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Maths-

General

Easy

Question

Let OPQR is a tetrahedron such that O is origin and $stack p with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack q with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack r with rightwards arrow on top$ are position vectors of P, Q, R respectively and a is the angle which OP makes with face PQR then

1
2
tana = $\frac{|[\vec{p} \vec{q} \vec{r}]|}{|\vec{q} \times \vec{r} + \vec{r} \times \vec{p} + \vec{p} \times \vec{r}| \cdot |\vec{p}|}$
none of these

The correct answer is: 2

To find the angle made by OP with face PQR
It is given that OP=p
OQ=q
OR=r
PQR=(OQ-OP) X(OR-OQ)=(q-p)X(r-q)
Angle= $cos left parenthesis theta right parenthesis equals 2 left parenthesis q minus p right parenthesis X left parenthesis r minus q right parenthesis divided by left parenthesis q minus p right parenthesis left parenthesis r minus q right parenthesis equals 2$

Hence required angle is 2.

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