ROCK ON.....@VECTORS & 3D

1 ) TRUE / FALSE

If the point (a,b,c) lies above the plane px + qy + rz + s = 0 then

( pa + qb + rc + s ) / r > 0

MULTIPLE ANSWER TYPE

2) Let a , b , c be 3 vectors such that a.a = b.b = c.c

& l a - b l² + l b - c l ² + l c - a l² = 27 ,then

a) a,b,c are necessarily coplanar

b) the 3 vectors represent sides of a triangle in magnitude & direction

c) a.b + b.c + c.a has the least value

d) the 3 vectors represent an orthogonal triad of vectors

In 1 ) does sign of r matter??///

Manish bhaiya ap bi questions dalo n answr mei help kijiye nah!!!

4. a line perpendiocular to x+2y+2z=0 and passes through the point (0,1,0).the perpendicular distance of this lone from the origin is ---------------------

for q4 the ans is (root45)/9 ....
its an easy board level q
to find eq of line perpendicular to the plane x+2y+2z=0 and passing through (0,1,0)
the direction ratios of line are <1,2,2>
since this line passes through point (0,1,0)
eq of line is...(x-0)/1 =(y-1)/2 =(z-0)/2 =k

so any arbitrary point on line interms of k r (k,2k+1,2k)

direction ratios of the perpendicular from origin to this line are <k,2k+1,2k>

the direction ratios of line are <1,2,2>

so k(1) +2(2k+1) +2(2k)=0

k=-2/9

we hav to find k to find the foot of perpendicular from origin to this line
and hence the dist

plz try question3

Q3) Let Q be (a,b,c). Then, since PQ is perpendicular to the plane x+y-2z=3, so the normal to the plane is parallel to PQ. So the direction ratio of the normal to the plane i.e. 1,1,-2 is proportional to the direction ratio of PQ which is a-2, b-1, c-6. So we have
a-21=b-11=c-6-2=k
Hence, a = k+2, b=k+1, c=6-2k
The mid point of PQ is (2+a2, 1+b2, 6+c2) i.e. (k+42, 2+k2,12-2k2) lies on the plane, so we have
k+42 + 2+k2 - 2 12-2k2 = 3
which gives k = 4. Hence the point Q is (6, 5, -2)