Chapter 10: Vector Algebra

Ans: The given vectors are,

\[\overrightarrow {\text{a}} {\text{ = }}\widehat {\text{i}}\,{\text{ + }}\widehat {\text{j}}\,{\text{ + }}\widehat {\text{k}}{\text{;}}\,\overrightarrow {\text{b}} {\text{ = 2}}\widehat {\text{i}}\,{\text{ - 7}}\widehat {\text{j}}\,{\text{ - 3}}\widehat {\text{k}}{\text{;}}\,\overrightarrow {\text{c}} {\text{ = }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{i}}\,{\text{ + }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{j}}\,{\text{ - }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{k}}\,\]

Magnitude of a vector say,\[\overrightarrow {{{\text{a}}_{\text{1}}}} {\text{ = }}\widehat {{{\text{i}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{j}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{k}}_{\text{1}}}}\] is,

$\left| {\overrightarrow {{{\text{a}}_{\text{1}}}} } \right|{\text{ = }}\sqrt {{{\left( {{{\text{i}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{j}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{k}}_{\text{1}}}} \right)}^{\text{2}}}} $

Therefore, the magnitude of the following vectors are,

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

Now, for $\overrightarrow b $, 

\[\left| {\overrightarrow {\text{b}} } \right|{\text{ = }}\sqrt {{{\left( {\text{2}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 7}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 3}}} \right)}^{\text{2}}}} {\text{ = }}\sqrt {{\text{4 + 49 + 9}}} {\text{ = }}\sqrt {{\text{62}}} \]

Again, for $\overrightarrow c $,

$\left| {\overrightarrow {\text{c}} } \right|{\text{ = }}\sqrt {{{\left( {\frac{{\text{1}}}{{\sqrt {\text{3}} }}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\frac{{\text{1}}}{{\sqrt {\text{3}} }}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\frac{{{\text{ - 1}}}}{{\sqrt {\text{3}} }}} \right)}^{\text{2}}}} {\text{ = 1}}$

Ans: Let us assume the two vectors to be,

and \[\overrightarrow {\text{b}} {\text{ = 2}}\widehat {\text{i}}\,{\text{ - }}\widehat {\text{j}}\,{\text{ - 3}}\widehat {\text{k}}\,\]

Magnitude of a vector say, \[\overrightarrow {{{\text{a}}_{\text{1}}}} {\text{ = }}\widehat {{{\text{i}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{j}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{k}}_{\text{1}}}}\] is,

$\left| {\overrightarrow {{{\text{a}}_{\text{1}}}} } \right|{\text{ = }}\sqrt {{{\left( {{{\text{i}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{j}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{k}}_{\text{1}}}} \right)}^{\text{2}}}} \overrightarrow {\text{a}} {\text{ = }}\widehat {\text{i}}\,{\text{ + 2}}\widehat {\text{j}}\,{\text{ + 3}}\widehat {\text{k}}\,$\[\]

Therefore, the magnitude of the following vectors are,

$\left| {\overrightarrow {\text{a}} } \right|{\text{ = }}\sqrt {{{\left( {\text{1}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\text{2}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\text{3}} \right)}^{\text{2}}}} {\text{ = }}\,\sqrt {{\text{1 + 4 + 9}}} {\text{ = }}\sqrt {{\text{14}}} $

Now, for $\overrightarrow b $, 

\[\left| {\overrightarrow {\text{b}} } \right|{\text{ = }}\sqrt {{{\left( {\text{2}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 1}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 3}}} \right)}^{\text{2}}}} \,{\text{ = }}\sqrt {{\text{4 + 1 + 9}}} {\text{ = }}\sqrt {{\text{14}}} \]

Both, vectors have got an equal magnitude that is $\sqrt {14} $ units but are completely different and not equal vectors since they are going in different directions.

Ans: Let us assume the two vectors as,

\[\overrightarrow p {\text{ = }}\widehat {\text{i}}\,{\text{ + }}\widehat {\text{j}}\,{\text{ + }}\widehat {\text{k}}\,\]and \[\overrightarrow q  = 2\widehat i\, + 2\widehat j\, + 2\widehat k\,\]

The direction cosines of the two vectors can be calculated as,

In case of $\overrightarrow {\text{p}} $,

$l = \frac{1}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }} = \frac{1}{{\sqrt 3 }}$

$m = \frac{1}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }} = \frac{1}{{\sqrt 3 }}$

$n = \frac{1}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }} = \frac{1}{{\sqrt 3 }}$

Now, for \[\overrightarrow {\text{q}} \], 

$l = \frac{2}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{2}{{2\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

$m = \frac{2}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{2}{{2\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

$n = \frac{2}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{2}{{2\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

It can be clearly observed that the direction cosines obtained in case of both the vectors that is, $\overrightarrow p $ and $\overrightarrow {\text{q}} $ are same and therefore they have got the same direction but different vectors.

Ans: Observe that it is required for the corresponding coefficients of $\widehat i$ and $\widehat j$to be equal in order to have two equal vectors.

Therefore, 

\[{\text{x = 2,}}\,{\text{y = 3}}\]

Ans: The given initial and terminal point are respectively, $\left( {{\text{2,}}\,{\text{1}}} \right)$ and \[\left( {{\text{ - 5,}}\,{\text{7}}} \right)\]. 

Therefore, the vector is,

$\overrightarrow {PQ}  = \left( {{\text{ - }}5{\text{ - }}2} \right)\widehat i{\text{ + }}\left( {{\text{7 - 1}}} \right)\widehat j{\text{  =  - 7}}\widehat i + 6\widehat {\text{j}}$

Ans: The given vectors are respectively, $\overrightarrow a  = \widehat i - 2\widehat j + \widehat k,\,\overrightarrow b  =  - 2\widehat i + 4\widehat j + 5\widehat k\,$ and $\overrightarrow c  = \widehat i - 6\widehat j - 7\widehat k$. 

Therefore, the sum of the following vectors can be calculated as,  \[\,\,\,\,\,\,\,\overrightarrow a  + \overrightarrow b  + \overrightarrow c  = \widehat i - 2\widehat j + \widehat k + ( - 2\widehat i) + 4\widehat j + 5\widehat k\, + \widehat i - 6\widehat j - 7\widehat k\]

\[ \Rightarrow \overrightarrow a  + \overrightarrow b  + \overrightarrow c  = \left( {1 - 2 + 1} \right)\widehat i + \left( { - 2 + 4 - 6} \right)\widehat j + \left( {1 + 5 - 7} \right)\widehat k\]

\[ \Rightarrow \overrightarrow a  + \overrightarrow b  + \overrightarrow c  =  - 4\widehat j - \widehat k\]

Ans: The given vectors is, $\overrightarrow a  = \widehat i + \widehat j + 2\widehat k$. 

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow a  = \widehat i + \widehat j + 2\widehat k$ is represented as $\widehat a$ and can be calculated as,

\[\left| {\overrightarrow a } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}}  = \sqrt 6 \,\,\,\,\,\]

\[\therefore \,\widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \frac{{\widehat i + \widehat j + 2\widehat k}}{{\sqrt 6 }} = \frac{1}{{\sqrt 6 }}\widehat i + \frac{1}{{\sqrt 6 }}\widehat j + \frac{2}{{\sqrt 6 }}\widehat k\]

Ans: The given points are$P{\text{ }}$and $Q$ which are $\left( {1,\,2,\,3} \right)$and $\left( {4,\,5,\,6} \right)$ respectively.

Now, the vector $\overrightarrow {PQ} $ can be calculated as,  

\[\overrightarrow {PQ}  = \left( {4 - 1} \right)\widehat i + \left( {5 - 2} \right)\widehat j + \left( {6 - 3} \right)\widehat k = 3\widehat i + 3\widehat j + 3\widehat k\]

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow {PQ}  = 3\widehat i + 3\widehat j + 3\widehat k$ is represented as $\widehat {PQ}$ and can be calculated as,

\[\left| {\overrightarrow {PQ} } \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 3 \right)}^2}}  = \sqrt {27} \, = 3\sqrt 3 \,\,\,\,\]

\[\therefore \,\widehat {PQ} = \frac{{\overrightarrow {PQ} }}{{\left| {\overrightarrow {PQ} } \right|}} = \frac{{3\widehat i + 3\widehat j + 3\widehat k}}{{3\sqrt 3 }} = \frac{1}{{\sqrt 3 }}\widehat i + \frac{1}{{\sqrt 3 }}\widehat j + \frac{1}{{\sqrt 3 }}\widehat k\]

Ans: It is given that the vectors are $\overrightarrow a  = 2\widehat i - \widehat j + 2\widehat k$and $\overrightarrow b  =  - \widehat i + \widehat j + \widehat k$respectively.

Now, the vector $\overrightarrow {a + b} $ can be calculated as,  

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

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow {a + b}  = \widehat i + \widehat k$ is represented as $\widehat {a + b}$ and can be calculated as,

\[\left| {\overrightarrow {a + b} } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}}  = \sqrt 2 \, = \sqrt 2 \,\,\,\,\]

\[\therefore \,\widehat {a + b} = \frac{{\overrightarrow {a + b} }}{{\left| {\overrightarrow {a + b} } \right|}} = \frac{{\widehat i + \widehat k}}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\widehat i + \frac{1}{{\sqrt 2 }}\widehat k\]

Ans: Assume the given vector as, $\overrightarrow a  = 5\widehat i - \widehat j + 2\widehat k$. 

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow a  = 5\widehat i - \widehat j + 2\widehat k$ is represented as $\widehat a$ and can be calculated as,

\[\left| {\overrightarrow a } \right| = \sqrt {{{\left( 5 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}}  = \sqrt {30} \,\,\,\,\,\]

\[\therefore \,\widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \frac{{5\widehat i - \widehat j + 2\widehat k}}{{\sqrt {10} }} = \frac{5}{{\sqrt {10} }}\widehat i - \frac{1}{{\sqrt {10} }}\widehat j + \frac{2}{{\sqrt {10} }}\widehat k\]

Now, a vector in the direction of the vector $5\widehat i - \widehat j + 2\widehat k$which has magnitude $8$ units can be calculated as,

\[\therefore \,8\widehat a = 8\left( {\frac{5}{{\sqrt {10} }}\widehat i - \frac{1}{{\sqrt {10} }}\widehat j + \frac{2}{{\sqrt {10} }}\widehat k} \right) = \frac{{40}}{{\sqrt {10} }}\widehat i - \frac{8}{{\sqrt {10} }}\widehat j + \dfrac{{16}}{{\sqrt {10} }}\widehat k\]