The image of the line x−$$13=y$$−$$31=z$$−$$4$$−$$5$$ in the plane $$2x$$−$$y+z+3=0$$ is the line

Question

The image of the line x−$$13=y$$−$$31=z$$−$$4$$−$$5$$ in the plane $$2x$$−$$y+z+3=0$$ is the line

Infinity Learn · Accepted Answer

$$3(2)+1($$−$$1)+($$−$$5)(1)=0Given$$ line and given plane are parallel∴ Image line is also parallel to the given lineImage of A (1$$,$$3$$,$$4) w.r.to given plane lies on the image line.Equation of the normal to the plane is x−$$12=y$$−$$3$$−$$1=z$$−$$41Any$$ point on the line $$B=(2r+1$$,−$$r+3$$,$$r+4)If$$ B is the image of A (1$$,$$3$$,$$4) then mid point of AB lies on the planeMid point $$=2r+22$$,−$$r+62$$,$$r+82Mid$$ point lies in the given plane ⇒$$22r+22$$−−$$r+62+r+82+3=0$$⇒$$r=$$−$$2$$,   $$B($$−$$3$$,$$5$$,$$2)Image$$ lines is $$x+33=y$$−$$51=z$$−$$2$$−$$5$$

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane $2 x - y + z + 3 = 0$ is the line

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The image of the line x−13=y−31=z−4−5 in the plane 2x−y+z+3=0 is the line

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The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane $2 x - y + z + 3 = 0$ is the line