IIT JEE , AIEEE Forum - Mathematics , Vectors

neeraj_agarwal_1990

3 poles of height x,x+y,x+z are posted at the vertices A,B,C of a triangular park of sides a,b,c respectively.A plane sheet is mounted at the tops of the poles ,If the plane of the sheet is inclined at an angle @ to the horizontal plane,prove(using vector):
tan@=   [ y²/c² +z²/b²- 2yzcosA/bc]²
             -----------------------------------------
                             sinA

avinash.sharma

In this solution Bold&underlined are vectors.

ABC is a triangular park in XY plane with lengths a,b and c of sides BC, AC and AB respectively. Let take 'A' at the origin and B at X axis. So the coordinates of A and B are (0,0,0) and (c,0,0) respectively. Now as it is clear from the figure (See the figure below for clarity) A,B and C are in XY plane so their Z coordinates are zeroes. C is somewhere in the middle of XY plane from 'b' distance to A, so the coordinate of C is (bCosA, bSinA,0). AA' =x , BB'= x + y, CC'= x + z are three poles at A,B and C points (marked with blue colour). The coordinates of A', B' and C' can be obtained by adding only Z component in A, B and C respectively.

A'(0,0,x) B'(c,0,x+y) C'(bCosA, bSinA,x+z)

To get the angle between these two planes (ABC and A'B'C') it's the simple work to get the angle between normals to these planes. Let n1 and n2 are normal to ABC and A'B'C' planes.

By vector algebra n1 = AB x AC =

i	j	k
c	0	0
bCosA	bSinA	0

n1= bc SinA k

Same way we can calculate the normal to the plane A'B'C' = n2 = A'B' x A'C' =

i	j	k
c	0	y
bCosA	bSinA	z

(Please solve it yourself) and you will get

n2 = bySinA i - (cz-byCosA) j + bcSinA k

Now angle between these two normals is

Cosq = (n1 . n2 ) / (|n1| |n2|)

= (b²c²Sin²A)/[Ö( b²c²Sin²A) Ö(b²y²Sin²A + c²z²+b²y²cos²A-2bcyzCosA+ b²c²Sin²A)]

As cos²A+Sin²A =1 so the above quantity is equal to

Cosq = (b²c²Sin²A)/[Ö( b²c²Sin²A) Ö(b²y² + c²z² -2bcyzCosA+ b²c²Sin²A)]

Or

Cosq = (bcSinA)/[Ö(b²y² + c²z² -2bcyzCosA+ b²c²Sin²A)]

Now Sinq = Ö(1- Cos²q)

= Ö[ 1- (b²c²Sin²A)/(b²y² + c²z² -2bcyzCosA+ b²c²Sin²A)

= Ö[ (b²y² + c²z² -2bcyzCosA+ b²c²Sin²A- b²c²Sin²A)/( b²y² + c²z² -2bcyzCosA+ b²c²Sin²A)]

=Ö[( b²y² + c²z² -2bcyzCosA)/ ( b²y² + c²z² -2bcyzCosA+ b²c²Sin²A)]

so Tanq = Sinq/ Cosq = Ö( b²y² + c²z² -2bcyzCosA)/ bcSinA

dividing both numerator and denominator by bc we get

Tanq = Ö(y²/ c² + z²/ b² -2yzCosA/bc) / SinA

Or

Tanq = (y²/ c² + z²/ b² -2yzCosA/bc)^½ / SinA

neeraj_agarwal_1990

awesome work sir..

Ask & Discuss Questions with Community & Experts