4In 1 ) does sign of r matter??///
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41 ) 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
1357good thread
these are the most simple topics, don't leave them by any means.
62If the point (a,b,c) lies above the plane px + qy + rz + s = 0 then
( pa + qb + rc + s ) / r > 0
Is wale me..
think of it like this.. if z increases then teh point is higher..
when z increases then (px+qy+rz+s)/r will also increase...
So the statement is true..
33. if the point P(2,1,6) then find the point Q such that PQ is perpendicular to the plane x+y-2z=3 and the mid point of PQ lies on it
34. 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 ---------------------
62Q 3 is hardly clear...
Q4: Hint and observation (origin lies on the plane that you have given ) can you solve it now?
21for 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
66Q3) 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)