SATHEE: Vectors 2 Question 6

Vectors 2 Question 6

6. 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) $\cos^{- 1} (\frac{7}{31})$

(b) $\cos^{- 1} (\frac{9}{35})$

(d) $\cos^{- 1} (\frac{17}{31})$

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Answer:

Correct Answer: 6. (c)

Solution:

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 $O P Q$

$\begin{aligned} = O P \times O Q = (\hat{i} + 2 \hat{i} + \hat{k}) \times (2 \hat{i} + \hat{j} + 3 \hat{k}) \\ = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array} | = 5 \hat{i} - \hat{j} - 3 \hat{k} \end{aligned}$

and the normal vector to the face $P Q R$

$\begin{aligned} = P Q \times P R = (\hat{i} - \hat{j} + 2 \hat{k}) \times (- 2 \hat{i} - \hat{j} + \hat{k}) \\ = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 2 \\ - 2 & - 1 & 1 \end{array} | \\ = \hat{i} (- 1 + 2) - \hat{j} (1 + 4) + \hat{k} (- 1 - 2) = \hat{i} - 5 \hat{j} - 3 \hat{k} \end{aligned}$

Now, the angle between the faces $O P Q$ and $P Q R$ is the angle between their normals

$= \cos^{- 1} \frac{| 5 + 5 + 9 |}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \cos^{- 1} (\frac{19}{35})$

Vectors 2 Question 6

6. 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

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Vectors 2 Question 6

6. 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 OPQ and PQR is

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6. 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