A problem involving direction cosines (Vector Algebra)

[sahilmm15]

In the below figure how triangle OAP is right angled. I have imagined everything but I cannot imagine angle A as right angled. Thanks!

 

[jedishrfu]

Notice they show a rectangular prism and OA, OB, and OC are the edges of this prism.

Angles AOB, AOC, and BOC are all right angles as a given of the three unit vector OA, OB and OC all being perpendicular to one another.

AP is a diagonal of one of the faces and the OAP is a right angle since that face is parallel to the plane containing the angle BOC.

Try building a model using a cardboard box and define one corner on the base as O and the opposing corner on the top as P. By opposing I mean follow the vertical edge and then at the top of the box follow a diagonal to the opposite corner.

 

[Gaussian97]

Yes, in fact, notice that OA is parallel to the base vector $\vec x$, while AP is parallel to the plane defined by the vectors $\vec y$ and $\vec z$ (i.e. AP is a linear combination of $\vec y$ and $\vec z$).
So, if by assumption you are in a orthogonal base, then OA must be orthogonal to AP.

 

[Lnewqban]

In the below figure how triangle OAP is right angled. I have imagined everything but I cannot imagine angle A as right angled. Thanks!

Note that both points A and P are contained in the same vertical plane.
That vertical lane is perpendicular to axis X.
That vertical plane is also perpendicular to the horizontal plane containing points A, B and O, as well to the vertical plane containing points A, C and O.

For the above reasons, a horizontal projection of the triangle OAP on the horizontal plane ABO will show a right angle, just like a projection of the triangle on the vertical plane ACO will.