Question 1
If \vec{a},\vec{b},\vec{c} be such that |\vec{a} + \vec{b} + \vec{c}|=1, \vec{c} = \lambda\vec{a} x \vec{b} and |a|=1√2, |b|=1√3, |c|=1√6 then the angle between \vec{a} and \vec{b} is:
A)Î 6
B)Î 4
C)Î 3
D)Î 2
Question 2
Consider the plane (x,y,z)=(0,1,1)+λ(1,-1,1)+m(2,-1,0). The distance of this plane from the origin is
A)13
B)√32
C)√3√2
D)2√3
Answers:
Q1>D)
Q2>C)
39In Q1, vector c is in a plane perpendicular to both a and b. So conveniently a.c = b.c = 0.
|\vec{a} + \vec{b} + \vec{c}|^2 = 1
=> |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{a}.\vec{c})
=> \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + 2(\vec{a}.\vec{b} + 0 + 0)
=> 1 + 2\vec{a}.\vec{b} = 1
=> \vec{a}.\vec{b} = 0
But
|\vec{a}| \neq |\vec{b}| \neq 0
So
cos\theta = 0
This implies theta is a multiple of pi/2.
So D.
1q1> right thx pritish
q2>@smriti: right but cud u elaborate as to how u read the LC of vectors
1the plane passes thru (0,1,1) and the lines with dc's (1,-1,1) & (2,-1,0)
lie in the plane.
39Solve the determinant(which is equal to zero) :
|x-0 y-1 z-1 |
|1 -1 1 |
|2 -1 0 |
First row of determinant is x-x1, y-y1, z-z1 or the point which satisfies the plane. Next two rows are the position vectors/direction ratios of the lines.
x(+1) - (y-1)(-2) + (z-1)(1) = 0
=> x + 2y - 2 + z - 1 = 0
=> x + 2y + z -3 = 0
Distance of origin from plane is given by :
\frac{|0 + 0 + 0 -3|}{\sqrt{1 + 4 + 1}} = \frac{3}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{2}}