maths. +2 higher secondary exam model question paper pls give answer
Answers
Answer:
Hope this helps.
a) The point (0,0,1) does not already lie in the plane 2x-z=0 (that would have made the question easy!).
So the plane is
( 3x - y + 5 ) + b ( 2x - z ) = 0
for some b.
Since (x,y,z) = (0,0,1) lies on the plane, we get
( 3(0) - 0 + 5 ) + b ( 2(0) - 1 ) = 0
=> 5 - b = 0
=> b = 5
Therefore the plane P is given by
( 3x - y + 5 ) + 5 ( 2x - z ) = 0
<=> 13x - y - 5z + 5 = 0 ... (1)
b) Points on the y-axis have the form (0, y, 0).
Since we want the one also on the plane P, put this into (1) to get
13(0) - y - 5(0) + 5 = 0
=> -y + 5 = 0
=> y = 5
=> plane P meets y-axis at (0, 5, 0)
c) Points in yz-plane have the form (0, y, z).
Since we want those on the plane P, put this into (1) to get
13(0) - y - 5z + 5 = 0
=> y + 5z - 5 = 0
How to express "the equation" for the line depends on what's expected in your class. You might put of the following:
- y + 5z - 5 = 0 and x = 0
- y / 5 = ( z - 1 ) / -1 and x = 0
- x = 0, y = -5t+5, z = t