Question

1/2 i + 1/2 j + c k then the value of c is​

Answer 1

$\bigstar\large \underline \pink{ \pmb{ \sf Correct \: Question }} \bigstar$

if $\sf \dfrac{1}{2}\hat{i} + \dfrac{1}{2} \hat{j} + c \hat{k}$ is a unit vector, then find the value of c.

$\: \: \: \: \: \: \dag \underline{ \sf According \: to \: the \: question : - }$

● $\sf \dfrac{1}{2}\hat{i} + \dfrac{1}{2} \hat{j} + c \hat{k}$ is a unit vector.

● which means that the magnitude of this vector is 1.

● We are asked to calculate the value of c.

● We know that, the magnitude of a vector represented in its component form is calculated by:-

$\leadsto \underline {\boxed {\pink{{ \bf | \overrightarrow{A} | = \sqrt{ {A_x}^{2} + {A_y}^{2} + {A_ z}^{2} } }}}} \bigstar$

● Where, Ax, Ay and Az are the components of A vector along x, y and z axis respectively.

• Ax = $\sf\dfrac{1}{2}$

• Ay = $\sf \dfrac{1}{2}$

• Az = c

• |A| = 1

$\: \: \: \: \: \: \dag\underline{ \frak{substituting \: these \: values \: in \: formula:-}}$

$: \implies \sf 1 = \sqrt{ \bigg({ \dfrac{1}{2} } \bigg)^{2} + \bigg({ \dfrac{1}{2} } \bigg)^{2} + {c}^{2} }$

$: \implies \sf 1 = \dfrac{1}{4} + \dfrac{1}{4} + c²$

$: \implies \sf 1 = \dfrac{1}{2} + c²$

$: \implies {\boxed {\pink{ {\frak c = \frac{1}{\sqrt{2}} }}}} \bigstar$

I hope u got what u were looking for :D

Answer 2

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$\underline{\blacksquare\:\:\:\footnotesize{\red{\text{SOLUTION:-}}}}$

$\footnotesize{ \maltese \: \sf Let, \underline{\overrightarrow{A} = \dfrac{1}{2} \hat{i} +\dfrac{1}{2} \hat{j} + c \hat{k} } } \: \maltese$

$\footnotesize{ \maltese \: \underline{ \boxed{ \sf Where \: \overrightarrow{A} \: is \: a \: unit \: vector .} }} \: \maltese$

$\footnotesize{ \maltese \: \underline{ \boxed{ \sf \overrightarrow{A} = 1} }} \: \maltese$

$\longrightarrow \: \sf |\overrightarrow{A} | = \sqrt{x^2 + y^2 + z^2}$

Where,

$\footnotesize{ \maltese \: \: \sf x : co-efficient \: of \: \hat{i}} \: \: \maltese$

$\footnotesize{ \maltese \: \: \sf y : co-efficient \: of \: \hat{j}} \: \: \maltese$

$\footnotesize{ \maltese \: \: \sf z : co-efficient \: of \: \hat{k}} \: \: \maltese$

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$\longrightarrow \: \sf1 = \sqrt{\left( \frac{1}{2} \right) ^2 +\left( \frac{1}{2} \right) ^2 + (C)^2}$

$\longrightarrow \: \sf1 = \sqrt{\frac{1}{4} +\frac{1}{4} + (C)^2}$

$\longrightarrow \: \sf1 = \frac{1}{4} +\frac{1}{4} + (C)^2$

$\longrightarrow \: \sf1 = \frac{4 + 4}{4\times 4} + (C)^2$

$\longrightarrow \: \sf1 = \frac{8}{16} + (C)^2$

$\longrightarrow \: \sf1 = \frac{8 \div 4}{16 \div 4} + (C)^2$

$\longrightarrow \: \sf1 = \frac{2}{4} + (C)^2$

$\longrightarrow \: \sf1 = \frac{1}{2} + (C)^2$

$\longrightarrow \: \sf (C)^2 = 1 - \frac{1}{2}$

$\longrightarrow \: \sf (C)^2 = \frac{2-1}{2}$

$\longrightarrow \: \sf (C)^2 = \frac{1}{2}$

$\longrightarrow \: \underline{ \underline{\sf C= \dfrac{1}{\sqrt{2}}}}$

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