Question

find the unit vector parallel to the resultant of the vectors A vector = 2 i^-6j^-3k^ and B vector=4i^+3j^-k^​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{\vec{a}} = 2\: \hat{i} - 6 \: \hat{j} - 3 \: \hat{k} \\ &\sf{\vec{b}} = 4\: \hat{i} + 3 \: \hat{j} - \: \hat{k}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{unit \: vector \: \parallel \: to \: \: \vec{a} \: + \: \vec{b}}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula\:Used - }}}} \end{gathered}$

We know,

$\tt \: unit \: vector \: in \: the \: direction \: of \: \vec{a} = \dfrac{\vec{a}}{ |\vec{a}| }$

$\large\underline{\bold{Solution-}}$

$\rm :\longmapsto\: \tt \: Let \: the \:required \: resultant \: vector \: be \: \vec{c}$

So,

$\rm :\longmapsto\:\vec{c} \: = \: \vec{a} \: + \: \vec{b}$

$\rm :\longmapsto\:\vec{c} = 2\: \hat{i} - 6 \: \hat{j} - 3\: \hat{k} + 4\: \hat{i} + 3 \: \hat{j} - \: \hat{k}$

$\rm :\longmapsto\:\vec{c} = 6\: \hat{i} - 3 \: \hat{j} - 4\: \hat{k}$

Hence,

$\rm :\longmapsto\: |\vec{c}| = \sqrt{ {(6)}^{2} + {( - 3)}^{2} + {( - 4)}^{2} }$

$\rm :\longmapsto\: |\vec{c}| = \sqrt{36 + 9 + 16}$

$\rm :\longmapsto\: |\vec{c}| = \sqrt{61}$

So,

• unit vector parallel to the resultant of vector a and vector b is given by

$\rm :\longmapsto\: \: \hat{c} \: = \: \dfrac{\vec{c}}{ |\vec{c}| }$

$\rm :\longmapsto\: \: \hat{c} \: = \: \dfrac{6\: \hat{i} - 3 \: \hat{j} - 4\: \hat{k}}{ \sqrt{61} }$

$\rm :\longmapsto\: \: \hat{c} \: = \: \dfrac{6}{ \sqrt{61} } \: \hat{i} - \dfrac{3}{ \sqrt{61} } \: \hat{j} - \dfrac{4}{ \sqrt{61} } \: \hat{k}$

Basic Concepts :-

Properties of Unit Vector :-

• The length of unit vector is 1 unit.

• The scalar components of unit vector gives direction cosines.

$\sf \: Let \: us \: consider \: a \: unit \: vector \: \: \hat{a} \: = l\: \hat{i} + m \: \hat{j} + n\: \hat{k}$

then,

$1. \: \: \tt \: {l}^{2} + {m}^{2} + {n}^{2} = 1$

$(2). \: \tt \: if \: \: \hat{a} \: makes \: angle \alpha \: and \: \beta \: and \: \gamma \: with \: axis \: then$

$\rm :\longmapsto\:l \: = \: \cos( \alpha) \\ \rm :\longmapsto\:m \: = \: \cos( \beta ) \\ \rm :\longmapsto\:n \: = \: \cos( \gamma )$

and

$\rm :\longmapsto\: {cos}^{2} \alpha + {cos}^{2} \beta + {cos}^{2} \gamma = 1$

Answer 2

Answer:

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{\vec{a}} = 2\: \hat{i} - 6 \: \hat{j} - 3 \: \hat{k} \\ &\sf{\vec{b}} = 4\: \hat{i} + 3 \: \hat{j} - \: \hat{k}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{unit \: vector \: \parallel \: to \: \: \vec{a} \: + \: \vec{b}}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula\:Used - }}}} \end{gathered}$

We know,

$\tt \: unit \: vector \: in \: the \: direction \: of \: \vec{a} = \dfrac{\vec{a}}{ |\vec{a}| }$

$\large\underline{\bold{Solution-}}$

$\rm :\longmapsto\: \tt \: Let \: the \:required \: resultant \: vector \: be \: \vec{c}$

So,

$\rm :\longmapsto\:\vec{c} \: = \: \vec{a} \: + \: \vec{b}$

$\rm :\longmapsto\:\vec{c} = 2\: \hat{i} - 6 \: \hat{j} - 3\: \hat{k} + 4\: \hat{i} + 3 \: \hat{j} - \: \hat{k}$

$\rm :\longmapsto\:\vec{c} = 6\: \hat{i} - 3 \: \hat{j} - 4\: \hat{k}$

Hence,

$\rm :\longmapsto\: |\vec{c}| = \sqrt{ {(6)}^{2} + {( - 3)}^{2} + {( - 4)}^{2} }$

$\rm :\longmapsto\: |\vec{c}| = \sqrt{36 + 9 + 16}$

$\rm :\longmapsto\: |\vec{c}| = \sqrt{61}$

So,

unit vector parallel to the resultant of vector a and vector b is given by

$\rm :\longmapsto\: \: \hat{c} \: = \: \dfrac{\vec{c}}{ |\vec{c}| }$

$\rm :\longmapsto\: \: \hat{c} \: = \: \dfrac{6\: \hat{i} - 3 \: \hat{j} - 4\: \hat{k}}{ \sqrt{61} }$

$\rm :\longmapsto\: \: \hat{c} \: = \: \dfrac{6}{ \sqrt{61} } \: \hat{i} - \dfrac{3}{ \sqrt{61} } \: \hat{j} - \dfrac{4}{ \sqrt{61} } \: \hat{k}$

Basic Concepts :-

Properties of Unit Vector :-

The length of unit vector is 1 unit.

The scalar components of unit vector gives direction cosines.

$\sf \: Let \: us \: consider \: a \: unit \: vector \: \: \hat{a} \: = l\: \hat{i} + m \: \hat{j} + n\: \hat{k}$

then,

$1. \: \: \tt \: {l}^{2} + {m}^{2} + {n}^{2} = 1$

$(2). \: \tt \: if \: \: \hat{a} \: makes \: angle \alpha \: and \: \beta \: and \: \gamma \: with \: axis \: then$

$\rm :\longmapsto\:l \: = \: \cos( \alpha) \\ \rm :\longmapsto\:m \: = \: \cos( \beta ) \\ \rm :\longmapsto\:n \: = \: \cos( \gamma )$

and

$\rm :\longmapsto\: {cos}^{2} \alpha + {cos}^{2} \beta + {cos}^{2} \gamma = 1$