Q. 285.0( 2 Votes )

Find the distance

Answer :


Let the point be A = (-2, 3, -4)


The line is


Compare line with


Where is point on the line and is the direction of line


and


The point on line is (1, 2, -1) and direction is <1, 3, -9>


Rewriting the line equation in cartesian from



Let it be equal to k so that we can write the general point on the line



x – 1 = k and y – 2 = 3k and z + 1 = -9k


x = k + 1 and y = 3k + 2 and z = -9k – 1


So, any general point on the line say B has coordinates (k + 1, 3k + 2, -9k – 1)


We have to measure the distance from point A to the line parallel to the plane x – y + 2z – 3 = 0


Comparing x – y + 2z – 3 = 0 with general plane equation ax + by + cz + d = 0 where <a, b, c> is the normal to the plane


let the normal is


Say we have to find the distance AB such that AB is parallel to plane


AB parallel to plane which means AB is perpendicular to the normal of plane that is CD




As is perpendicular to their dot product is zero because the right hand side will have cos90° which is 0




k + 3 + (3k – 1) (-1) + (-9k + 3)2 = 0


k + 3 – 3k + 1 – 18k + 6 = 0


-20k + 10 = 0


20k = 10


k = 1/2


Putting k = 1/2 in B we will get the point B


B = (1/2 + 1, 3(1/2) + 2, -9(1/2) – 1)



Now the distance AB





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