Chapter 10: Vector Algebra

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}. $

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} $

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 $\left( \frac{1}{2},\,\frac{1}{\sqrt{2}},\frac{1}{2} \right)$.

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}. $

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 $\mu =\frac{27}{2}$.

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 $\left| \overrightarrow{b} \right|=0$.

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.

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.

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.

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.