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

If \overrightarrow{\math{a}}=\hat{\math{i}}+\hat{\math{j}}+\hat{\math{k}} \text { and } \overrightarrow{\math{b}}=\hat{\math{j}}-\hat{\math{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{\math{a}}=\hat{\math{i}}+\hat{\math{j}}+\hat{\math{k}} \text { and } \overrightarrow{\math{b}}=\hat{\math{j}}-\hat{\math{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)

\\ \begin{aligned} &\text { Also, for } \overrightarrow{\text { a. }} \overrightarrow{\mathrm{c}}=3\\ &\overrightarrow{\text { à. }} \overrightarrow{\mathrm{c}}=(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})\\ &\Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=3\\ &\text { since }_{2}(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\mathrm{x}+\mathrm{y}+\mathrm{z}, \text { as } \hat{1} . \hat{\imath}=\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{k}} . \hat{\mathrm{k}}=1 \text { and other dot }\\ &\text { multiplication is zero. We get, } \end{aligned}
\\x + y + z = 3 $ \ldots $ (iv)\\ \\

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

\\ (x - z) + (x - y) = 1 + 1\\ \\ $ \Rightarrow $ x + x - y - z = 2\\ \\ $ \Rightarrow $ 2x - y - z = 2 $ \ldots $ (v)\\ \\

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

\\ (x + y + z) + (2x - y - z) = 3 + 2\\ \\ $ \Rightarrow $ x + 2x + y - y + z - z = 5\\ \\ $ \Rightarrow $ 3x = 5\\ \\

\\ \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}

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

\\ \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}

 

 

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