Chapter 10: Vector Algebra

Ans: Assume the given vector as, $\overrightarrow a  = 2\widehat i - 3\widehat j + 4\widehat k$ and \[\overrightarrow b  =  - 4\widehat i + 6\widehat j - 8\widehat k\]respectively.

Therefore, it can be observed that $\overrightarrow b $ can be deduced in terms of$\overrightarrow a $as,

\[\overrightarrow b  =  - 4\widehat i + 6\widehat j - 8\widehat k =  - 2\left( {2\widehat i - 3\widehat j + 4\widehat k} \right) =  - 2\overrightarrow a \]

\[\therefore \overrightarrow b  = \lambda \overrightarrow a \]

The value of $\lambda  =  - 2$ in this case and therefore the given two vectors collinear.

Ans: Assume the given vector as, $\overrightarrow a  = \widehat i + 2\widehat j + 3\widehat k$.

Therefore, the magnitude of the vector, that is, $\overrightarrow a  = \widehat i + 2\widehat j + 3\widehat k$ can be calculated as,

$\left| {\overrightarrow a } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}}  = \sqrt {14} $

Therefore, the direction cosines of the vector, that is, $\overrightarrow a  = \widehat i + 2\widehat j + 3\widehat k$ can be calculated as, $\left( {\dfrac{1}{{\sqrt {14} }},\,\frac{2}{{\sqrt {14} }},\,\frac{3}{{\sqrt {14} }}} \right)$ .

Ans: Observe that the given points are,$A\left( {1,\,2,\, - 3} \right)$and $B\left( { - 1,\, - 2,\,1} \right)$.

Hence, the vector $\overrightarrow {AB} $ can be calculated as,  

\[\therefore \overrightarrow {AB}  = \left( { - 1 - 1} \right)\widehat i + \left( { - 2 - 2} \right)\widehat j + \left( {1 - \left( { - 3} \right)} \right)\widehat k =  - 2\widehat i - 4\widehat j + 4\widehat k\]

Therefore, the magnitude of the vector, that is, $\overrightarrow {AB}  =  - 2\widehat i - 4\widehat j + 4\widehat k$ can be calculated as,

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

Therefore, the direction cosines of the vector, that is, $\overrightarrow {AB}  =  - 2\widehat i - 4\widehat j + 4\widehat k$ can be calculated as, $\left( {\frac{-1}{3},\,\frac{-2}{3},\,\frac{2}{3}} \right)$.

Ans: Consider the given vector as,\[\overrightarrow a  = \widehat i + \widehat j + \widehat k\].

Therefore, the magnitude and the direction cosines of the vector, that is, \[\overrightarrow a  = \widehat i + \widehat j + \widehat k\] can be calculated as,

$\left| {\overrightarrow a } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}}  = \sqrt 3 $

$\therefore \left( {\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}} \right)$.

Assume, $\alpha ,\,\beta $ and $\gamma $ to be the angles formed by $\overrightarrow a $ with the axes $OX,\,OY$ and$OZ$.$\therefore \cos \alpha  = \frac{1}{{\sqrt 3 }},\,\cos \beta  = \frac{1}{{\sqrt 3 }},\,\cos \gamma  = \frac{1}{{\sqrt 3 }}$

This shows that the given vector is equally inclined to the  the positive direction of all the three axes.

Ans: Observe that the given position vectors of the two points, $P$ and $Q$are,\[\overrightarrow {OP}  = \widehat i + 2\widehat j - \widehat k\]and \[\overrightarrow {OQ}  =  - \widehat i + \widehat j + \widehat k\].

Therefore, the position vector of $R$ dividing the line joining the two given points through an internal division in the ratio of $2:1$ can be calculated using the formula, $\frac{{m\overrightarrow b  + n\overrightarrow a }}{{m + n}}$ (where the ratio is represented by $m:n$), as shown below, 

\[\overrightarrow {OR}  = \frac{{2\left( { - \widehat i + \widehat j + \widehat k} \right) + 1\left( {\widehat i + 2\widehat j - \widehat k} \right)}}{{2 + 1}} = \frac{{ - \widehat i + 4\widehat j + \widehat k}}{3} = \frac{{ - 1}}{3}\widehat i + \frac{4}{3}\widehat j + \frac{1}{3}\widehat k\]

Ans: Observe that the given position vectors of the two points, $P$ and $Q$are,\[\overrightarrow {OP}  = \widehat i + 2\widehat j - \widehat k\]and \[\overrightarrow {OQ}  =  - \widehat i + \widehat j + \widehat k\].

Therefore, the position vector of $R$ dividing the line joining the two given points through an external division in the ratio of $2:1$ can be calculated using the formula, $\frac{{m\overrightarrow b  - n\overrightarrow a }}{{m - n}}$ (where the ratio is represented by $m:n$), as shown below,  

\[\overrightarrow {OR}  = \frac{{2\left( { - \widehat i + \widehat j + \widehat k} \right) - 1\left( {\widehat i + 2\widehat j - \widehat k} \right)}}{{2 - 1}} = \frac{{ - 3\widehat i - \widehat k}}{1} =  - 3\widehat i + 3\widehat k\].

Ans: Consider the midpoint of the vector joining the position vectors of the two points,\[\overrightarrow {OP}  = 2\widehat i + 3\widehat j + 4\widehat k\]and \[\overrightarrow {OQ}  = 4\widehat i + \widehat j - 2\widehat k\] to be $R$. 

Therefore, the position vector of $R$ dividing the line joining the two given points through an internal division in the ratio of $1:1$ can be calculated as,  

\[\overrightarrow {OR}  = \frac{{\left( {2\widehat i + 3\widehat j + 4\widehat k} \right) + \left( {4\widehat i + \widehat j - 2\widehat k} \right)}}{{1 + 1}} = \frac{{6\widehat i + 4\widehat j + 2\widehat k}}{2} = 3\widehat i + 2\widehat j + \widehat k\]

Ans: Observe that the position vectors of the points\[A,\,B\] and\[C\]are,$\overrightarrow a  =3 \widehat i - 4\widehat j - 4\widehat k,\,\overrightarrow b  = 2\widehat i - \widehat j + \widehat k$ and$\overrightarrow c  = \widehat i - 3\widehat j - 5\widehat k$. 

Now, \[\overrightarrow {AB} ,\,\overrightarrow {BC} \]and \[\overrightarrow {CA} \]can be calculated as,  

\[\therefore \overrightarrow {AB}  = \overrightarrow b  - \overrightarrow a  = \left( {2 - 3} \right)\widehat i + \left( { - 1 + 4} \right)\widehat j + \left( {1 + 4} \right)\widehat k =  - \widehat i + 3\widehat j + 5\widehat k\]

\[\overrightarrow {BC}  = \overrightarrow c  - \overrightarrow b  = \left( {1 - 2} \right)\widehat i + \left( { - 3 + 1} \right)\widehat j + \left( { - 5 - 1} \right)\widehat k =  - \widehat i - 2\widehat j - 6\widehat k\]

\[\overrightarrow {CA}  = \overrightarrow a  - \overrightarrow c  = \left( {3 - 1} \right)\widehat i + \left( { - 4 + 3} \right)\widehat j + \left( { - 4 + 5} \right)\widehat k = 2\widehat i - \widehat j + \widehat k\]

Again, calculate the squares of magnitudesas shown below,

\[\therefore {\left| {AB} \right|^2} = {\left( { - 1} \right)^2} + {\left( 3 \right)^2} + {\left( 5 \right)^2} = 35\]

\[\,\,\therefore \,{\left| {BC} \right|^2} = {\left( { - 1} \right)^2} + {\left( { - 2} \right)^2} + {\left( { - 6} \right)^2} = 41\]

\[\therefore \,\,{\left| {CA} \right|^2} = {\left( 2 \right)^2} + {\left( { - 1} \right)^2} + {\left( 1 \right)^2} = 6\]

Now, use Pythagoras theorem it can be shown that this triangle is right-angled triangle as shown below,

${\left| {AB} \right|^2} + {\left| {CA} \right|^2} = 35 + 6 = 41 = {\left| {BC} \right|^2}$

Ans:

On considering the triangle law of addition for this triangle it can be observed that,

\[\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC} \]

\[ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  =  - \overrightarrow {CA} \]

\[ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow 0 \]

Therefore, the equation in (A) is true.

Again,

\[\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC} \]

\[ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  - \overrightarrow {AC}  = \overrightarrow 0 \]

Therefore, the equation in (B) is true.

Again,

\[\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC} \]

\[ \Rightarrow \overrightarrow {AB}  - \overrightarrow {CB}  + \overrightarrow {CA}  = \overrightarrow 0 \]

Therefore, the equation in (D) is true.

Now, for the equation in (C) also,

\[\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC} \]

\[ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  =  - \overrightarrow {CA} \]

\[ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow 0 \]

\[ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  - \overrightarrow {CA}  \ne \overrightarrow 0 \]

Therefore, the equation in (C) is incorrect and henceforth, it is the correct answer.

Ans: It can be observed that if \[\overrightarrow a \]and\[\overrightarrow b \] are collinear vectors then, they are parallel.

\[\overrightarrow b  = \lambda \overrightarrow a \], for some scalar \[\lambda \].

Again, if \[\lambda  =  \pm 1\], then $\overrightarrow a  =  \pm \overrightarrow b $.

Again if is considered that, $\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ and $\,\overrightarrow b  = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$, then \[\overrightarrow b  = \lambda \overrightarrow a \].

\[\,\,\,\,\,\,\,{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = \lambda \left( {{a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k} \right)\]

\[ \Rightarrow {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = \lambda {a_1}\widehat i + \lambda {a_2}\widehat j + \lambda {a_3}\widehat k\]

\[ \Rightarrow {b_1} = \lambda {a_1},\,{b_2} = \lambda {a_2},\,{b_3} = \lambda {a_3}\]

\[ \Rightarrow \lambda  = \frac{{{b_1}}}{{{a_1}}} = \frac{{{b_2}}}{{{a_2}}} = \frac{{{b_3}}}{{{a_3}}}\]

This shows that the respective components of $\overrightarrow a $ and $\overrightarrow b $, are proportional but the vectors can have different directions. 

Therefore, the statement in (D) is incorrect and henceforth, it is the correct answer.