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Question Number 108931 by mohammad17 last updated on 20/Aug/20

Commented by mohammad17 last updated on 20/Aug/20

$s i r c a n y o u h e l p m e p l e a s$
	Answered by Aziztisffola last updated on 20/Aug/20
	$let A(1;2;3) and (△) { ((x=2+2t)),((y=1+6t)),((z=3)) :}⇒u^(→) ((2),(6),(0) ) B(x_0 ;y_0 ;z_0 )∈(△) such that (AB)⊥(△) AB^(→) (((x_0 −1)),((y_0 −2)),((z_0 −3)) ) AB^(→) .u^(→) =0 2x_0 −2+6y_0 −12=0 x_0 +3y_0 −7=0 we know B(x_0 ;y_0 ;z_0 )∈(△) ⇒ { ((x_0 =2+2t)),((y_0 =1+6t)),((z_0 =3)) :}⇒2+2t+3(1+6t)−7=0 ⇒20t=2⇒t=0.1 then B(2.2;1.6;3) Hence AB^(→) (((2.2−1)),((1.6−2)),((3−3)) ) ⇔AB^(→) (((1.2)),((−0.4)),(0) ) d(A;(△))=AB=(√((1.2)^2 +(−0.4)^2 )) =1.36$
	$let A (1; 2; 3) and (△) {\begin{cases} x = 2 + 2 t \\ y = 1 + 6 t \\ z = 3 \end{cases} \Rightarrow \vec{u} (\begin{matrix} 2 \\ 6 \\ 0 \end{matrix})$ $B (x_{0}; y_{0}; z_{0}) \in (△) such that (AB) ⊥ (△)$ $\vec{AB} (\begin{matrix} x_{0} - 1 \\ y_{0} - 2 \\ z_{0} - 3 \end{matrix}) \vec{AB} . \vec{u} = 0$ $2 x_{0} - 2 + 6 y_{0} - 12 = 0$ $x_{0} + 3 y_{0} - 7 = 0$ $we know B (x_{0}; y_{0}; z_{0}) \in (△)$ $\Rightarrow {\begin{cases} x_{0} = 2 + 2 t \\ y_{0} = 1 + 6 t \\ z_{0} = 3 \end{cases} \Rightarrow 2 + 2 t + 3 (1 + 6 t) - 7 = 0$ $\Rightarrow 20 t = 2 \Rightarrow t = 0.1$ $then B (2.2; 1.6; 3)$ $Hence \vec{AB} (\begin{matrix} 2.2 - 1 \\ 1.6 - 2 \\ 3 - 3 \end{matrix}) \Leftrightarrow \vec{AB} (\begin{matrix} 1.2 \\ - 0.4 \\ 0 \end{matrix})$ $d (A; (△)) = AB = \sqrt{{(1.2)}^{2} + {(- 0.4)}^{2}}$ $= 1.36$
	Answered by Aziztisffola last updated on 20/Aug/20


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