Vectors, finding acute angle between VB and plane OVA on a square base pyramid

Question

A pyramid has a square base OABC and Vertex V . The position vector of A,B,C and V with reference to O are given by OA = $2i$, OB= $2i + 2j$ , OC= $2j$ and OV= $ i + j + 3k$ ,

state the acute angle between VB and the plane OVA .

I actually have trouble visualising this. Is there any diagrams that I can visualise actually what angle I am trying to find ?

I attempted some steps to help me attempt it the question

I find the vector that is perpendicular to the plane OVA using $OV \times AV$ and I got $-6j + 2k$, let’s call this vector $x$.

I find the angle between $x$ and $VB$ which is $124.9$ .

Since I cant visualise what angle I’m finding, I cannot carry on the question from here.

Answer 1

HINT

1. Find a vector orthogonal to plane OVA by $\vec n= \vec {VO}\times \vec {VA}$ (directed towards the side of B)
2. Find the angle $\alpha$ between $\vec n$ and $\vec {VB}$ by $\cos \alpha=\frac{\vec n \cdot \vec {VB}}{|\vec n|\,|\vec {VB}|}$
3. The desidered angle between $\vec {VB}$ and the plane is $\theta=\frac{\pi}2-\alpha$