Find the distance of the point (1,2,0) from the point where the line joining A(2,-3,1) and B(3,-4,5) into the plane 2x+y+z=7?

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Find the distance of the point (1,2,0) from
the point where the line joining A(2,-3,1) and B(3,-4,5) into the
plane 2x+y+z=7?

Ajanta Trivedi · Accepted Answer

equation of the line joining the points A(2,-3,1) and B(3,-4,5) is $\frac{x - 2}{3 - 2} = \frac{y + 3}{- 4 + 3} = \frac{z - 1}{5 - 1} \frac{x - 2}{1} = \frac{y + 3}{- 1} = \frac{z - 1}{4} (1)$ suppose the line $\frac{x - 2}{1} = \frac{y + 3}{- 1} = \frac{z - 1}{4}$ cuts the plane 2x+y+z=7 at point A (r+2, -r-3, 4r+1)

therefore 2(r+2)+(-r-3)+(4r+1)=7 2r+4-r-3+4r+1=7 5r=5 r=5/5=1 thus the coordinates of A is (1+2,-1-3,4+1) i e A(3, -4, 5) therefore the required distance $P A = \sqrt{(3 - 1)^{2} + (- 4 - 2)^{2} + (5 - 0)^{2}} = \sqrt{4 + 36 + 25} P A = \sqrt{65} u n i t s$