((★BeMath⊚)/⊓) (1) ∫ x tan^(−1) (x) dx ? (2) Find the distance of the point (3,3,1) from the plane Π with equation (r^→ −i^→ −j^→ )•(i^→ −j^→ +k^→ ) = 0 , also find the point on the plane that is nearest to (3,3,1).


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Question Number 108099 by bemath last updated on 14/Aug/20

$\frac{★ B e M a t h ⊚}{⊓}$ $(1) \int x \tan^{- 1} (x) d x ?$ $(2) F i n d t h e d i s t a n c e o f t h e p o i n t$ $(3, 3, 1) f r o m t h e p l a n e Π w i t h e q u a t i o n$ $(\vec{r} - \vec{i} - \vec{j}) ∙ (\vec{i} - \vec{j} + \vec{k}) = 0, a l s o$ $f i n d t h e p o i n t o n t h e p l a n e t h a t i s$ $n e a r e s t t o (3, 3, 1) .$
Commented by bemath last updated on 14/Aug/20

$t h a n k y o u b o t h$
	Answered by Dwaipayan Shikari last updated on 14/Aug/20

	$\int {x t a n}^{- 1} x$ $\frac{x^{2}}{2} {t a n}^{- 1} x - \frac{1}{2} \int \frac{x^{2}}{1 + x^{2}}$ $\frac{x^{2}}{2} {t a n}^{- 1} x - \frac{x}{2} + \frac{1}{2} {t a n}^{- 1} x + C$
	Answered by mathmax by abdo last updated on 14/Aug/20

	$1) A = \int x \arctan (x) dx by parts$ $A = \frac{x^{2}}{2} arctanx - \int \frac{x^{2}}{2} \frac{dx}{1 + x^{2}}$ $= \frac{x^{2}}{2} \arctan (x) - \frac{1}{2} \int \frac{1 + x^{2} - 1}{1 + x^{2}} dx$ $= \frac{x^{2}}{2} \arctan (x) - \frac{x}{2} + \frac{1}{2} \arctan (x) + C$ $= \frac{1}{2} (x^{2} + 1) arctanx - \frac{x}{2} + C$
	Answered by 1549442205PVT last updated on 15/Aug/20
	$i^→ =(1,0,0),j^(→) =(0,1,0),k^(→) =(0,0,1) ⇒i ^(→) −j ^(→) +k ^(→) =(1,−1,1),r ^(→) =(x,y,z) ⇒r ^(→) −i ^(→) −j ^(→) =(x−1,y−1,z) (r ^(→) −i ^(→) −j ^(→) )•(i ^(→) −j ^(→) +k ^(→) ) =(x−1,y−1,z)•(1,−1,1)=0⇔x−1−y+1+z=0 ⇔x−y+z=0 d[(3,3,1),Π]=((∣3−3+1∣)/( (√(1^2 +(−1)^2 +1^2 ))))=(1/( (√3))) ii)Let M(x,y,y−x)∈Πbe any point of the plane Π which has the equation: x−y+z=0 and A(3,3,1).Then d(A,M)=(√((x−3)^2 +(y−3)^2 +(y−x−1)^2 )) we need to find smallest of the expression P=(x−3)^2 +(y−3)^2 +(y−x−1)^2 =2x^2 +2y^2 −4x−8y−2xy+19 =2x^2 −(2y+4)x+2y^2 −8y+19 =2[(x−((y+2)/2))^2 +2y^2 −8y+19−(((y+2)^2 )/2) =2[(x−((y+2)/2))^2 +((3y^2 −20y+34)/2) =2[(x−((y+2)/2))^2 +((3(y−((10)/3))^2 +(2/3))/2) 2[(x−((y+2)/2))^2 +((3(y−2)^2 )/2)+(1/3)≥(1/3) ⇒d(A,M)≥(1/( (√3))) The equality ocurrs if and only if { ((y−((10)/3)=0)),((x−((y+2)/2)=0)) :} ⇔ { ((x=8/3)),((y=10/3)) :}⇒z=2/3 Thus the nearest distance between the point A(3,3,1)and Π be (1/( (√3))) and the point M lie on the plane Πwhich is nearest to A having coordinate be M((8/3),((10)/3),(2/3)) other way Let H(x,y,y−x)∈Π.Then A(3,3,1)is nearest to the plane Π if and only if AH^(→) ⊥Π ⇔AH^(→) //n_Π ^(→) =(1,−1,1),since AH^(→) =(x−3,y−3,y−x−1), so AH^(→) //n_Π ^(→) ⇔((x−3)/1)=((y−3)/(−1))=((y−x−1)/1) ⇔ { ((x+y=6)),((2x−y=2)) :}⇔ { ((x=8/3)),((y=10/3)),((z=2/3)) :}⇒H((8/3),((10)/3),(2/3)) The distance between A and the plane Π equal to d(A,H)=(√((3−(8/3))^2 +(3−((10)/3))^2 +(1−(2/3))^2 )) =(√(3((1/3))^2 ))=(1/( (√3)))$
	$\vec{i} = (1, 0, 0), \vec{j} = (0, 1, 0), \vec{k} = (0, 0, 1)$ $\Rightarrow \vec{i} - \vec{j} + \vec{k} = (1, - 1, 1), \vec{r} = (x, y, z)$ $\Rightarrow \vec{r} - \vec{i} - \vec{j} = (x - 1, y - 1, z)$ $(\vec{r} - \vec{i} - \vec{j}) ∙ (\vec{i} - \vec{j} + \vec{k})$ $= (x - 1, y - 1, z) ∙ (1, - 1, 1) = 0 \Leftrightarrow x - 1 - y + 1 + z = 0$ $\Leftrightarrow x - y + z = 0$ $d [(3, 3, 1), Π] = \frac{∣ 3 - 3 + 1 ∣}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$ $ii) Let M (x, y, y - x) \in Π be any point of$ $the plane Π which has the equation :$ $x - y + z = 0 and A (3, 3, 1) . Then$ $d (A, M) = \sqrt{{(x - 3)}^{2} + {(y - 3)}^{2} + {(y - x - 1)}^{2}}$ $we need to find smallest of the expression$ $P = {(x - 3)}^{2} + {(y - 3)}^{2} + {(y - x - 1)}^{2}$ $= {2 x}^{2} + {2 y}^{2} - 4 x - 8 y - 2 xy + 19$ $= {2 x}^{2} - (2 y + 4) x + {2 y}^{2} - 8 y + 19$ $= 2 [{(x - \frac{y + 2}{2})}^{2} + {2 y}^{2} - 8 y + 19 - \frac{{(y + 2)}^{2}}{2}$ $= 2 [{(x - \frac{y + 2}{2})}^{2} + \frac{{3 y}^{2} - 20 y + 34}{2}$ $= 2 [{(x - \frac{y + 2}{2})}^{2} + \frac{3 {(y - \frac{10}{3})}^{2} + \frac{2}{3}}{2}$ $2 [{(x - \frac{y + 2}{2})}^{2} + \frac{3 {(y - 2)}^{2}}{2} + \frac{1}{3} ⩾ \frac{1}{3}$ $\Rightarrow d (A, M) ⩾ \frac{1}{\sqrt{3}}$ $The equality ocurrs if and only if$ ${\begin{cases} y - \frac{10}{3} = 0 \\ x - \frac{y + 2}{2} = 0 \end{cases} \Leftrightarrow {\begin{cases} x = 8 / 3 \\ y = 10 / 3 \end{cases} \Rightarrow z = 2 / 3$ $Thus the nearest distance between the$ $point A (3, 3, 1) and Π be \frac{1}{\sqrt{3}} and the point$ $M lie on the plane Π which is nearest to$ $A having coordinate be M (\frac{8}{3}, \frac{10}{3}, \frac{2}{3})$ $other way$ $Let H (x, y, y - x) \in Π . Then A (3, 3, 1) is nearest$ $to the plane Π if and only if \vec{AH} ⊥ Π$ $\Leftrightarrow \vec{AH} / / \vec{n_{Π}} = (1, - 1, 1), since \vec{AH} = (x - 3, y - 3, y - x - 1),$ $so \vec{AH} / / \vec{n_{Π}} \Leftrightarrow \frac{x - 3}{1} = \frac{y - 3}{- 1} = \frac{y - x - 1}{1}$ $\Leftrightarrow {\begin{cases} x + y = 6 \\ 2 x - y = 2 \end{cases} \Leftrightarrow {\begin{cases} x = 8 / 3 \\ y = 10 / 3 \\ z = 2 / 3 \end{cases} \Rightarrow H (\frac{8}{3}, \frac{10}{3}, \frac{2}{3})$ $The distance between A and the plane$ $Π equal to d (A, H) = \sqrt{{(3 - \frac{8}{3})}^{2} + {(3 - \frac{10}{3})}^{2} + {(1 - \frac{2}{3})}^{2}}$ $= \sqrt{3 {(\frac{1}{3})}^{2}} = \frac{1}{\sqrt{3}}$
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