Q. 165.0( 2 Votes )

Find the points o

Answer :

Given, A line at a distance of 5 units from the point P(1, 3, 3)

To find : Find the points on the given line.


Explanation: We have line given



Then, x = 3μ - 2 , y = 2μ - 1 and z = 2μ + 3


So, The points on the given line are


(x, y, z) = ( 3μ - 2 , 2μ - 1 , 2μ + 3)


Now, By using distance formula between two points









Therefore, The points are (3(0) - 2 , 2(0) - 1 , 2(0) + 3) and (3(2) - 2 , 2(2) - 1 , 2(2) + 3)


Hence, Points are ( - 2, - 1, 3) and (4, 3, 7).


OR


Let the equation of the plane passing through A(3, - 1, 2) is A(x - 3) + B(y + 1) + C(z - 2)


And, It also passes through the points B(5, 2, 4) and C( - 1, - 1, 6)


Then, Take point B and equation will be


A(5 - 3) + B(2 - ( - 1)) + C(4 - 2) = 0


So, 2A + 3B + 2C = 0 - - - - - - (i)


And, From point C , the equation will be


A( - 1 - 3) + B( - 1 - ( - 1)) + C(6 - 2) = 0


So, - 4A + 0B + 4C = 0 - - - - - (ii)


On Solving equation (i) and (ii), we get





Now, The Equation of plane is given by


3(x - 3) - 4(y + 1) + 3(z - 2) = 0


3x - 4y + 3z = 19


And, Distance of point P(6, 5, 9) from the given plane





Hence, The distance from the given plane is


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