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

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).

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

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NCERT Solutions for Class 12 Maths

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability