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Question: A tetrahedron has vertices \(P(1,2,1),Q(2,1,3),R( - 1,1,2)\) and \(O(0,0,0)\). The angle between the...

A tetrahedron has vertices P(1,2,1),Q(2,1,3),R(−1,1,2)P(1,2,1),Q(2,1,3),R( - 1,1,2) and O(0,0,0)O(0,0,0). The angle between the faces OPQ and PQR is:
A cos⁡−1(935){\cos ^{ - 1}}\left( {\dfrac{9}{{35}}} \right)
B cos⁡−1(1935){\cos ^{ - 1}}\left( {\dfrac{{19}}{{35}}} \right)
C cos⁡−1(1731){\cos ^{ - 1}}\left( {\dfrac{{17}}{{31}}} \right)
D cos⁡−1(731){\cos ^{ - 1}}\left( {\dfrac{7}{{31}}} \right)

Explanation

Solution

In this question we have been given a tetrahedron PQRO with vertices, P(1,2,1),Q(2,1,3),R(−1,1,2)P(1,2,1),Q(2,1,3),R( - 1,1,2)and O(0,0,0)O(0,0,0) we need to find out the angle between the faces OPQ and PQR. For that we need to find the cross product, and after that use the formula:cos⁡θ=(a.b∣a∣.∣b∣)\cos \theta = \left( {\dfrac{{a.b}}{{\left| a \right|.\left| b \right|}}} \right), where a=OP→×OQ→a = \mathop {OP}\limits^ \to \times \mathop {OQ}\limits^ \to and b=PQ→×PR→\mathop {b = PQ}\limits^ \to \times \mathop {PR}\limits^ \to

Complete step-by-step answer:
We have been provided with a tetrahedron PQRO with vertices P(1,2,1),Q(2,1,3),R(−1,1,2)P(1,2,1),Q(2,1,3),R( - 1,1,2)and O(0,0,0)O(0,0,0),
So, according to the question we need to find the angle between the faces OPQ and PQR. So, for that firstly we need to find the vectors OP and OQ.
OP→=(i^(1−0)+j^(2−0)+k^(1−0)) OQ→=(i^(2−0)+j^(1−0)+k^(3−0)) \begin{gathered} \mathop {OP}\limits^ \to = (\hat i(1 - 0) + \hat j(2 - 0) + \hat k(1 - 0)) \\\ \mathop {OQ}\limits^ \to = (\hat i(2 - 0) + \hat j(1 - 0) + \hat k(3 - 0)) \\\ \end{gathered} ,
Simplifying it further we get,
OP→=(1i^+2j^+1k^) OQ→=(2i^+1j^+3k^) \begin{gathered} \mathop {OP}\limits^ \to = (1\hat i + 2\hat j + 1\hat k) \\\ \mathop {OQ}\limits^ \to = (2\hat i + 1\hat j + 3\hat k) \\\ \end{gathered}
Now we will be calculating the cross product of:
OP→=(1i^+2j^+1k^) OQ→=(2i^+1j^+3k^) \begin{gathered} \mathop {OP}\limits^ \to = (1\hat i + 2\hat j + 1\hat k) \\\ \mathop {OQ}\limits^ \to = (2\hat i + 1\hat j + 3\hat k) \\\ \end{gathered} ,
So, the cross product: \mathop {OP}\limits^ \to \times \mathop {OQ}\limits^ \to = \left| {\left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 1&2&1 \\\ 2&1&3 \end{array}} \right)} \right|,
Now we will be finding the cross product by solving the above determinant: OP→×OQ→=i^(6−1)−j^(3−2)+k^(1−4)\mathop {OP}\limits^ \to \times \mathop {OQ}\limits^ \to = \hat i(6 - 1) - \hat j(3 - 2) + \hat k(1 - 4),
The cross product comes out to be: OP→×OQ→=5i^−j^−3k^\mathop {OP}\limits^ \to \times \mathop {OQ}\limits^ \to = 5\hat i - \hat j - 3\hat k,
Now we will be finding the vectors PQ and PR.
PQ→=(i^(2−1)+j^(1−2)+k^(3−1)) PR→=(i^(−1−2)+j^(1−2)+k^(2−1)) \begin{gathered} \mathop {PQ}\limits^ \to = (\hat i(2 - 1) + \hat j(1 - 2) + \hat k(3 - 1)) \\\ \mathop {PR}\limits^ \to = (\hat i( - 1 - 2) + \hat j(1 - 2) + \hat k(2 - 1)) \\\ \end{gathered}
Now we will be simplifying them further,
PQ→=(i^−j^+2k^) PR→=(−2i^−j^+k^) \begin{gathered} \mathop {PQ}\limits^ \to = (\hat i - \hat j + 2\hat k) \\\ \mathop {PR}\limits^ \to = ( - 2\hat i - \hat j + \hat k) \\\ \end{gathered}
Now we will be calculating the cross product of:
PQ→=(i^−j^+2k^) PR→=(−2i^−j^+k^) \begin{gathered} \mathop {PQ}\limits^ \to = (\hat i - \hat j + 2\hat k) \\\ \mathop {PR}\limits^ \to = ( - 2\hat i - \hat j + \hat k) \\\ \end{gathered} ,

So, the cross product: b=PQ→×PR→b = \mathop {PQ}\limits^ \to \times \mathop {PR}\limits^ \to
\mathop {PQ}\limits^ \to \times \mathop {PR}\limits^ \to = \left| {\left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 1&{ - 1}&2 \\\ { - 2}&{ - 1}&1 \end{array}} \right)} \right|,
Now we will be finding the cross product by solving the above determinant:PQ→×PR→=i^(−1+2)−j^(1+4)+k^(−1−2)\mathop {PQ}\limits^ \to \times \mathop {PR}\limits^ \to = \hat i( - 1 + 2) - \hat j(1 + 4) + \hat k( - 1 - 2),
The cross product comes out to be: PQ→×QR→=i^−5j^−3k^\mathop {PQ}\limits^ \to \times \mathop {QR}\limits^ \to = \hat i - 5\hat j - 3\hat k,
Now as we want to calculate the angle between OPQ and PQR,
So, we will be using the formula:cos⁡θ=(a.b∣a∣.∣b∣)\cos \theta = \left( {\dfrac{{a.b}}{{\left| a \right|.\left| b \right|}}} \right),
In this question, a=OP→×OQ→a = \mathop {OP}\limits^ \to \times \mathop {OQ}\limits^ \to and b=PQ→×QR→b = \mathop {PQ}\limits^ \to \times \mathop {QR}\limits^ \to ,
So, we will be putting the values of a and b in cos⁡θ=(a.b∣a∣.∣b∣)\cos \theta = \left( {\dfrac{{a.b}}{{\left| a \right|.\left| b \right|}}} \right),
So, the equation becomes: cos⁡θ=((OP→×OQ→).(PQ→×QR→)∣OP→×OQ→∣.∣PQ→×QR→∣)\cos \theta = \left( {\dfrac{{\left( {\mathop {OP}\limits^ \to \times \mathop {OQ}\limits^ \to } \right).\left( {\mathop {PQ}\limits^ \to \times \mathop {QR}\limits^ \to } \right)}}{{\left| {\mathop {OP}\limits^ \to \times \mathop {OQ}\limits^ \to } \right|.\left| {\mathop {PQ}\limits^ \to \times \mathop {QR}\limits^ \to } \right|}}} \right),
Putting the values, we will get: cos⁡θ=((5i^−j^−3k^).(i^−5j^−3k^)∣5i^−j^−3k^∣.∣i^−5j^−3k^∣)\cos \theta = \left( {\dfrac{{(5\hat i - \hat j - 3\hat k).\left( {\hat i - 5\hat j - 3\hat k} \right)}}{{\left| {5\hat i - \hat j - 3\hat k} \right|.\left| {\hat i - 5\hat j - 3\hat k} \right|}}} \right),
Now we need to calculate the dot product of: (5i^−j^−3k^).(i^−5j^−3k^)(5\hat i - \hat j - 3\hat k).\left( {\hat i - 5\hat j - 3\hat k} \right),
So, the dot product: 5i^−j^−3k^).(i^−5j^−3k^)=5(1)−1(−5)−3(−3)5\hat i - \hat j - 3\hat k).\left( {\hat i - 5\hat j - 3\hat k} \right) = 5(1) - 1( - 5) - 3( - 3),
The dot product comes out to be: 5i^−j^−3k^).(i^−5j^−3k^)=195\hat i - \hat j - 3\hat k).\left( {\hat i - 5\hat j - 3\hat k} \right) = 19,
Now we need to calculate the value of ∣5i^−j^−3k^∣\left| {5\hat i - \hat j - 3\hat k} \right|,
So, it will be: ∣5i^−j^−3k^∣\left| {5\hat i - \hat j - 3\hat k} \right|=(52)+(−12)+(−32)=35\sqrt {({5^2}) + ( - {1^2}) + ( - {3^2})} = \sqrt {35} ,
Now we need to calculate the value of ∣i^−5j^−3k^∣\left| {\hat i - 5\hat j - 3\hat k} \right|,
So, it will be: ∣i^−5j^−3k^∣\left| {\hat i - 5\hat j - 3\hat k} \right|=(12)+(−52)+(−32)=35\sqrt {({1^2}) + ( - {5^2}) + ( - {3^2})} = \sqrt {35}
Keeping this value in cos⁡θ=((5i^−j^−3k^).(i^−5j^−3k^)∣5i^−j^−3k^∣.∣i^−5j^−3k^∣)\cos \theta = \left( {\dfrac{{(5\hat i - \hat j - 3\hat k).\left( {\hat i - 5\hat j - 3\hat k} \right)}}{{\left| {5\hat i - \hat j - 3\hat k} \right|.\left| {\hat i - 5\hat j - 3\hat k} \right|}}} \right),
The angle between OPQ and PQR: cos⁡θ=(1935.35)\cos \theta = \left( {\dfrac{{19}}{{\sqrt {35} .\sqrt {35} }}} \right)
So, the angle comes out to be: cos⁡θ=(1935) θ=cos⁡−1(1935) \begin{gathered} \cos \theta = \left( {\dfrac{{19}}{{35}}} \right) \\\ \theta = {\cos ^{ - 1}}\left( {\dfrac{{19}}{{35}}} \right) \\\ \end{gathered}

So, the correct answer is “Option b”.

Note: In this question do remember that during dot product does not include the direction vectors, because it is a scalar product. While finding the cross product of OP and OQ, PQ and PR find the vectors OP→,OQ→,PQ→,PR→\mathop {OP}\limits^ \to ,\mathop {OQ}\limits^ \to ,\mathop {PQ}\limits^ \to ,\mathop {PR}\limits^ \to first do avoid any kind of mistakes.