Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle a with the positive x-axis, then cos a equals
(A) 1/√3
(B) 1/2
(C) 1
(D) 1/√2
Answer is (A) 1/√3
If direction cosines of L be l, m and n, then
2l + 3m + n = 0
l + 3m + 2n = 0
On solving, we get
l/3 = m/-3 = n/3
Therefore,
l: m:n = 1/√3 : - 1/√3 : 1/√3 ⇒ cosα = 1/√3
