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

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NCERT Solutions for Maths Exercise 10.4 Class 12 Chapter 10 Vector Algebra - FREE PDF Download

In Class 12 Maths NCERT Solutions for Chapter 10 Exercise 10.4 - Vector Algebra, we explore the application of vector operations in solving various mathematical problems. Ex 10.4 Class 12  focuses on understanding and applying concepts such as the dot product, cross product, and their properties. We will learn how to use these operations to find angles between vectors, determine the area of parallelograms, and solve problems involving vector projections. By working through these problems, we will enhance our ability to analyze and manipulate vectors, which is crucial for higher-level mathematics and physics. Access the CBSE Class 12 Maths Syllabus here.

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Table of Content
1. NCERT Solutions for Maths Exercise 10.4 Class 12 Chapter 10 Vector Algebra - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 10 Exercise 10.4 Class 12 | Vedantu
3. Formulas Used in Class 12 Maths Ex 10.4
4. Access NCERT Solutions for Maths Class 12 Chapter 10 - Vector Algebra
5. Class 12 Maths Chapter 10: Exercises Breakdown
6. CBSE Class 12 Maths Chapter 10 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 10 Exercise 10.4 Class 12 | Vedantu

  • In Class 12 Maths Ex 10.4 covers the topic named Vector (or cross) product of two vectors.

  • The vector product of two vectors, denoted by a×b (read as "a cross b"), is a new vector resulting from the perpendicularity and magnitude of the original vectors.

  • Direction - The direction of the resultant vector is perpendicular to both the original vectors a and b. Imagine the two vectors like the arms of a cross, and the resulting cross product points out of the plane formed by the two vectors.

  • Right-Hand Rule - Determining the direction of the cross product can be visualized using the right-hand rule. Curl the fingers of your right hand such that they follow the direction from a to b. Your thumb then points in the direction of the resulting cross-product vector.

  • The magnitude of the cross product depends on the lengths (magnitudes) of the original vectors and the angle between them. It's calculated as ∣∣a×b∣∣=∣∣a∣∣∣∣b∣∣sin(θ), where θ is the angle between a and b (less than 180 degrees).

  • Ex 10.4 Class 12 contains 12 Questions and Solutions.


Formulas Used in Class 12 Maths Ex 10.4

  • The magnitude of the Cross Product:

∣∣a×b∣∣=∣∣a∣∣∣∣b∣∣sin(θ)

Where:

  • ∣∣a×b∣∣ represents the magnitude of the resulting cross product vector.

  • ∣∣a∣∣ and ∣∣b∣∣ represent the magnitudes of the original vectors a and b respectively.

  • θ represents the angle between vectors a and b (less than 180 degrees).

Competitive Exams after 12th Science
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Access NCERT Solutions for Maths Class 12 Chapter 10 - Vector Algebra

Exercise 10.4

1. Find$\left| \overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}} \right|$, if $\overrightarrow{\mathbf{a}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{-7}\widehat{\mathbf{j}}\mathbf{+7}\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}\mathbf{=3}\widehat{\mathbf{i}}\mathbf{-2}\widehat{\mathbf{j}}\mathbf{+2}\widehat{\mathbf{k}}$

Ans: The given vectors are 

$\overrightarrow{\text{a}}\text{=}\widehat{\text{i}}\text{-7}\widehat{\text{j}}\text{+7}\widehat{\text{k}}$ and $\overrightarrow{\text{b}}\text{=3}\widehat{\text{i}}\text{-2}\widehat{\text{j}}\text{+2}\widehat{\text{k}}$.

Then the cross-product between the vectors is given by

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

 $ =\widehat{i}\left( -14+14 \right)+-\widehat{j}\left( 2-21 \right)+\widehat{k}\left( -2+21 \right) $ 

 $ =19\widehat{j}+19\widehat{k}. $ 

Therefore,

$ \left| \overrightarrow{a}\times \overrightarrow{b} \right|=\sqrt{{{19}^{2}}+{{19}^{2}}} $ 

 $ =\sqrt{2\times {{19}^{2}}} $ 

 $=19\sqrt{2}. $ 


2. Find a unit vector perpendicular to each of the vector $\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}$, where $\overrightarrow{\mathbf{a}}\mathbf{=3}\widehat{\mathbf{i}}\mathbf{+2}\widehat{\mathbf{j}}\mathbf{+2}\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{+2}\widehat{\mathbf{j}}\mathbf{-2}\widehat{\mathbf{k}}$.

Ans: The given vectors are

$\overrightarrow{\text{a}}\text{=3}\widehat{\text{i}}\text{+2}\widehat{\text{j}}\text{+2}\widehat{\text{k}}$ and $\overrightarrow{\text{b}}\text{=}\widehat{\text{i}}\text{+2}\widehat{\text{j}}\text{-2}\widehat{\text{k}}$.

Then, adding and subtracting the vectors successively, we have

$\overrightarrow{a}+\overrightarrow{b}=4\widehat{i}+4\widehat{j}$ and $\overrightarrow{a}-\overrightarrow{b}=2\widehat{i}+4\widehat{k}$.

Therefore, their cross-product,

$\left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right)=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \\ \end{matrix} \right| $

 $ =16\widehat{i}-16\widehat{j}-8\widehat{k}. $ 

Then, its magnitude,

$ \left| \left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}-\overrightarrow{b} \right) \right|=\sqrt{{{16}^{2}}+{{\left( -16 \right)}^{2}}+{{\left( -8 \right)}^{2}}} $ 

 $=\sqrt{{{2}^{2}}\times {{8}^{2}}+{{2}^{2}}\times {{8}^{2}}+{{8}^{2}}} $ 

 $ =8\sqrt{{{2}^{2}}+{{2}^{2}}+1} $ 

 $ =8\sqrt{9} $ 

 $ =8\times 3 $ 

 $ =24. $ 

Thus, the unit vector perpendicular to each of the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$

is provided as,

$ =\pm \frac{\left( \overrightarrow{a}+\overrightarrow{b} \right)\left( \overrightarrow{a}-\overrightarrow{b} \right)}{\left| \left( \overrightarrow{a}+\overrightarrow{b} \right)\left( \overrightarrow{a}-\overrightarrow{b} \right) \right|} $ 

 $ =\pm \frac{16\widehat{i}-16\widehat{j}-8\widehat{k}}{24} $ 

 $ =\pm \frac{2\widehat{i}-2\widehat{j}-\widehat{k}}{3} $ 

 $ =\pm \frac{2}{3}\widehat{i}\,\mp \frac{2}{3}\widehat{j}\,\,\mp \frac{1}{3}\widehat{k} $ 


3. If a unit vector $\overrightarrow{\mathbf{a}}$ makes angles $\frac{\mathbf{\pi }}{\mathbf{3}}$  with $\overrightarrow{\mathbf{i}}\mathbf{,}\,\frac{\mathbf{\pi }}{\mathbf{4}}$ with $\overrightarrow{\mathbf{j}}$ and an acute angle $\mathbf{\theta }$ with $\widehat{\mathbf{k}}$, then find $\mathbf{\theta }$ and hence, the components of $\overrightarrow{\mathbf{a}}$.

Ans: Suppose the components of the given unit vector $\overrightarrow{a}$ is $\left( {{a}_{1}},{{a}_{2}},{{a}_{3}} \right)$.

Then, $\overrightarrow{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}$.

Also, $\left| \overrightarrow{a} \right|=1$ as $\overrightarrow{a}$ is a unit vector.

Again, we are provided that, the vector $\overrightarrow{\text{a}}$ makes angles $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  with $\overrightarrow{\text{i}}\text{,}\,\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ with $\overrightarrow{\text{j}}$ and an acute angle $\text{ }\!\!\theta\!\!\text{ }$ with $\widehat{\text{k}}$. Therefore, it gives

\[\cos \frac{\pi }{3}=\frac{{{a}_{1}}}{\left| \overrightarrow{a} \right|}\]

\[\Rightarrow \frac{1}{2}={{a}_{1}}\], since $\overrightarrow{a}$ is a unit vector.

Also, 

$\cos \frac{\pi }{4}=\frac{{{a}_{2}}}{\left| \overrightarrow{a} \right|}$

$\Rightarrow \frac{1}{\sqrt{2}}={{a}_{2}}$, since $\overrightarrow{a}$ is a unit vector.

Now, let

$\cos \theta =\frac{{{a}_{3}}}{\left| \overrightarrow{a} \right|}$.

$\Rightarrow {{a}_{3}}=\cos \theta $

Since, $\left| \overrightarrow{a} \right|=1$, so

$ \Rightarrow \sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}=1 $ 

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

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

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

 $ \Rightarrow {{\cos }^{2}}\theta =1-\frac{3}{4}=\frac{1}{4} $ 

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

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

Thus, ${{a}_{3}}=\cos \frac{\pi }{3}=\frac{1}{2}$.

Therefore, $\theta =\frac{\pi }{3}$ and the components of the vector $\overrightarrow{a}$ are $\left( \frac{1}{2},\,\frac{1}{\sqrt{2}},\frac{1}{2} \right)$.


4. Show that

$\left( \overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}} \right)\mathbf{\times }\left( \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right)\mathbf{=2}\left( \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right)$.

Ans: The given cross-product can be written as

$\left( \overrightarrow{\text{a}}\text{-}\overrightarrow{\text{b}} \right)\text{ }\!\!\times\!\!\text{ }\left( \overrightarrow{\text{a}}\text{+}\overrightarrow{\text{b}} \right)$

$=\left( \overrightarrow{a}-\overrightarrow{b} \right)\times \overrightarrow{a}+\left( \overrightarrow{a}-\overrightarrow{b} \right)\times \overrightarrow{b}$, (using the distributive property of vector product over addition)

$=\overrightarrow{a}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{b}\times \overrightarrow{b}$, (Distributive property of vector product over addition)

$=\overrightarrow{0}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{0} $ 

 $ =2\overrightarrow{a}\times \overrightarrow{b}. $ 


5. Find $\mathbf{\lambda }$ and $\mathbf{\mu }$ if $\left( \mathbf{2}\widehat{\mathbf{i}}\mathbf{+6}\widehat{\mathbf{j}}\mathbf{+27}\widehat{\mathbf{k}} \right)\mathbf{\times }\left( \widehat{\mathbf{i}}\mathbf{+\lambda }\widehat{\mathbf{j}}\mathbf{+\mu }\widehat{\mathbf{k}} \right)\mathbf{=}\overrightarrow{\mathbf{0}}$.

Ans: The given vector equation can be written as,

$\left( \text{2}\widehat{\text{i}}\text{+6}\widehat{\text{j}}\text{+27}\widehat{\text{k}} \right)\text{ }\!\!\times\!\!\text{ }\left( \widehat{\text{i}}\text{+ }\!\!\lambda\!\!\text{ }\widehat{\text{j}}\text{+ }\!\!\mu\!\!\text{ }\widehat{\text{k}} \right)\text{=}\overrightarrow{\text{0}}$

$\Rightarrow \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{matrix} \right|=0\widehat{i}+0\widehat{j}+0\widehat{k}$

$\Rightarrow \widehat{i}\left( 6\mu -27\lambda  \right)-\widehat{j}\left( 2\mu \,-27 \right)+\widehat{k}\left( 2\lambda -6 \right)=0\widehat{i}+0\widehat{j}+0\widehat{k}$

Now, compare the scalar components both sides of the equation.

Then, it yields

$6\mu -27\lambda =0$                              …… (i)

$2\mu -27=0$                                …… (ii)

$2\lambda -6=0$                                  …… (iii)

Hence, on solving the equations (i), (ii), and (iii), we obtain

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


6. Given that $\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}\mathbf{=0}$ and $\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}\mathbf{=0}$.

What can you conclude about the vector $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$?

Ans: Since, $\overrightarrow{a}\cdot \overrightarrow{b}=0$, then we can say that

either $\left| \overrightarrow{a} \right|=0$, or $\left| \overrightarrow{b} \right|=0$, or the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular to each other.

Again, since $\overrightarrow{a}\times \overrightarrow{b}=0$, so it can be said that

either $\left| \overrightarrow{a} \right|=0$, or $\left| \overrightarrow{b} \right|=0$, or the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel to each other.

But, since the vectors cannot be perpendicular and parallel at simultaneous, so the only possibility is either $\left| \overrightarrow{a} \right|=0$ or $\left| \overrightarrow{b} \right|=0$.


7. Let the vectors $\overrightarrow{\mathbf{a}}\mathbf{,}\,\,\overrightarrow{\mathbf{b}}\mathbf{,}\,\,\overrightarrow{\mathbf{c}}$ given as ${{\mathbf{a}}_{\mathbf{1}}}\widehat{\mathbf{i}}\mathbf{+}{{\mathbf{a}}_{\mathbf{2}}}\widehat{\mathbf{j}}\mathbf{+}{{\mathbf{a}}_{\mathbf{3}}}\widehat{\mathbf{k}}\mathbf{,}\,\,\,{{\mathbf{b}}_{\mathbf{1}}}\widehat{\mathbf{i}}\mathbf{+}{{\mathbf{b}}_{\mathbf{2}}}\widehat{\mathbf{j}}\mathbf{+}{{\mathbf{b}}_{\mathbf{3}}}\widehat{\mathbf{k}}\mathbf{,}\,\,{{\mathbf{c}}_{\mathbf{1}}}\widehat{\mathbf{i}}\mathbf{+}{{\mathbf{c}}_{\mathbf{2}}}\widehat{\mathbf{j}}\mathbf{+}{{\mathbf{c}}_{\mathbf{3}}}\widehat{\mathbf{k}}$

Then show that $\overrightarrow{\mathbf{a}}\mathbf{\times }\left( \overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{c}} \right)\mathbf{=}\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{c}}$.

Ans: The given vectors are

$\overrightarrow{\text{a}}\text{=}{{\text{a}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{a}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{a}}_{\text{3}}}\widehat{\text{k}}$, 

$\overrightarrow{\text{b}}\text{=}{{\text{b}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{b}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{b}}_{\text{3}}}\widehat{\text{k}}$,

$\overrightarrow{\text{c}}\text{=}{{\text{c}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{c}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{c}}_{\text{3}}}\widehat{\text{k}}$.

So, $\left( \overrightarrow{b}+\overrightarrow{c} \right)=\left( {{b}_{1}}+{{c}_{1}} \right)\widehat{i}+\left( {{b}_{2}}+{{c}_{2}} \right)\widehat{j}+\left( {{b}_{3}}+{{c}_{3}} \right)\widehat{k}$

Then,

$ \overrightarrow{a}\times \left( \overrightarrow{b}+\overrightarrow{c} \right)=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k}\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}}+{{c}_{1}} & {{b}_{2}}+{{c}_{2}} & {{b}_{3}}+{{c}_{3}} \end{matrix} \right| $

 $=\widehat{i}\left[ {{a}_{2}}\left( {{b}_{3}}+{{c}_{3}} \right)-{{a}_{3}}\left( {{b}_{2}}+{{c}_{2}} \right) \right]-\widehat{j}\left[ {{a}_{1}}\left( {{b}_{3}}+{{c}_{3}} \right)-{{a}_{3}}\left( {{b}_{1}}+{{c}_{1}} \right) \right]+\widehat{k}\left[ {{a}_{1}}\left( {{b}_{2}}+{{c}_{2}} \right)-{{a}_{2}}\left( {{b}_{1}}+{{c}_{1}} \right) \right]$ $ =\widehat{i}\left( {{a}_{2}}{{b}_{3}}+{{a}_{2}}{{c}_{3}}-{{a}_{3}}{{b}_{2}}-{{a}_{3}}{{c}_{2}} \right)+\widehat{j}\left( -{{a}_{1}}{{b}_{3}}-{{a}_{1}}{{c}_{3}}+{{a}_{3}}{{b}_{1}}+{{a}_{3}}{{c}_{1}} \right) $ 

 $ \,\,\,\,+\widehat{k}\left( {{a}_{1}}{{b}_{2}}+{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{b}_{1}}-{{a}_{2}}{{c}_{1}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\text{ (i)} $ 

Therefore, 

\[\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \end{matrix} \right|\]

\[=\widehat{i}\left( {{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}} \right)+\widehat{j}\left( {{a}_{3}}{{b}_{1}}-{{a}_{1}}{{b}_{3}} \right)+\widehat{k}\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)\]          …… (ii)

Also,

$\overrightarrow{a}\times \overrightarrow{c}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \end{matrix} \right|$

$=\widehat{i}\left( {{a}_{2}}{{c}_{3}}-{{a}_{3}}{{c}_{2}} \right)+\widehat{j}\left( {{a}_{3}}{{c}_{1}}-{{a}_{1}}{{c}_{3}} \right)+\widehat{k}\left( {{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}} \right)$             …… (iii)

Now, add the equations (ii) and (iii). Then, it yields

$\left( \overrightarrow{a}\times \overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{c} \right)=\widehat{i}\left( {{a}_{2}}{{b}_{3}}+{{a}_{2}}{{c}_{3}}-{{a}_{3}}{{b}_{2}}-{{a}_{3}}{{c}_{2}} \right)+\widehat{j}\left( {{b}_{1}}{{a}_{3}}+{{a}_{3}}{{c}_{1}}-{{a}_{1}}{{b}_{3}}-{{a}_{1}}{{c}_{3}} \right) $ 

 $ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\widehat{k}\left( {{a}_{1}}{{b}_{2}}+{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{b}_{1}}-{{a}_{2}}{{c}_{1}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\text{ (iv)} $ 

 

Then, the equations (i) and (iv) together implies that

$\overrightarrow{a}\times \left( \overrightarrow{b}+\overrightarrow{c} \right)=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}$.

Thus, the required result has been proved.


8. If either $\overrightarrow{\mathbf{a}}\mathbf{=}\overrightarrow{\mathbf{0}}$ or $\overrightarrow{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{0}}$, then $\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{0}}$.

Is the converse true? Justify your answer with an example.

Ans: Consider any parallel non-zero vectors so that $\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}\text{=}\overrightarrow{\text{0}}$.

So, take the nonzero vectors \[\overrightarrow{a}=2\widehat{i}+3\widehat{j}+4\widehat{k}\] and $\overrightarrow{b}=4\widehat{i}+6\widehat{j}+8\widehat{k}$.

Therefore, the cross-product between the vectors,

$ \overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{matrix} \right| $ 

 $=\widehat{i}\left( 24-24 \right)-\widehat{j}\left( 16-16 \right)+\widehat{k}\left( 12-12 \right) $ 

 $ =0\widehat{i}+0\widehat{j}+0\widehat{k}. $ 

Then, the magnitudes of the vectors are given by

$\left| \overrightarrow{a} \right|=\sqrt{{{2}^{2}}+{{3}^{2}}+{{4}^{2}}}=\sqrt{29}$ and

$\left| \overrightarrow{b} \right|=\sqrt{{{4}^{2}}+{{6}^{2}}+{{8}^{2}}}=\sqrt{116}.$

Thus, observing the above results, it is found that $\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}\text{=}\overrightarrow{\text{0}}$, while $\overrightarrow{a}\ne \overrightarrow{0}$ and $\overrightarrow{b}\ne \overrightarrow{0}$.

Therefore, it is justified that the converse of the given statement need not be true.

 

9. Find the area of the triangle with vertices $\mathbf{A}\left( \mathbf{1,1,2} \right)\mathbf{,}\,\,\mathbf{B}\left( \mathbf{2,3,5} \right)$ and $\mathbf{C}\left( \mathbf{1,5,5} \right)$.

Ans: The triangle $ABC$ has the vertices $\text{A}\left( \text{1,1,2} \right)\text{,}\,\,\text{B}\left( \text{2,3,5} \right)$ and $\text{C}\left( \text{1,5,5} \right)$.

The adjacent sides of the triangle $\Delta \,ABC$ are $\overrightarrow{AB}$ and $\overrightarrow{BC}$ such that

\[\overrightarrow{AB}=\left( 2-1 \right)\widehat{i}+\left( 3-1 \right)\widehat{j}+\left( 5-2 \right)\widehat{k}=\widehat{i}+2\widehat{j}+3\widehat{k}\] and

$\overrightarrow{BC}=\left( 1-2 \right)\widehat{i}+\left( 5-3 \right)\widehat{j}+\left( 5-5 \right)\widehat{k}=-\widehat{i}+2\widehat{j}$.

Now, the cross-product between the vectors is given by

$\overrightarrow{AB}\times \overrightarrow{BC}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 3 \\ -1 & 2 & 0 \end{matrix} \right| $

 $ =\widehat{i}\left( -6 \right)-\widehat{j}\left( 3 \right)+\widehat{k}\left( 2+2 \right) $ 

 $ =-6\widehat{i}-3\widehat{j}+4\widehat{k}. $ 

Therefore,

$ \left| \overrightarrow{AB}\times \overrightarrow{BC} \right|=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( -3 \right)}^{2}}+{{4}^{2}}} $ 

 $ =\sqrt{36+9+16} $ 

 $ =\sqrt{61}. $ 

Thus, the area of the triangle $\Delta \,ABC$ 

$=\frac{1}{2}\times \left| \overrightarrow{AB}\times \overrightarrow{BC} \right|$

$=\frac{1}{2}\times \sqrt{61}$

$=\frac{\sqrt{61}}{2}$ square units.


10. Find the area of the parallelogram whose adjacent sides are determined by the vector \[\overrightarrow{\mathbf{a}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{-}\widehat{\mathbf{j}}\mathbf{+3}\widehat{\mathbf{k}}\] and $\overrightarrow{\mathbf{b}}\mathbf{=2}\widehat{\mathbf{i}}\mathbf{-7}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}$.

Ans: The given vectors are \[\overrightarrow{\text{a}}\text{=}\widehat{\text{i}}\text{-}\widehat{\text{j}}\text{+3}\widehat{\text{k}}\] and $\overrightarrow{\text{b}}\text{=2}\widehat{\text{i}}\text{-7}\widehat{\text{j}}\text{+}\widehat{\text{k}}$ such that they are the adjacent sides of the parallelogram.

Therefore, the cross-product between the vectors,

$\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{matrix} \right| $

 $ =\widehat{i}\left( -1+21 \right)-\widehat{j}\left( 1-6 \right)+\widehat{k}\left( -7+2 \right) $ 

 $ =20\widehat{i}+5\widehat{j}-5\widehat{k}. $ 

Therefore, its magnitude,

$ \left| \overrightarrow{a}\times \overrightarrow{b} \right|=\sqrt{{{20}^{2}}+{{5}^{2}}+{{5}^{2}}} $ 

 $ =\sqrt{400+25+25} $ 

 $ =15\sqrt{2}. $ 

Thus, the area of the parallelogram is $15\sqrt{2}$ square units.


11. Let the vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be such that $\left| \overrightarrow{\mathbf{a}} \right|\mathbf{=3}$ and $\left| \overrightarrow{\mathbf{b}} \right|\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{3}}$, then $\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}$ is a unit vector, if the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is

(a) $\frac{\mathbf{\pi }}{\mathbf{6}}$        (b) $\frac{\mathbf{\pi }}{\mathbf{4}}$          (c) $\frac{\mathbf{\pi }}{\mathbf{3}}$             (d) $\frac{\mathbf{\pi }}{\mathbf{2}}$

Ans: We are provided that, $\left| \overrightarrow{a} \right|=3$ and $\left| \overrightarrow{b} \right|=\frac{\sqrt{2}}{3}$.

Now, it is generally known that $\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin \theta \cdot \widehat{n}$, where $\widehat{n}$ is a unit vector that is perpendicular to both the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$; $\theta $ is the angle the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. 

Since, $\overrightarrow{a}\times \overrightarrow{b}$ is a unit vector, so $\left| \overrightarrow{a}\times \overrightarrow{b} \right|=1$

$ \Rightarrow \left| \left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin \theta \widehat{n} \right|=1 $ 

$ \Rightarrow \left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\left| \sin \theta  \right|=1 $ 

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

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

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

Thus, the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\pi }{4}$.

Hence, the correct answer is option (b).


12. Area of a rectangle having vertices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, and $\mathbf{D}$ with position vectors $\mathbf{-}\widehat{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\widehat{\mathbf{j}}\mathbf{+4}\widehat{\mathbf{k}}\mathbf{,}\,\,\widehat{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\widehat{\mathbf{j}}\mathbf{+4}\widehat{\mathbf{k}}\mathbf{,}\,\,\widehat{\mathbf{i}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\widehat{\mathbf{j}}\mathbf{+4}\widehat{\mathbf{k}}$ and $\mathbf{-}\widehat{\mathbf{i}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\widehat{\mathbf{j}}\mathbf{+4}\widehat{\mathbf{k}}$ respectively is

(a) $\frac{\mathbf{1}}{\mathbf{2}}$           (b) $\mathbf{1}$               (c) $\mathbf{2}$                  (d) $\mathbf{4}$

Ans: The position vectors of the vertices $A,\,\,B,\,\,C,\,\,D$ of the rectangle $ABCD$ are such that 

$\overrightarrow{OA}=-\widehat{i}+\frac{1}{2}\widehat{j}+4\widehat{k}$,

$\overrightarrow{OB}=\widehat{i}+\frac{1}{2}\widehat{j}+4\widehat{k}$,

$\overrightarrow{OC}=\widehat{i}-\frac{1}{2}\widehat{j}+4\widehat{k}$, and

$\overrightarrow{OD}=-\widehat{i}-\frac{1}{2}\widehat{j}+4\widehat{k}$.

The adjacent sides of the rectangle $ABCD$ are $\overrightarrow{AB}$ and $\overrightarrow{BC}$ such that

$\overrightarrow{AB}=\left( 1+1 \right)\widehat{i}+\left( \frac{1}{2}-\frac{1}{2} \right)\widehat{j}+\left( 4-4 \right)\widehat{k}=2\widehat{j}$

$\overrightarrow{BC}=\left( 1-1 \right)\widehat{i}+\left( -\frac{1}{2}-\frac{1}{2} \right)\widehat{j}+\left( 4-4 \right)\widehat{k}=-\widehat{j}$.

Therefore, the cross-product between these two vectors,

$\overrightarrow{AB}\times \overrightarrow{BC}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & 0 & 0 \\ 0 & -1 & 0 \end{matrix} \right| $

 $ =\widehat{k}\left( -2 \right) $ 

 $ =-2\widehat{k}. $ 

Since, the area of a parallelogram having the adjacent sides $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\left| \overrightarrow{a}\times \overrightarrow{b} \right|$, so the area of the rectangle $ABCD$ is $\left| \overrightarrow{AB}\times \overrightarrow{CD} \right|=2$ square units.

Thus, the correct answer is the option (c).


Conclusion

In Class 12 Maths Ex 10.4 Solutions, we focus on the vector (or cross) product of two vectors, a fundamental concept in vector algebra. By working through the problems, we have learned how to calculate the cross product and understand its significance, including its magnitude, direction, and various properties. In class 12 ex 10.4, we have seen how the cross product can be used to solve practical problems, such as finding the area of parallelograms and determining the torque in physics. Mastering these concepts enhances our understanding of three-dimensional geometry and prepares us for more advanced applications in mathematics, physics, and engineering. 


Class 12 Maths Chapter 10: Exercises Breakdown

Exercise

Number of Questions

Exercise 10.1

5 Questions & Solutions

Exercise 10.2

19 Questions & Solutions

Exercise 10.3

18 Questions & Solutions

Miscellaneous Exercise 

19 Questions & Solutions


CBSE Class 12 Maths Chapter 10 Other Study Materials


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 Chapter 10 - Vector Algebra Exercise 10.4

1. What is the concept on which Exercise 10.4 of Class 12 Maths is based?

Exercise 10.4 of NCERT Class 12 Maths is primarily based on the concepts of cross product of 2 vectors. What do you get when you take the cross product of a couple of vectors? A third vector is perpendicular to the two original ones. Using the right-hand thumb rule, you may calculate its magnitude by measuring the area of the parallelogram between them.

2. Which question from Class 12 Maths Exercise 10.4 is frequently asked in CBSE Board exams?

Randomly, a question or exercise's frequency or weight is determined based on the chapter and the structure. Due to the fact that the sums of any chapter might be randomly picked in board examinations, it is vital that you practice all of the sums to be prepared for any question that is set. Besides this, students are recommended to review the chapter-by-chapter weightage to have a better idea of what is expected.

3. Is Class 12 Maths Exercise 10.4 easy to solve or practice?

Problems in the NCERT Class 12 Math Exercise 10.4 may be easily solved by students with a clear understanding. Those questions aren't really hard. Practice them all to refresh your memory and clarify any doubts you may have. Visit Vedantu if you need assistance comprehending the chapter. Aid regarding concept understanding along with few other modules will be provided on the official website of Vedantu.

4. Why is Vector Algebra used in Class 12?

Vector Algebra is used to measure angles and distances between panels in satellites, in the design of pipe networks in many sectors, and in computing angles and distances between beams and structures in civil engineering. Apart from these, Vector Algebra has multiple uses across different domains as well. Hence it is very important to learn vector algebra and grasp the fundamentals.

5. Where do we apply Vector Algebra?

Vectors are used in a variety of real-world scenarios, including those involving force or velocity. Consider the forces at work on a boat crossing a river. The boat's motor creates a force in one direction, while the river current creates a force in the opposite direction. Both of these forces are vectors. These are few of the many examples or applications of vector algebra.

6. What is the cross product of two vectors explained in  Class 12 Maths Ex 10.4 Solutions?

The cross product, or vector product, of two vectors is a vector that is perpendicular to the plane containing the original vectors. It is calculated using the formula A×B and has both magnitude and direction.

7. How do you calculate the magnitude of the cross product?

The magnitude of the cross product A×B is given by ∣A∣∣B∣sin⁡θ∣, where ∣A∣ and ∣B∣ are the magnitudes of the vectors and θθ is the angle between them.

8. What is the direction of the cross product?

The direction of the cross product is perpendicular to the plane formed by the original vectors. It follows the right-hand rule: if the fingers of your right hand point in the direction of the first vector (A) and curl towards the second vector (B), your thumb points in the direction of the cross product (A×B).

9. What are some applications of the cross product?

As we studied in class 12 ex 10.4, The applications of the cross product include:

  • Calculating the area of a parallelogram formed by two vectors.

  • Determining torque in physics, which is the rotational equivalent of force.

  • Finding the normal vector to a plane in three-dimensional space.

10. What are the properties of the cross product in class 12 ex 10.4?

Key properties of the cross product include:

  • Anti-commutative: A×B=−(B×A).

  • Distributive over addition: A×(B+C)=A×B+A×C.