I CAN'T SEEM TO GET THE ANSWER THAT IS CONSISTENT WITH MY UNDERSTANDING OF THE USE OF DOT AND CROSS PRODUCTS AND THE USE OF THE PARAMETRIC EQUATION OF THE LINE.
LOOK AT THIS PLEASE:
the parametric equation of the line is:
x = 2 + 3t
y = -4t
z = 5 + t
the plane is 4x + 5y - 2z = 18...
let P be a point NOT on line L that passes through points Q and R.
\vec{A} = QR
\vec{B} = QP
prove that distance from point P to anywhere on line L is
d = |\vec{A} x \vec{B}| divided by |\vec{A}|
so, I've tried doing the cross product after assigning variables for the A and B...
LOOK...determinants work because there are orthogonal relationships between two vectors in the xyz coordinate and the area vector that the parallelogram or the rectangle the two vectors can form.
what is the importance of a 90 degree angle in geometry? sin, cos, tan all work within a 90...
I just find it a bit fascinating that values can be interpreted in a geometric way. For Instance, the area vector is orthogonal to two vectors in the xy plane after we adjust these two vectors in the xy plane to be orthogonal by projecting one vector over the other...thus, we again form an...
what is the meaning of orthogonal relationships in addition to right angles in the xyz coordinate system?
for instance, if a 3 piece rocket separated in space in an orthogonal way...will there be any significance when compared to the 3 piece rocket that does not separate in an orthogonal...
Given:
\vec A \cdot \vec B = non zero
and
\theta does not equal 0
I can't seem to prove that Vector B minus the Projection of B onto A makes the orthogonal projection of B onto A.
Can you help?
\vec C \cdot \vec C = |\vec C|^2
|\vec C|^2 = |\vec A|^2 + |\vec B|^2 + 2|\vec A||\vec B|cos\theta
i'm not really sure how to put up the absolute value sign in proper laTeX notation. And, I'm stuck here...I don't know how these steps came about.