Class 12 Maths Chapter 10 Miscellaneous Exercise Solutions

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

The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.5: Miscellaneous Exercise provides a thorough review of Vector Algebra. This exercise is essential for the term 2 CBSE Board exam as it includes various types of questions from Exercises 10.1, 10.2, 10.3, and 10.4. Studying these solutions will help you review and master the concepts covered in the chapter. You can access all Class 12 Maths NCERT solutions for free, and subscribe for updates on your study materials.

Download the Class 12 Maths Chapter 10 Miscellaneous Exercise PDF for Vector Algebra. This helpful resource includes detailed solutions for all problems in the Miscellaneous Exercise of Chapter 10. It will assist you with your homework and exam preparation. The PDF format allows you to study offline, so you can review the material at your convenience, anytime and anywhere.

Q1.Write down a unit vector in the XY plane, making an angle of 30° with the positive direction of the x-axis.

Solution. Let us cosider $\overrightarrow{r}$ as a unit vector in XY-plane aking an angle θ with the positive direction of x-axis

So, $\overrightarrow{r}=cos \theta \hat{i}+ sin \theta \hat{j}$

Here θ = 30°

$\overrightarrow{r}=cos 30^{0}\hat{i}+ sin 30^{0}\hat{j}$

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

Hece the required unit vector is $\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$

Q2. Find the scalar components and magnitude of the vector joining the points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂).

Solution. The vector $\overrightarrow{PQ}$ joining the points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is

$\overrightarrow{PQ}=\left ( x_{2}- x_{1}\right )\hat{i}+\left ( y_{2}- y_{1}\right )\hat{j}+\left ( z_{2}- z_{1}\right )\hat{k}$

The magnitude of the vector $\overrightarrow{PQ}$ is

$\overrightarrow{PQ}=\sqrt{\left ( x_{2}- x_{1}\right )^{2}+\left ( y_{2}- y_{1}\right )^{2}+\left ( z_{2}-z_{1}\right )^{2}}$

Hence the scaler components are {(x₂– x₁ ),(y₂– y₁ ),(z₂– z₁ )} and magnitude of $\overrightarrow{PQ}$ is

$\sqrt{\left ( x_{2}- x_{1}\right )^{2}+\left ( y_{2}- y_{1}\right )^{2}+\left ( z_{2}-z_{1}\right )^{2}}$

Q3. A girl walks 4 km towards the west, then she walks 3 km in a direction 30° east of the north and stops. Determine the girl’s displacement from her initial point of departure.

Solution. Let the initial point of the girl is O, from O she walks 4 km towards the west up to the point P, then she walks 3 km in a direction 30° east of the north up to the point Q, it can be shown in the figure below.

As per the figure

$\overrightarrow{OP}=-4\hat{i}$

Let’s find the displacement during the given journey,i.e $\overrightarrow{OQ}$

According to the triangular rule of vector addition

$\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{OQ}$

$\overrightarrow{PQ}=\left | PQ \right |cos60^{0}\hat{i}+\left | PQ \right |sin60^{0}\hat{j}$

$\overrightarrow{PQ}=3.cos60^{0}\hat{i}+3.sin60^{0}\hat{j}$

$\overrightarrow{PQ}=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$

$\overrightarrow{OQ}=-4\hat{i}+\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{k}$

$\overrightarrow{OQ}=-\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{k}$

Hence, the the girl’s displacement from her initial point of departure is $-\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{k}$

Q4.If $\overrightarrow{a} =\overrightarrow{b} +\overrightarrow{c}$ ,then it is true that $\left | \overrightarrow{a} \right |=\left | \overrightarrow{b} \right |+\left | \overrightarrow{c} \right |$ ?

Solution. The given condition is $\overrightarrow{a} =\overrightarrow{b} +\overrightarrow{c}$

The given vectors can be drawn as in the following figure

In ΔABC, $\overrightarrow{CB}=\overrightarrow{a},\overrightarrow{CA}=\overrightarrow{b},\overrightarrow{AB}=\overrightarrow{c}$

By the triangular rule of the addition of the vectors,we have

$\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$

The magnitudes of the vectors $\left | \overrightarrow{a} \right |,\: \left | \overrightarrow{b} \right |,\left |\overrightarrow{c} \right |$ represents the sides of the triangle

Since side of a triangle is always less than the sum of other two sides

$\left | \overrightarrow{a} \right |< \: \left | \overrightarrow{b} \right |+\left |\overrightarrow{c} \right |$

Therefore it the given situation $\left | \overrightarrow{a} \right |=\left | \overrightarrow{b} \right |+\left | \overrightarrow{c} \right |$ is not true.

Q5.Find the value of x for which $x\left ( \hat{i}+\hat{j} +\hat{k} \right )$ is a unit vector.

Solution. It is given to us that $x\left ( \hat{i}+\hat{j} +\hat{k} \right )$ is a unit vector

$\therefore \left | x\left ( \hat{i}+\hat{j} +\hat{k} \right ) \right |=1$

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

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

±x√3 = 1

x = ±1/√3

Q6.Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

$\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}\: and\: \hat{i}-2\hat{j}+\hat{k}$

Solution. The given vectors are

$\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}\: and\: \hat{i}-2\hat{j}+\hat{k}$

Let the resultant vector of the given vectors is $\overrightarrow{c}$

$\therefore \overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}$

$\overrightarrow{c}=(2\hat{i}+3\hat{j}-\hat{k}) + (\hat{i}-2\hat{j}+\hat{k})=3\hat{i}+\hat{j}$

The unit vector in the direction of $\overrightarrow{c}$

$=\frac{\overrightarrow{c}}{\left | \overrightarrow{c} \right |}$

$\left | \overrightarrow{c} \right |=\sqrt{3^{2}+1^{2}}=\sqrt{10}$

$\Rightarrow \frac{\overrightarrow{c}}{\left | \overrightarrow{c} \right |}=\frac{3\hat{i}+\hat{j}}{\sqrt{10}}$

Therefore the vector parallel to the resultant vectors $\overrightarrow{a}\: and\: \overrightarrow{b}$ of the magnitude 5 units

= ±5×unit vector

$\Rightarrow \pm 5.\frac{3\hat{i}+\hat{j}}{\sqrt{10}}=\pm \frac{15\hat{i}+5\hat{j}}{\sqrt{10}}$

Rationalizing denominator

$=\pm \frac{3\sqrt{10}\hat{i}}{2}\pm \frac{\sqrt{10}}{2}\hat{j}$

Q7. If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=2\hat{i}-\hat{j}+3\hat{k}\: and\: \overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$ , find a unit vector parallel to the vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}$

Solution. The given vectors are

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

A unit vector parallel to the given vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}$ is

$=\frac{2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}}{\left |2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c} \right |}$

$2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}=2(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})+3(\hat{i}-2\hat{j}+\hat{k})$

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

$\left | 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c} \right |=\sqrt{3^{2}+\left ( -3 \right )^{2}+2^{2}}=\sqrt{22}$

Therefore unit vector parallel to the given vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}$ is

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

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

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

Class 12 Maths Chapter 10 Miscellaneous Exercise - Vector Algebra NCERT Solutions with PDF

Download Class 12 Maths Chapter 10 Miscellaneous Exercise - Vector Algebra NCERT Solutions PDF

Class 12 Maths Chapter 10 - Vector Algebra: Find Links to All Exercises NCERT Solutions

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

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