Find the normal to plane O A B OAB O A B
The vectors in plane O A B OAB O A B
O A ⃗ = ( 1 , 2 , 1 ) , O B ⃗ = ( 2 , 1 , 3 ) \vec{OA}=(1,2,1), \qquad \vec{OB}=(2,1,3) O A = ( 1 , 2 , 1 ) , O B = ( 2 , 1 , 3 ) So a normal vector is
n ⃗ 1 = O A ⃗ × O B ⃗ \vec{n}_1=\vec{OA}\times\vec{OB} n 1 = O A × O B Compute:
n ⃗ 1 = ∣ i ^ j ^ k ^ 1 2 1 2 1 3 ∣ = i ^ ( 2 ⋅ 3 − 1 ⋅ 1 ) − j ^ ( 1 ⋅ 3 − 1 ⋅ 2 ) + k ^ ( 1 ⋅ 1 − 2 ⋅ 2 ) \vec{n}_1=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
1 & 2 & 1\\
2 & 1 & 3
\end{vmatrix}
=\hat i(2\cdot 3-1\cdot 1)-\hat j(1\cdot 3-1\cdot 2)+\hat k(1\cdot 1-2\cdot 2) n 1 = i ^ 1 2 j ^ 2 1 k ^ 1 3 = i ^ ( 2 ⋅ 3 − 1 ⋅ 1 ) − j ^ ( 1 ⋅ 3 − 1 ⋅ 2 ) + k ^ ( 1 ⋅ 1 − 2 ⋅ 2 ) n ⃗ 1 = 5 i ^ − 1 j ^ − 3 k ^ = ( 5 , − 1 , − 3 ) \vec{n}_1=5\hat i-1\hat j-3\hat k=(5,-1,-3) n 1 = 5 i ^ − 1 j ^ − 3 k ^ = ( 5 , − 1 , − 3 )
Find the normal to plane A B C ABC A B C
Take two vectors in plane A B C ABC A B C
A B ⃗ = B − A = ( 2 − 1 , 1 − 2 , 3 − 1 ) = ( 1 , − 1 , 2 ) \vec{AB}=B-A=(2-1,1-2,3-1)=(1,-1,2) A B = B − A = ( 2 − 1 , 1 − 2 , 3 − 1 ) = ( 1 , − 1 , 2 ) A C ⃗ = C − A = ( − 1 − 1 , 1 − 2 , 2 − 1 ) = ( − 2 , − 1 , 1 ) \vec{AC}=C-A=(-1-1,1-2,2-1)=(-2,-1,1) A C = C − A = ( − 1 − 1 , 1 − 2 , 2 − 1 ) = ( − 2 , − 1 , 1 ) So a normal vector is
n ⃗ 2 = A B ⃗ × A C ⃗ \vec{n}_2=\vec{AB}\times\vec{AC} n 2 = A B × A C Compute:
n ⃗ 2 = ∣ i ^ j ^ k ^ 1 − 1 2 − 2 − 1 1 ∣ = i ^ ( ( − 1 ) ⋅ 1 − 2 ⋅ ( − 1 ) ) − j ^ ( 1 ⋅ 1 − 2 ⋅ ( − 2 ) ) + k ^ ( 1 ⋅ ( − 1 ) − ( − 1 ) ⋅ ( − 2 ) ) \vec{n}_2=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
1 & -1 & 2\\
-2 & -1 & 1
\end{vmatrix}
=\hat i((-1)\cdot 1-2\cdot(-1)) - \hat j(1\cdot 1-2\cdot(-2)) + \hat k(1\cdot(-1)-(-1)\cdot(-2)) n 2 = i ^ 1 − 2 j ^ − 1 − 1 k ^ 2 1 = i ^ (( − 1 ) ⋅ 1 − 2 ⋅ ( − 1 )) − j ^ ( 1 ⋅ 1 − 2 ⋅ ( − 2 )) + k ^ ( 1 ⋅ ( − 1 ) − ( − 1 ) ⋅ ( − 2 )) n ⃗ 2 = i ^ ( 1 ) − j ^ ( 5 ) + k ^ ( − 3 ) = ( 1 , − 5 , − 3 ) \vec{n}_2=\hat i(1)-\hat j(5)+\hat k(-3)=(1,-5,-3) n 2 = i ^ ( 1 ) − j ^ ( 5 ) + k ^ ( − 3 ) = ( 1 , − 5 , − 3 )
Angle between the planes
The angle between two planes equals the angle between their normal vectors.
So if θ \theta θ O A B OAB O A B A B C ABC A B C
cos θ = ∣ n ⃗ 1 ⋅ n ⃗ 2 ∣ ∣ n ⃗ 1 ∣ ∣ n ⃗ 2 ∣ \cos\theta=\frac{|\vec{n}_1\cdot\vec{n}_2|}{|\vec{n}_1|\,|\vec{n}_2|} cos θ = ∣ n 1 ∣ ∣ n 2 ∣ ∣ n 1 ⋅ n 2 ∣ Now,
n ⃗ 1 ⋅ n ⃗ 2 = ( 5 ) ( 1 ) + ( − 1 ) ( − 5 ) + ( − 3 ) ( − 3 ) = 5 + 5 + 9 = 19 \vec{n}_1\cdot\vec{n}_2=(5)(1)+(-1)(-5)+(-3)(-3)=5+5+9=19 n 1 ⋅ n 2 = ( 5 ) ( 1 ) + ( − 1 ) ( − 5 ) + ( − 3 ) ( − 3 ) = 5 + 5 + 9 = 19 Also,
∣ n ⃗ 1 ∣ = 5 2 + ( − 1 ) 2 + ( − 3 ) 2 = 25 + 1 + 9 = 35 |\vec{n}_1|=\sqrt{5^2+(-1)^2+(-3)^2}=\sqrt{25+1+9}=\sqrt{35} ∣ n 1 ∣ = 5 2 + ( − 1 ) 2 + ( − 3 ) 2 = 25 + 1 + 9 = 35 ∣ n ⃗ 2 ∣ = 1 2 + ( − 5 ) 2 + ( − 3 ) 2 = 1 + 25 + 9 = 35 |\vec{n}_2|=\sqrt{1^2+(-5)^2+(-3)^2}=\sqrt{1+25+9}=\sqrt{35} ∣ n 2 ∣ = 1 2 + ( − 5 ) 2 + ( − 3 ) 2 = 1 + 25 + 9 = 35 Thus,
cos θ = 19 35 35 = 19 35 \cos\theta=\frac{19}{\sqrt{35}\sqrt{35}}=\frac{19}{35} cos θ = 35 35 19 = 35 19 Therefore,
θ = cos − 1 ( 19 35 ) \theta=\cos^{-1}\left(\frac{19}{35}\right) θ = cos − 1 ( 35 19 )
Match with options
This corresponds to Option B .