A) 3 dice are rolled and 3 numbers a,b,c are selected, then probability that the 3 planes :
ax+by+cz=0
bx+cy+az=0
cx+ay+bz=0 form
1) exactly 1 line is 0
2) a point is 105/108
3) a prism is 0
4) whole space is 1
(Multiple Choice question)
B) Consider a plane P=x+2y+3z-6=0 and a point A(0,0,0). True statement is/are
1)Distance of A from P in perpendicular direction of plane x+y+z+1=0 is √3
2)Distance of A from P in direction parellel to line x-16=y-36=z2 is 72
3)A` is the image of A through P, then distance AA` is 12√2
4) Disance of A from P which make 45° with P is 97
1well bhabani my answer to ur 1st qstn would be 1, 2 and 3...
the reason is very simple... the point (0,0,0) will always satisfy the equatns.. Now if a=b=c...then the planes coincide to form only plane not a straight line or a prism or blah blah blah...... So these planes always meet at the origin. Now if u roll 3 dice then the prob. that a=b=c is 6/63=6/216... thus prob that they meet at only 1 point is 1-6/216=105/108..
1Ur second problem is all about calculations... I will give u the outline of the solution.. Find the distance of A from P, which is 6/(14)1/2... Now find the angles between the normal vector of P and the given vectors, and after that just calculate the projection of 6/(14)1/2 on them... apparently it seems only 1 is correct.. I may b wrong..
