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The statement (p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))  is equivalent to ________.

Option: 1

(\sim \mathrm{p}) \vee(\sim \mathrm{q})


Option: 2

(\sim \mathrm{p}) \wedge(\sim \mathrm{q})


Option: 3

p \vee(\sim q)


Option: 4

\mathrm{p} \vee \mathrm{q}


Answers (1)

best_answer

(\mathrm{p} \wedge \sim \mathrm{q}) \vee(\sim \mathrm{p} \wedge \mathrm{q}) \vee(\sim \mathrm{p} \wedge \sim \mathrm{q})

\mathrm{p} \wedge \sim \mathrm{q} \Rightarrow             

\sim p \wedge q=                 

\sim \mathrm{p} \wedge \sim \mathrm{q}=            

                                                                      

\begin{aligned} & (\mathrm{p} \wedge \sim \mathrm{q}) \vee(\sim \mathrm{p} \wedge \mathrm{q})(\sim \mathrm{p} \wedge \sim \mathrm{q}) \\ & (\alpha, \beta) \\ & (\sim \mathrm{p}) \vee(\sim \mathrm{q}) \end{aligned}

Plane passing through (0,-1,2)

and (-1,2,1)
then vector in plane \langle-1,3,-1\rangle

vector parallel to plane is \langle 4,2,-6\rangle

normal vector to plane \mathrm{L}_2
$$ \begin{aligned} & (\overrightarrow{\mathrm{n}})=\left|\begin{array}{ccc} \hat{\mathrm{l}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 3 & -1 \\ 4 & 2 & -6 \end{array}\right| \\ & =\mathrm{i}(-16)-\hat{\mathrm{j}}(10)+\hat{\mathrm{k}}(-14) \\ & \overrightarrow{\mathrm{n}}=<8,5,7> \end{aligned}
Equation of plane
$$ \begin{aligned} & 8(x-0)+5(y+1)+7(z-2)=0 \\ & \Rightarrow 8 x+5 y+7 z=9 \end{aligned}
From given options point (-2,5,0)

lies on plane.

Posted by

Nehul

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