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

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 10 Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra includes solutions to all Miscellaneous Exercise problems. Class 12 Maths Chapter 10 Miscellaneous Exercise Solutions are based on the ideas presented in Maths Chapter 10. This practice is crucial for competitive exams as well as the CBSE Board exams. Get the Class 12 Maths Vector Algebra Miscellaneous Exercise NCERT Solutions in PDF format and practise them offline to do well in the board exam. Students can download the revised Class 12 Maths NCERT Solutions from our page, which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 10 Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra
2.1Miscellaneous Exercise
3. Class 12 Maths Chapter 10: Exercises Breakdown
4. CBSE Class 12 Maths Chapter 10 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Miscellaneous Exercise Class 12 Chapter 10 Solutions are aligned with the updated CBSE guidelines for Class 12, ensuring students are well-prepared for exams. Access the Class 12 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

### Miscellaneous Exercise

1. Write down a unit vector in the XY-plane, making an angle of ${30^\circ }$ with the positive direction of the x-axis.

Ans: Unit vector is $\vec r = \cos \theta \hat i + \sin \theta \hat j$, where $\theta$ is angle with positive ${\text{X}}$ axis. $\vec r = \cos {30^\circ }\hat i + \sin {30^\circ }\hat j = \dfrac{{\sqrt 3 }}{2}\hat i + \dfrac{1}{2}\hat j$

2. Find the scalar components and magnitude of the vector joining the points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$

Ans: $\overrightarrow {{\text{PQ}}} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k$

$|\overrightarrow {{\text{PQ}}} |\; = \;\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$

3. A girl walks 4 km towards the west, then she walks 3 km in a direction ${30^\circ }$ east of north and stops. Determine the girl's displacement from her initial point of departure.

Ans:

$\overrightarrow {{\text{OA}}} = - 4\hat i$

$\overrightarrow {{\text{AB}}} = \hat i|\overrightarrow {{\text{AB}}} |\cos {60^\circ } + \hat j|\overrightarrow {{\text{AB}}} |\sin {60^\circ }$

$= \hat i3 \times \dfrac{1}{2} + \hat j3 \times \dfrac{{\sqrt 3 }}{2}$

$= \dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$\overrightarrow {{\text{OB}}} = \overrightarrow {{\text{OA}}} + \overrightarrow {{\text{AB}}}$

$= ( - 4\hat i) + \left( {\dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}j} \right)$

$= \left( { - 4 + \dfrac{3}{2}} \right)\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$= \left( {\dfrac{{ - 8 + 3}}{9}} \right)\hat i + \dfrac{{3\sqrt 3 }}{9}\hat j$

$= \dfrac{{ - 5}}{2}\vec i + \dfrac{{3\sqrt 3 }}{2}\hat j$

4. If $\overrightarrow a = \vec b + \vec c$, then is it true that $|\overrightarrow a | = |\vec b| + |\vec c|$ ? Justify your answer.

Ans:  $\operatorname{In} \Delta {\text{ABC}},\overrightarrow {\;CB} = \vec a,\;\overrightarrow {CA} = b,\;\overrightarrow {AB} = \vec c$

$\vec a = \vec b + \vec c$,  by triangle law of addition for vectors.

$|\vec a|\; < \;|\vec b| + |\vec c|$,  by triangle inequality law of lengths.

Hence, it's not true that $|\vec a| = |\vec b| + |\vec c|$

5. Find the value of $x$ for which $x(\hat i + \hat j + \hat k)$ unit vector.

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

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

$\Rightarrow \sqrt {3{x^2}} = 1$

$\Rightarrow \sqrt 3 x = 1$

$\Rightarrow x = \pm \dfrac{1}{{\sqrt 3 }}$

6.  Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec a = 2\hat i + 3\hat j - \hat k$ and $\vec b = \hat i - 2\hat j + \hat k$

Ans: $\vec c = \vec a + \vec b = (2 + 1)\hat i + (3 - 2)\hat j + ( - 1 + 1)\hat k = 3\hat i + \hat j$

$|\vec c| = \sqrt {{3^2} + {1^2}} = \sqrt {9 + 1} = \sqrt {10}$

$\therefore \hat c = \dfrac{{\vec c}}{{|\vec c|}} = \dfrac{{(3\hat i + \hat j)}}{{\sqrt {10} }}$

So, a vector of magnitude 5 and parallel to the resultant of $\vec a$ and $\vec b$ is $\pm 5(\hat c) = \pm 5\left( {\dfrac{1}{{\sqrt {10} }}(3\hat i + \hat j)} \right) = \pm \dfrac{{3\sqrt {10} }}{2}\hat i \pm \dfrac{{\sqrt {10} }}{2}\hat j$

7.  If $\vec a = \hat i + \hat j + \hat k\;,\;\vec b = 2\hat i - \hat j + 3\hat k$ and $\vec c = \hat i - 2\hat j + \hat k$, find a unit vector parallel to the vector $2\vec a - \vec b + 3\vec c$.

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

$= 2\hat i + 2\hat j + 2\hat k - 2\hat i + j - 3\vec k + 3\hat i - 6\hat j + 3\hat k$

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

$|2\vec a - \vec b + 3c| = \sqrt {{3^2} + {{( - 3)}^2} + {2^2}} = \sqrt {9 + 9 + 4} = \sqrt {22}$

Thus ,required unit vector is $\dfrac{{2\vec a - \vec b + 3\vec c}}{{|2\vec a - \vec b + 3\vec c|}} = \dfrac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {22} }} = \dfrac{3}{{\sqrt {22} }}\hat i - \dfrac{3}{{\sqrt {22} }}\hat j + \dfrac{2}{{\sqrt {22} }}\hat k$

8. Show that the points ${\text{A}}(1, - 2, - 8),{\text{B}}(5,0, - 2)$ and ${\text{C}}(11,3,7)$ are collinear, and find the ratio in which B divides Ac.

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

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

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

$|\overrightarrow {{\text{AB}}} | = \sqrt {{4^2} + {2^2} + {6^2}} = \sqrt {16 + 4 + 36} = \sqrt {56} = 2\sqrt {14}$

$|\overrightarrow {{\text{BC}}} | = \sqrt {{6^2} + {3^2} + {9^2}} = \sqrt {36 + 9 + 81} = \sqrt {126} = 3\sqrt {14}$

$|\overrightarrow {AC} | = \sqrt {{{10}^2} + {5^2} + {{15}^2}} = \sqrt {100 + 25 + 225} = \sqrt {350} = 5\sqrt {14}$

$\therefore \;\;|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$

So, the points are collinear.

Let B divide AC in ratio $\lambda :1$. $\overrightarrow {{\text{OB}}} = \dfrac{{\lambda \overrightarrow {{\text{OC}}} + \overrightarrow {{\text{OA}}} }}{{(\lambda + 1)}}$

$\Rightarrow 5\hat i - 2\hat k = \dfrac{{\lambda (11\hat i + 3\hat j + 7\hat k) + (\hat i - 2\hat j - 8\hat k)}}{{\lambda + 1}}$

$\Rightarrow (\lambda + 1)(5\hat i - 2\hat k) = 11\lambda \hat i + 3\lambda \hat j + 7\lambda \hat k + \hat i - 2\hat j - 8\hat k$

$\Rightarrow 5(\lambda + 1)\hat i - 2(\lambda + 1)\hat k = (11\lambda + 1)\hat i + (3\lambda - 2)\hat j + (7\lambda - 8)\hat k$

$\Rightarrow \lambda = \dfrac{2}{3}$

So, the required ratio is 2:3

9. Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec a + \vec b)$ and $(\vec a - 3\vec b)$ externally in the ratio 1: 2. Also, show that $P$ is the midpoint of the line segment RQ.

Ans: $\overrightarrow {{\text{OP}}} = 2\vec a + \vec b,\overrightarrow {{\text{OQ}}} = \overrightarrow {\text{a}} - 3\overrightarrow {\text{b}}$

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

So, the position vector of ${\text{R}}$ is $3\overrightarrow {\text{a}} + 5\overrightarrow {\text{b}}$

Position vector of midpoint of ${\text{RQ}} = \dfrac{{\overrightarrow {{\text{OQ}}} + \overrightarrow {{\text{OR}}} }}{2}$

$= \dfrac{{(a\sqrt 6 ) + (3\vec a + 5\bar b)}}{2}$

$= 2\vec a + \vec b$

$= \overrightarrow {OP}$

Thus, ${\text{P}}$ is midpoint of line segment ${\text{RQ}}$

10. The two adjacent sides of a parallelogram are $2\hat i - 4\hat j + 5\hat k$ and $\hat i - 2\hat j - 3\hat k$. Find the unit vector parallel to its diagonal. Also, find its area.

Ans:  Diagonal of a parallelogram is $\vec a + \vec b$

$\vec a + \vec b = (2 + 1)\hat i + ( - 4 - 2)\hat j + (5 - 3)\hat k = 3\hat i - 6\hat j + 2\hat k$

So, the unit vector parallel to diagonal is $\dfrac{{\vec a + \vec b}}{{|\vec a + \vec b|}} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {9 + 36 + 4} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k$

$\vec{a}\times \vec{b} = \begin{vmatrix} \hat i& \hat j &\hat k \\ 2& -4 & 3\\ 1& -2 & -3 \end{vmatrix}$

$= \hat i(12 + 10) - \hat j( - 6 - 5) + \bar k( - 4 + 4)$

$= 22\hat i + 11\hat j$

$= 11(2\hat i + \hat j)$

$\therefore |\vec a \times \vec b| = 11\sqrt {{2^2} + {1^2}} = 11\sqrt 5$

So, the area of the parallelogram is $11\sqrt 5$ sq units

11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ  are $\dfrac{1}{{\sqrt 3 }}$ $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.

Ans: Let a vector be equally inclined to ${\text{OX}},{\text{OY}}$, and ${\text{OZ}}$ at an angle $\alpha$.

So, the Direction Cosines of the vector are $\cos \alpha ,\cos \alpha$and $\cos \alpha$.

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

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

$\Rightarrow \cos \alpha = \dfrac{1}{{\sqrt 3 }}$

So, the DCs of the vector are $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.

12. Let $\vec a = \hat i + 4\hat j + 2\hat k$ and $\vec b = 3\hat i - 2\hat j + 7\hat k$ and $\vec c = 2\hat i - \hat j + 4\hat k$. Find a vector $\vec d$ which is

perpendicular to both $\vec a$ and $\vec b$ and $\vec c \cdot \vec d = 15$

Ans: $\vec d = {d_1}\hat i + {d_2}\hat \jmath + {d_3}\hat k$

$\vec d\;.\;\vec a = 0 \Rightarrow {d_1} + 4{d_2} + 2{d_3} = 0$

$\vec d\;.\;\vec b = 0 \Rightarrow 3{d_1} - 2{d_2} + 7{d_3} = 0$

$\vec c \cdot \vec d = 15 \Rightarrow 2{d_1} - {d_2} + 4{d_3} = 15$

Solving these equations, we get ${d_1} = \dfrac{{160}}{3},{d_2} = - \dfrac{5}{3},{d_3} = - \dfrac{{70}}{3}$

$\therefore \vec d = \dfrac{{160}}{3}\hat i + \dfrac{5}{3}\widehat j + \dfrac{{70}}{3}\hat k = \dfrac{1}{3}(160\hat i + 5\hat j + 70\hat k)$

13. The scalar product of the vector $\hat i + \hat j + \hat k$ with a unit vector along the sum of vectors $2\hat i + 4\hat j - 5\hat k$ änd $\lambda \hat i + 2\hat j + 3\hat k$ is equal to one. Find the value of $\lambda$.

Ans: $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k) = (2 + \lambda )\hat i + 6\hat j - 2\hat k$

So, unit vector along $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k)$ is $\left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda + 44} }}} \right)$

$(\hat i + \hat j + \hat k) \cdot \left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda + 44} }}} \right) = 1$

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

$\Rightarrow \sqrt {{\lambda ^2} + 4\lambda + 44} = \lambda + 6$

$\Rightarrow {\lambda ^2} + 4\lambda + 44 = {(\lambda + 6)^2}$

$\Rightarrow {\lambda ^2} + 4\lambda + 44 = {\lambda ^2} + 12\lambda + 36$

$\Rightarrow 8\lambda = 8$

$\Rightarrow \lambda = 1$

14. If $\vec a,\vec b,\vec c$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec a + \vec b + \vec c$ is equally inclined to $\vec a\;,\;\vec b$ and $\vec c$.

Ans: $\vec a\;.\;\vec b = \vec b\;.\;\vec c = \vec c\;.\;\vec a = 0$

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

Let $\vec a + \vec b + \vec c$ be inclined to $\vec a,\vec b,\vec c$ at angles ${\theta _1},{\theta _2},{\theta _3}$ respectively. $\cos {\theta _1} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{\vec a \cdot \vec a + \vec b \cdot \vec a + \vec c \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{|\vec a{|^2}}}{{|\vec a + \vec b + \vec c||\vec a|}} = \dfrac{{|\vec a|}}{{|\vec a + \vec b + \vec c|}}$

$\cos {\theta _2} = \dfrac{{(\vec a + \vec b + \vec c)\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{\vec a\vec b + \vec b\vec b + \vec c\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{|\vec b{|^2}}}{{|\vec a + \vec b + \vec c||\vec b|}} = \dfrac{{|\vec b|}}{{|\vec a + \vec b + \vec c|}}$

$\operatorname{os} {\theta _3} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec c}}{{|\vec a + \vec b + \vec c||\bar c|}} = \dfrac{{\vec a\vec c + \vec b\vec c + \vec c\vec c}}{{|\vec a + \vec b + \vec c||\vec c\mid }} = \dfrac{{|\vec c{|^2}}}{{|\vec a + \vec b + \vec c||\vec c|}} = \dfrac{{|\vec c|}}{{|\vec a + \vec b + \vec c|}}$

Since, $|\vec a| = |\vec b| = |\vec c|\; \Rightarrow \;\cos {\theta _1} = \cos {\theta _2} = \cos {\theta _3}$, So, ${\theta _1} = {\theta _2} = {\theta _3}$

15. Prove that, $(\vec a + \vec b) \cdot (\vec a + \vec b) = \;|\vec a{|^2} + |\vec b{|^2}$ if and only if $\vec a,\vec b$ are perpendicular, given $\vec a \ne \vec 0,\vec b \ne 0$

Ans: $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$

$\Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b \cdot \vec a + \vec b\;.\;\vec b = |\vec a{|^2} + |\vec b{|^2}$

$\Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\vec b{|^2} = |a{|^2} + |\vec b{|^2}$

$\Rightarrow 2\vec a\;.\;\vec b = 0$

$\Rightarrow \vec a\;.\;\vec b = 0$

So $\vec a$ and $\vec b$ are perpendicular.

16. If $\theta$ is the angle between two vectors $\vec a$ and $\vec b$, then $\vec a\vec b \geqslant 0$ only when

a. $0 < \theta < \dfrac{\pi }{2}$

b. $0 \leqslant \theta \leqslant \dfrac{\pi }{2}$

c. $0 < \theta < \pi$

d. $0 \leqslant \theta \leqslant \pi$

Ans: $\therefore \vec a\vec b \geqslant 0$

$\Rightarrow |\vec a||\vec b|\cos \theta \geqslant 0$

$\Rightarrow \cos \theta \geqslant 0\quad \because [|\vec a| \geqslant 0$ and $|\vec b| \geqslant 0]$

$\Rightarrow 0 \leqslant \theta \leqslant \dfrac{\pi }{2}$

$\vec a \cdot \vec b \geqslant 0$ if $0 \leqslant \theta \leqslant \dfrac{\pi }{2}$

So the right answer is B

17. Let $\vec a$ and $\vec b$ be two unit vectors and $\theta$ is the angle between them. Then $\vec a + \vec b$ is a unit vector if

a. $\theta = \dfrac{\pi }{4}$

b. $\theta = \dfrac{\pi }{3}$

c. $\theta = \dfrac{\pi }{2}$

d. $\theta = \dfrac{{2\pi }}{3}$

Ans: $|\vec a| = |\vec b| = 1$

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

$\Rightarrow (\vec a + \overrightarrow b )(\vec a + \overrightarrow b ) = 1$

$\Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b\;.\;\vec a + \vec b\;.\;\vec b = 1$

$\Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\overrightarrow b {|^2} = 1$

$\Rightarrow {1^2} + 2|\vec a||\vec b|\cos \theta + {1^2} = 1$

$\Rightarrow 1 + 2.1.1\cos \theta + 1 = 1$

$\Rightarrow \cos \theta = - \dfrac{1}{2}$

$\Rightarrow \theta = \dfrac{{2\pi }}{3}$

So, $\vec a + \vec b$ is unit vector if $\theta = \dfrac{{2\pi }}{3}$

18. The value of $\hat i.(\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$ is

$\begin{array}{*{20}{l}} {{\text{ a. }}0}&{{\text{ b. }} - 1}&{{\text{ c. }}1}&{{\text{ d. }}3} \end{array}$

Ans:

$\hat i \cdot (\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$

$= \hat i\;.\;\hat i + \hat j \cdot ( - \hat j) + \hat k\;.\;\hat k$

$= 1 - 1 + 1$

$= 1$

19. If $\theta$ is the angle between any two vectors $\vec a$ and $\vec b$, then $|\vec a\vec b| = |\vec a \times \vec b|$ when $\theta$ is equal to

a. 0

b. $\dfrac{\pi }{4}$

c. $\dfrac{\pi }{2}$

d. n

Ans: $|\vec a\vec b| = |\vec a \times \vec b|$

$\Rightarrow |\vec a|\vec b|\cos \theta = |\vec a||\vec b\mid \sin \theta$

$\Rightarrow \cos \theta = \sin \theta$

$\Rightarrow \tan \theta = 1$

$\Rightarrow \theta = \dfrac{\pi }{4}$

## Conclusion

Miscellaneous Exercise Class 12 Chapter 10 is important for understanding various concepts thoroughly. Class 12 Maths Chapter 10 Miscellaneous Exercise Solutions covers diverse problems that require the application of multiple formulas and techniques. Rather than simply learning answers by heart, it's important to concentrate on understanding the fundamental ideas behind each question.

## Class 12 Maths Chapter 10: Exercises Breakdown

 Exercise Number of Questions Exercise 10.1 5 Questions and Solutions Exercise 10.2 19 Questions and Solutions Exercise 10.3 18 Questions and Solutions Exercise 10.4 12 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 10 - Vector Algebra

1. What kind of problems are covered in the miscellaneous exercise?

The miscellaneous exercise in Chapter 10 deals with various concepts from vector algebra, including:

• Operations on vectors (magnitude, direction cosines/ratios, scalar product, vector product)

• Applications of vector product (finding the area of a parallelogram, finding the projection of a vector onto another)

• Analytical geometry involving vectors (finding the equation of a line or plane using vector methods)

• Identifying relationships between vectors (collinearity, perpendicularity)

2. Where can I find additional resources for practising miscellaneous exercise problems?

• The NCERT textbook itself might provide solutions to some problems within the miscellaneous exercise section.

• Vedantu offers comprehensive solutions and explanations for these problems. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 10 Miscellaneous Exercise Vector Algebra."

3. Vector Algebra sample questions?

• Find the area of the parallelogram formed by vectors A = 2i + 3j and B = 4i - j. (This question requires using the vector product to calculate the area)

• Find the equation of the plane passing through the points (1, 2, 3), (2, 1, 4), and (3, 0, 5). (This question involves using vectors to represent direction and solving for the plane equation)

4. How many questions are there in Vector Miscellaneous Class 12 Exercise?

The exercise contains 19 questions, varying in difficulty and type.

5. Are there any real-life applications of vector algebra covered in this exercise?

Some problems involve real-life applications like physics problems, force calculations, and directional vectors.

6. How can solving this exercise help in competitive exams?

It strengthens problem-solving skills and helps in understanding vector concepts, which are often tested in competitive exams like JEE and NEET.

7. What is the best way to approach solving the Miscellaneous Exercise?

Start by reviewing each concept, solve basic problems first, then gradually move to more complex ones. Consistent practice and revision are key.

8. How many marks are typically allocated to Vector Miscellaneous Class 12 questions in exams?

In board exams, vector algebra questions can range from 4 to 8 marks, depending on the complexity and type of question.

9. What is the significance of vector algebra in higher studies?

Vector algebra is fundamental in higher studies, especially in fields like physics, engineering, and computer science, where it is extensively used.

10. Can I find similar problems to practice apart from the NCERT textbook?

Yes, many reference books and online resources provide additional practice problems on vector algebra on the Vedantus website.