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On the diagram $M$ and $P$ are the midpoints of the edges $BC$ and $AB$ of the regular triangular prism $ABCA_1B_1C_1$.

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Which of the following lines is perpendicular to the plane $(BCC_1B_1):$ $AB;\quad PM;\quad A_1M;\quad PC_1?$

I am not really sure about the most correct way to approach the problem, but I was able to eliminate two of the answers.

The line $AB$ is perpendicular to $BB_1$ as $ABB_1A_1$ is a rectangle, but it cannot be parallel to $BC$ as $\measuredangle ABC=60^\circ$. Therefore $AB$ cannot be perpendicular to the plane $(BCC_1B_1)$. If that was the case, it would be perpendicular to all lines in that plane.

I think that the angle between $PM$ and $(BCC_1B_1)$ is the same as the angle between $AC$ and $(BCC_1B_1)$. Is this actually true? But $AC$ is not perpendicular to $(BCC_1B_1)$ as $\measuredangle ACB=60^\circ$.

I don't see what to do with the other two lines. I noted that as $AM\perp BC$, then $A_1M\perp BC$, but that was it.

How to find if the other two lines are perpendicular to the given plane?

yinivem462
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1 Answers1

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Your reasoning about the perpendicularity of the first two lines and the plane $(BCC_1B_1)$ is correct. For the third case suppose that $A_1M\bot (BCC_1B_1)$. Then $A_1M\bot MM_1$, where $M_1$ is middlepoint of $B_1C_1$. Consider triangle $AMM_1$. This triangle is a right triangle with right angle $\angle A_1M_1M=90^{\circ}$. So, then $\angle A_1MM_1<90^{\circ}$ and we get a contradiction. For the fourth case we will use the same argument and consider right triangle $PCC_1$ with right angle $\angle PCC_1=90^{\circ}$. Then $\angle PC_1C<90^{\circ}$ and therefore $PC_1$ is not perpendicular to $CC_1$ and hence, not perpendicular to the plane $(BCC_1B)$.

greyls
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  • Thank you for the response! Actually the authors say it's the line $A_1M$ that is perpendicular to the plane $(BCC_1B)$. I don't see why you think $\measuredangle A_1M_1M=90^\circ$? – yinivem462 Apr 17 '23 at 18:43
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    @yinivem462, I think that either the authors are mistaken, or there is a typo in the book. $\angle A_1M_1M=90^{\circ}$, because $MM_1 \parallel BB_1$ and $BB_1\bot (A_1B_1C_1)$. From this we have that $MM_1\bot (A_1B_1C_1)$ and hence $MM_1\bot A_1M_1$. – greyls Apr 17 '23 at 19:15