Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If a = i + j + k , b = j - k find a vector C such that a - c = b and a . c =.3

If \overrightarrow{\mathbf{a}} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} \text{ and } \overrightarrow{\mathbf{b}} = \hat{\mathbf{j}} - \hat{\mathbf{k}}  find a vector\begin{aligned} \\ &\overrightarrow{\mathrm{c}} \text { such that } &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}_{\text {and }} \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3\end{aligned} .

 

Answers (1)

Given that,

\overrightarrow{\mathbf{a}} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} \text{ and } \overrightarrow{\mathbf{b}} = \hat{\mathbf{j}} - \hat{\mathbf{k}}

\\ \begin{aligned} &\text { We need to find vector } \overrightarrow{\mathrm{C}} \text { . }\\ &\text { Let } \overrightarrow{\mathrm{c}}=\mathrm{x} \hat{1}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}, \text { where } \mathrm{x}, \mathrm{y}, \mathrm{z} \text { be any scalars. }\\ &\text { Now, for } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{x} \hat{\mathrm{l}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \end{aligned}

\\ \Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{xi}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \end{array}\right|=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{1}((1)(\mathrm{z})-(1)(\mathrm{y}))-\hat{\jmath}((1)(\mathrm{z})-(1)(\mathrm{x}))+\hat{\mathrm{k}}(1)(\mathrm{y})-(1)(\mathrm{x}))=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-\mathrm{y})-\hat{\mathrm{j}}(\mathrm{z}-\mathrm{x})+\hat{\mathrm{k}}(\mathrm{y}-\mathrm{x})=\hat{\mathrm{j}}-\hat{\mathrm{k}}

Comparing Left Hand Side and Right Hand Side, we get

From coefficient of  \hat{i}  ⇒ z-y = 0 …(i)

From coefficient of \hat{j}  ⇒ -(z-x) = 1

⇒ x-z = 1 …(ii)

From coefficient of \hat{k}  ⇒ y-x = -1

⇒ x-y = 1 …(iii)


\\x + y + z = 3 $ \ldots $ (iv)\\ \\

Now, add equations (ii) and (iii), we get

\begin{align*} &(x - z) + (x - y) &= 1 + 1\\ &\Rightarrow x + x - y - z &= 2\\ &\Rightarrow 2x - y - z &= 2 \quad \ldots \quad (v) \end{align*}

Add equations (iv) and (v), we get

\begin{align*} &(x + y + z) + (2x - y - z) &= 3 + 2\\ &\Rightarrow x + 2x + y - y + z - z &= 5\\ &\Rightarrow 3x &= 5 \end{align*}

\\ \Rightarrow x=\frac{5}{3}\\ $Put value of x in equation (iii), we get Equation (iii)$\\ \Rightarrow x-y=1\ \Rightarrow \frac{5}{3}-\mathrm{y}=1 $$ \Rightarrow \mathrm{y}=\frac{5}{3}-1

\\\begin{aligned} &\Rightarrow \mathrm{y}=\frac{5-3}{3}\\ &\Rightarrow \mathrm{y}=\frac{2}{3}\\ &\text { Put this value of } y \text { in equation (i), we get } \end{aligned}

\\ \begin{align*} &\text{Equation } (\mathrm{i}) \Rightarrow \mathrm{z} - \mathrm{y} & \\= 0 \Rightarrow \mathrm{z} - \frac{2}{3} = 0 \Rightarrow \mathrm{z} = \frac{2}{3}\\ &\text{since, } \overrightarrow{\mathrm{c}} = \mathrm{x}\hat{1} + \mathrm{y}\hat{\mathrm{j}} + \mathrm{z}\hat{\mathrm{k}}\\ &\text{By putting the values of } x, y, \text{ and } z, \text{ we get} \end{align*}

\\ \begin{aligned} &\overrightarrow{\mathrm{c}}=\frac{5}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}+\frac{2}{3} \hat{\mathrm{k}}\\ &\text { Thus, we have found the vector } \overrightarrow{\mathrm{c}} \text { . } \end{aligned}

 

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question