Equation of a line in the plane π≡$$2x$$−$$y+z$$−$$4=0$$ which is perpendicular to the line l whose equation is x−$$21=y$$−$$2$$−$$1=z$$−$$3$$−$$2$$ and which passes through the point of intersection of l and π is

Question

Equation of a line in the plane π≡$$2x$$−$$y+z$$−$$4=0$$ which is perpendicular to the line l whose equation is x−$$21=y$$−$$2$$−$$1=z$$−$$3$$−$$2$$ and which passes through the point  of intersection of l and π is

Infinity Learn · Accepted Answer

Let direction ratios of the line be (a$$, b, c,$$); then $$2a$$−$$b+c=0$$ and a−b−$$2c=0$$, i.e., $$a3=b5=c$$−$$1$$ Therefore, direction ratios of the line are (3$$,$$5$$,−$$1).  Any point on the given line is (2+$$λ,$$2$$−λ,$$3$$−$$2$$λ$$); it lies on  the given plane π if $$2(2+$$λ$$)$$−(2$$−λ$$)+(3$$−$$2$$λ$$)=4$$ or $$4+2$$λ−$$2+$$λ$$+3$$−$$2$$λ$$=4$$ or λ$$=$$−$$1Therefore$$, the point of intersection of the line and the plane is $$(1$$, $$3$$, $$5). Therefore, equation of the required line is x−$$13=y$$−$$35=z$$−$$5$$−$$1$$

$Equation of a line in the plane π \equiv 2 x - y + z - 4 = 0 which is perpendicular to the line l whose equation is$
$\frac{x - 2}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{- 2} and which passes through the point of intersection of l and π is$

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Equation of a line in the plane π≡2x−y+z−4=0 which is perpendicular to the line l whose equation is x−21=y−2−1=z−3−2 and which passes through the point of intersection of l and π is

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$Equation of a line in the plane π \equiv 2 x - y + z - 4 = 0 which is perpendicular to the line l whose equation is$
$\frac{x - 2}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{- 2} and which passes through the point of intersection of l and π is$