Step-by-Step Solutions: Class 12 Maths 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).

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.

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