Vectors Ques 54
- 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 faces $O P Q$ and $P Q R$ is
(2019 Main, 12 Jan I)
(a) $\cos ^{-1}\left(\frac{7}{31}\right)$
(b) $\cos ^{-1}\left(\frac{9}{35}\right)$
(c) $\cos ^{-1}\left(\frac{19}{35}\right)$
(d) $\cos ^{-1}\left(\frac{17}{31}\right)$
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Answer:
Correct Answer: 54.(c)
Solution:
Formula:
Scalar Product Of Two Vectors:
- The given vertices of tetrahedron $P Q R O$ are $P(1,2,1)$, $Q(2,1,3), R(-1,1,2)$ and $O(0,0,0)$.
The normal vector to the face $O P Q$
$$ \begin{aligned} & =O P \times O Q=(\hat{i}+2 \hat{i}+\hat{k}) \times(2 \hat{i}+\hat{j}+3 \hat{k}) \\ & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right|=5 \hat{i}-\hat{j}-3 \hat{k} \end{aligned} $$
and the normal vector to the face $P Q R$
$$ \begin{aligned} & \quad=P Q \times P R=(\hat{i}-\hat{j}+2 \hat{k}) \times(-2 \hat{i}-\hat{j}+\hat{k}) \\ & \quad=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{array}\right| \\ & =\hat{i}(-1+2)-\hat{j}(1+4)+\hat{k}(-1-2)=\hat{i}-5 \hat{j}-3 \hat{k} \end{aligned} $$
Now, the angle between the faces $O P Q$ and $P Q R$ is the angle between their normals
$$ =\cos ^{-1} \frac{|5+5+9|}{\sqrt{25+1+9} \sqrt{1+25+9}}=\cos ^{-1}\left(\frac{19}{35}\right) $$
BETA
