If θ is the angle between the planes 2x-y+z-1=0 and x-2y+z+2=0, then cos θ is equal to

Question

If θ is the angle between the planes 2x-y+z-1=0 and x-2y+z+2=0, then cos θ is equal to (A)  2/3  (B)  3/4  (C)  4/5  (D)  5/6 s

Tardigrade Admin · Accepted Answer

cos θ =| (a1a2+b1b2+c1c2/√a12+b12+c12√a22+b22+c22) | Here, a1=2, b1=-1, c1=1, d1=-1 and a2=1,b2=-2, c2=1, d2=2 ∴ cos θ =| ((2× 1)+(-1× -2)+(1× 1)/√4+1+1√1+4+1) | =| (2+2+1/√6√6) | cos θ =| (5/6) | s

Q. If $θ$ is the angle between the planes $2 x - y + z - 1 = 0$ and $x - 2 y + z + 2 = 0,$ then $cos θ$ is equal to

$2/3$

$3/4$

$4/5$

$5/6$ s

Q. If θ is the angle between the planes 2x−y+z−1=0 and x−2y+z+2=0, then cosθ is equal to

2/3

3/4

4/5

5/6 s

Q. If $θ$ is the angle between the planes $2 x - y + z - 1 = 0$ and $x - 2 y + z + 2 = 0,$ then $cos θ$ is equal to

$2/3$

$3/4$

$4/5$

$5/6$ s