Given points are A(2,0,−1),B(1,−2,0),C(1,2,−1) and D(0,−1,−2). ‌AB=B−A=(1,−2,0)−(2,0,−1) ‌=(−1,−2,+1) So, AC‌=C−A=(1,2,−1)−(2,0,−1) ‌=(−1,2,0) Now, normal vector ‌n1=AB×AC=|
∧
i
∧
j
∧
k
−1
−2
1
−1
2
0
| ‌=(0−2)
∧
i
−(0+1)
∧
j
+(−2−2)
∧
k
‌=−2
∧
i
−
∧
j
−4
∧
k
=(−2,−1,−4) ‌‌ Now, ‌AD=D−A=(0,−1,−2)−(2,0,−1) ‌=(−2,−1,−1) So, normal vector n2=AC×AD ‌=|
∧
i
∧
j
∧
k
−1
2
0
−2
−1
−1
| ‌=(−2−0)
∧
i
−(1+0)
∧
j
+(1+4)
∧
k
‌=−2
∧
i
−
∧
j
+5
∧
k
=(−2,−1,5) So, n1×n2=|
∧
i
∧
j
∧
k
−2
−1
−4
−2
−1
5
| =(−5−4)
∧
i
−(−10−8)
∧
j
+(2−2)
∧
k
=−9
∧
i
+18
∧
j
+0
∧
k
(−9,18,10) And n1⋅n2=(−2,−1,−4)⋅(−2,−1,5) =4+1−20=−15 So, the tangent of the angle between the two planes tan‌θ‌=‌
