{a} + \vec{b}) and (\vec{a} - \vec{b}) is given by the relation, \\ = \pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{| (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) |} \\ = \pm \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24} \\ = \pm \frac{2 \hat{i} - 2 \hat{j} - \hat{k}}{3} \end{array}$

Q.3

$\begin{array}{l} If a unit vector \vec{a} makes angles \frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j} and an \\ acute angle θ with \hat{k}, then find θ and hence, the components of \vec{a} . \end{array}$

Ans

$\begin{array}{l} Let unit vector \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} and | \vec{a} | = 1 . \\ It is given that \vec{a} makes angles \frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j} and an acute \\ angle θ with \hat{k} . \\ Then, we have: \\ cos \frac{π}{3} = \frac{a_{1}}{| \vec{a} |} \Rightarrow \frac{1}{2} = \frac{a_{1}}{1} \\ \Rightarrow a_{1} = \frac{1}{2} \\ cos \frac{π}{4} = \frac{a_{2}}{| \vec{a} |} \Rightarrow \frac{1}{\sqrt{2}} = \frac{a_{2}}{1} \\ \Rightarrow a_{2} = \frac{1}{\sqrt{2}} \\ cos θ = \frac{a_{3}}{| \vec{a} |} \Rightarrow cos θ = \frac{a_{3}}{1} \\ \Rightarrow a_{3} = cos θ \\ Since, | \vec{a} | = 1 \\ \Rightarrow \sqrt{{a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}} = 1 \\ \Rightarrow {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} = 1 \\ {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} + {cos}^{2} θ = 1 \\ {cos}^{2} θ = 1 - \frac{1}{4} - \frac{1}{2} \\ {cos}^{2} θ = \frac{1}{4} \\ \Rightarrow cos θ = \frac{1}{2} = \cos \frac{π}{3} \\ So, θ = \frac{π}{3} \\ Thus, θ = \frac{π}{3} and and the components of \vec{a} are (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}) . \end{array}$

Q.4 $Show that (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b})$

Ans

$\begin{array}{l} L . H . S . = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \\ = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \\ = 0 + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - 0 [∵ \vec{b} \times \vec{a} = \vec{a} \times \vec{b}] \\ = 2 (\vec{a} \times \vec{b}) = R . H . S . \end{array}$

Q.5 $Find λ and μ if (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0 .$

Ans

$\begin{array}{l} (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0 \\ \Rightarrow | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & λ & μ \end{matrix} | = 0 \\ (6 μ - 27 λ) \hat{i} - (2 μ - 27) \hat{j} + (2 λ - 6) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \\ On comparing the corresponding components, we have : \\ 6 μ - 27 λ = 0, 2 μ - 27 = 0 and 2 λ - 6 = 0 \\ \Rightarrow μ = \frac{27}{2} and λ = 3 \end{array}$

Q.6

$\begin{array}{l} Given that \vec{a} . \vec{b} = 0 and \vec{a} \times \vec{b} = 0 . What can you conclude \\ about the vectors \vec{a} and \vec{b} ? \end{array}$

Ans

$\begin{array}{l} (i) \vec{a} . \vec{b} = 0 \\ Either | \vec{a} | = 0 or | \vec{b} | = 0, \\ or \vec{a} ⊥ \vec{b} (if \vec{a} and \vec{b} are non-zero.) \\ (ii) \vec{a} \times \vec{b} = 0 \\ Either | \vec{a} | = 0 or | \vec{b} | = 0, \\ or \vec{a} ∥ \vec{b} (if \vec{a} and \vec{b} are non-zero.) \\ But, \vec{a} and \vec{b} cannot be perpendicular and parallel simultaneously. \\ Hence, | \vec{a} | = 0 or | \vec{b} | = 0 . \end{array}$

Q.7

$\begin{array}{l} Let the vectors \vec{a}, \vec{b}, \vec{c} be given as a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \\ b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . Then show that \\ \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} . \end{array}$

Ans

$\begin{array}{l} The given vectors are: a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . \\ L . H . S . = \vec{a} \times (\vec{b} + \vec{c}) \\ = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} + c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}) \\ = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times {(b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k}} \\ = | \begin{matrix} \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{matrix} | \\ = {a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2})} \hat{i} + {a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1})} \hat{j} \\ + {a_{1} (b_{2} + c_{2}) - a_{2} (b_{3} + c_{3})} \hat{k} \\ = {a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}} \hat{i} + {a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}} \hat{j} \\ + {a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{3} - a_{2} c_{3}} \hat{k} . .. (i) \\ \vec{a} \times \vec{b} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix} | \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ \vec{a} \times \vec{c} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} | \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} + (a_{2} b_{3} - a_{3} b_{2}) \hat{i} \\ + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ = (a_{2} b_{3} - a_{3} b_{2} + a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1} + a_{1} b_{3} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} - a_{2} b_{1} + a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ = (a_{2} b_{3} + a_{2} b_{3} - a_{3} b_{2} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} + a_{1} b_{3} - a_{3} b_{1} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} + a_{1} b_{2} - a_{2} b_{1} - a_{2} b_{1}) \hat{k} . .. (ii) \\ From equation (i) and equation (ii), we have \\ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} . \\ Hence, the given result is proved. \end{array}$

Q.8

$\begin{array}{l} If either vector \vec{a} = 0 or \vec{b} = 0, then \vec{a} . \vec{b} = 0 . But the converse \\ need not be true. Justify your answer with an example. \end{array}$

Ans

$\begin{array}{l} Let \vec{a} = 3 \hat{i} + 4 \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \\ Then, \\ \vec{a} . \vec{b} = (3 \hat{i} + 4 \hat{j} + 3 \hat{k}) (2 \hat{i} + 3 \hat{j} - 6 \hat{k}) \end{array}$

$\begin{array}{l} = 6 + 12 - 18 \\ = 0 \\ And \\ | \vec{a} | = | 3 \hat{i} + 4 \hat{j} + 3 \hat{k} | \\ = \sqrt{3^{2} + 4^{2} + 3^{2}} \\ = \sqrt{9 + 16 + 9} \\ = \sqrt{34} \neq 0 \\ | \vec{b} | = | 2 \hat{i} + 3 \hat{j} - 6 \hat{k} | \\ = \sqrt{2^{2} + 3^{2} + {(- 6)}^{2}} \\ = \sqrt{4 + 9 + 36} \\ = \sqrt{49} \\ = 7 \neq 0 \\ Hence, the converse of the given statement need not be true. \end{array}$

Q.9 Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Ans

$\begin{array}{l} The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5) \\ and C (1, 5, 5). \\ The adjacent sides \vec{AB} and \vec{BC} of Δ ABC are given as: \\ \vec{AB} = (2 - 1) \hat{i} + (3 - 1) \hat{j} + (5 - 2) \hat{k} \\ = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \vec{BC} = (1 - 2) \hat{i} + (5 - 3) \hat{j} + (5 - 5) \hat{k} \\ = - \hat{i} + 2 \hat{j} \\ ∴ \vec{AB} \times \vec{BC} \\ = \frac{1}{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 1 & 2 & 0 \end{matrix} | \\ = \frac{1}{2} {(0 - 6) \hat{i} - (0 + 3) \hat{j} + (2 + 2) \hat{k}} \\ = \frac{1}{2} (- 6 \hat{i} - 3 \hat{j} + 4 \hat{k}) \\ The area ΔABC \\ = \frac{1}{2} | \vec{AB} \times \vec{BC} | \\ = \frac{1}{2} | - 6 \hat{i} - 3 \hat{j} + 4 \hat{k} | \\ = \frac{1}{2} \sqrt{{(- 6)}^{2} + {(- 3)}^{2} + 4^{2}} \\ = \frac{1}{2} \sqrt{36 + 9 + 16} \\ = \frac{1}{2} \sqrt{61} square units \\ Thus, the area of Δ ABC is \frac{1}{2} \sqrt{61} square units . \end{array}$

Q.10

$\begin{array}{l} Find the area of the parallelogram whose adjacents sides \\ are determined by the vectors \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} . \end{array}$

Ans

$\begin{array}{l} The area of the parallelogram whose adjacent sides are \vec{a} and \vec{b} \\ is | \vec{a} \times \vec{b} | . \\ Adjacent sides are in vector form, which are as follows: \\ \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \\ \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \\ ∴ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 3 \\ 2 & - 7 & 1 \end{matrix} | \\ = (- 1 + 21) \hat{i} + (1 - 6) \hat{j} + (- 7 + 2) \hat{k} \\ = 20 \hat{i} - 5 \hat{j} - 5 \hat{k} \\ | \vec{a} \times \vec{b} | = | 20 \hat{i} - 5 \hat{j} - 5 \hat{k} | \\ = \sqrt{20^{2} + {(- 5)}^{2} + {(- 5)}^{2}} \\ = \sqrt{400 + 25 + 25} \\ = \sqrt{450} \\ = 15 \sqrt{2} \\ Thus, the area of the given parallelogram is 15 \sqrt{2} square units. \end{array}$

Q.11

$\begin{array}{l} Let the vectors \vec{a} and \vec{b} be such that | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{3}}{2}, then \\ \vec{a} \times \vec{b} is a unit vector, if the angle between \vec{a} and \vec{b} is \\ (A) \frac{π}{6} \\ (B) \frac{π}{4} \\ (C) \frac{π}{3} \\ (D) \frac{π}{2} \end{array}$

Ans

$\begin{array}{l} Given : | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{2}}{3} \\ Since, \vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | sinθ \hat{n}, where \hat{n} is a unit vector perpendicular \\ to both \vec{a} and \vec{b} and θ is the angle between \vec{a} and \vec{b} . \\ Since, \vec{a} \times \vec{b} is a unit vector, so | \vec{a} \times \vec{b} | = 1 . \\ | \vec{a} \times \vec{b} | = 1 \\ \Rightarrow | | \vec{a} | | \vec{b} | sinθ \hat{n} | = 1 \\ \Rightarrow | \vec{a} | | \vec{b} | | sinθ | = 1 \\ 3 \times \frac{\sqrt{2}}{3} \times sinθ = 1 \\ sinθ = \frac{1}{\sqrt{2}} = \sin \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \\ Hence, \vec{a} \times \vec{b} is a unit vector if the angle between \vec{a} and \vec{b} is \frac{π}{4} . \\ The correct answer is B. \end{array}$

Q.12

$\begin{array}{l} Area of a rectangle having vertices A, B, C and D with \\ position vector - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ and - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} respectively is \\ (A) \frac{1}{2} \\ (B) 1 \\ (C) 2 \\ (D) 4 \end{array}$

Ans

$\begin{array}{l} The position vectors of vertices A, B, C, and D of rectangle \\ ABCD are given as: \\ \vec{OA} = - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{OB} = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{OC} = \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} and \\ \vec{OD} = - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ The adjacent sides \vec{AB} and \vec{BC} of the given rectangle are given as: \\ \vec{AB} = (1 + 1) \hat{i} + (\frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = 2 \hat{i} \\ \vec{BC} = (1 - 1) \hat{i} + (- \frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = - \hat{j} \\ ∴ \vec{AB} \times \vec{BC} \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & - 1 & 0 \end{matrix} | \\ = (- 2 - 0) \hat{k} \\ = - 2 \hat{k} \\ | \vec{AB} \times \vec{BC} | \\ = | - 2 \hat{k} | \\ = 2 \\ Now, it is known that the area of a parallelogram whose adjacent \\ sides are \vec{a} and \vec{b} is | \vec{a} \times \vec{b} | . \\ Hence, the area of the given rectangle is | \vec{AB} \times \vec{BC} | = 2 square units. \\ The correct answer is C. \end{array}$

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FAQs (Frequently Asked Questions)

1. How many problems and solutions are there in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 has 12 questions all of which have solutions provided by Extramarks.

2. In what ways is the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 useful for Class 12 Mathematics preparation?

Students should the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 as a guide while solving the chapter on Vector Algebra. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 can be used during the zero-hour preparation of the examination or be used as revision material before the examination. Class 12 Mathematics can be challenging; having the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 by the side can be helpful when students need help and do not have time to waste. This keeps the students’ flow while solving Exercise 10.4 Class 12 intact.

3. What is the best procedure to prepare for CBSE Class 12 Mathematics?

The preparation for CBSE Class 12 Mathematics is as follows:

Analyse the NCERT(CBSE) Syllabus thoroughly.
Create a plan according to one’s study pattern.
Use past years’ papers and study material like the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 to understand the level of preparation.
Gain strength over weak areas by continuous revision.
Constant self-assessment by using the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 and other revision material to enhance preparation.

4. What topics are dealt with in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 deals with the chapter on Vector Algebra. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 has clear step-by-step solutions with detailed explanations for all the questions.

5. Is the NCERT Solutions useful for Class 12 students?

Class 12 serves as a bridge between a student’s high school and college experiences. NCERT Solutions are a very useful resource to perform remarkably. Extramarks’ NCERT Solutions are accurate, easy and reliable. The preparations will benefit evidently from using study tools provided by Extramarks like the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4

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3. What is the best procedure to prepare for CBSE Class 12 Mathematics?

4. What topics are dealt with in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

5. Is the NCERT Solutions useful for Class 12 students?

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

Access NCERT Solutions for Class 12 Maths Chapter-10 – Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4

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FAQs (Frequently Asked Questions)

1. How many problems and solutions are there in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

2. In what ways is the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 useful for Class 12 Mathematics preparation?

3. What is the best procedure to prepare for CBSE Class 12 Mathematics?

4. What topics are dealt with in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

5. Is the NCERT Solutions useful for Class 12 students?

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