But PRQ will be an equilateral triangle

And the other two angles which are 45 degrees are not in the same plane

Updated with the figure

The angles in green is what we are looking for

Carcass see similar question here: https://gmatclub.com/forum/for-the-cube ... 13841.html

The explanatiion on GC is wrong



Where is BEG ?

also another user pointed out that the two angles are 45° each

The question and the OA and the explanation provided are correct

Thanks you

Carcass it is option C

the angle PQR ,when you measure it, you'll measure it in the plane of PQR

even Scott from TTP in the Gmat club said its option C

We are given a cube $P Q R$ and want to find the degree measure of $\(\angle P Q R\)$.
Step 1: Understand the points and the cube
- $\(P, Q, R\)$ are points on the cube.
- $\(P Q R\)$ forms an angle at $Q$.

Step 2: Assign coordinates
Assume the cube has side length 1 for simplicity.
- Let $Q$ be at the origin $\((0,0,0)\)$.
- Let $P$ along the x-axis: $\(P=(1,0,0)\)$.
- Let $R$ along the z-axis: $\(R=(0,0,1)\)$.

Step 3: Vectors from $Q$
- Vector $\(\overrightarrow{Q P}=P-Q=(1,0,0)\)$.
- Vector $\(\overrightarrow{Q R}=R-Q=(0,0,1)\)$.

Step 4: Calculate $\(\angle P Q R\)$
Use the dot product formula:

$$
\(\cos \theta=\frac{\overrightarrow{Q P} \cdot \overrightarrow{Q R}}{\|\overrightarrow{Q P}\|\|\overrightarrow{Q R}\|}\)
$$

- Dot product: $\((1,0,0) \cdot(0,0,1)=0\)$.
- Magnitudes: $\(\|\overrightarrow{Q P}\|=1,\|\overrightarrow{Q R}\|=1\)$.

Hence,

$$
\(\cos \theta=\frac{0}{1 \times 1}=0 \Longrightarrow \theta=90^{\circ}\) .
$$


Answer:

$\(90^{\circ}\)$

So, $\(\angle P Q R=90^{\circ}\)$, which corresponds to option (E).

why is P(1,0,0) and R(0,0,1)
it should be P(1,1,0) and R(0,1,1) point P and R arent on purely x and z axes right ?

so then PQ vector is = 1i + 1j
RQ vector is 1j + 1k

dot product of these two vectors is 0 +1 +0 = 1

|PQ vector| is \sqrt{2}
|RQ vector| is \sqrt{2}

cos(x) = 1/2
so x = 60

I hate this job

Assume the cube has side length 1, and vertices at $\((0,0,0),(1,0,0),(0,1,0),(0,0,1)\)$, etc.

Suppose $Q$ is at $\((0,0,0)\)$. Let $P$ be at $\((1,0,0)\)$ (an edge along x), and $R$ be at $\((0,1,0)\)$ (an edge along y). Then vectors $\(Q P=(1,0,0)\)$ and $\(Q R=(0,1,0)\)$, and the angle is $90^{\circ}$.

But if we take $P$ at $\((1,1,0)\)$ (face diagonal) and $R$ at $\((1,0,1)\)$ Then compute the angle.

$$
\(\begin{aligned}
& \overrightarrow{Q P}=(1,1,0) \\
& \overrightarrow{Q R}=(1,0,1)
\end{aligned}\)
$$


The dot product: $\(1 * 1+1 * 0+0 * 1=1\)$

Magnitudes: $\(|Q P|=\sqrt{1^2+1^2+0^2}=\sqrt{2}\)$, similarly $\(|Q R|=\sqrt{2}\)$

So $\(\cos \theta=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}\)$, so $\(\theta=60^{\circ}\)$.

This is a well-known result: the angle between face diagonals from the same vertex is $\(60^{\circ}\)$.

yup

i was flabbergasted for few seconds when you explained it using 3D vectors. its like bringing a nuke to street fight
mistakes happen. i now understand your thought process (sort off), it's right on point but on a rare occasions it overshoots a bit.
my thought process is highly unstable, it overshoots and undershoots.

I don't get it. Each side length of the triangle is equivalent to to sqrt(2)*length of cube. If all sides are equal in magnitude that implies angle is 60 degs each. I am confused about the nuance? Seems trivial.

but it is not a good question. when it gets so complicated it never is a good question

90 degress is wrong - please change this OA. it is 60 degress as others have mentioned. the triangle PQR exists in 3 dimensional space and is an equilateral triangle. there is no need for vectors.