These notes are intended to accompany your copy of the handwritten notes.
Two vectors v and w are parallel if there exists a constant number c such that v = c * w. If you are given the components of v and w this condition can be written out as a vector equation. This equation is satisfied if and only if each component of the left-hand side is equal to the corresponding component on the right. So for example if the vectors are in 3-dimensional space, you will get three simultaneous equations in the single variable c. If the vectors are parallel, the three equations will all have the same solution for c. If they don't, then the vectors aren't parallel.
If we are given three points, say P, Q and R, in the plane, they form a triangle whose sides have lengths equal to the magnitudes of the vectors PQ, PR and QR. If the coordinates of P, Q and R are given, then we can write out the three vectors in terms of these coordinates, and we can then write down the magnitudes of the three. The triangle will be a right triangle if and only if the Pythagorean Theorem is satisfied, i.e., if and only if the square of the longest side is equal to the sum of the squares of the two shorter sides.
An independent test for a right triangle: If the dot product of two of the vectors is zero, the angle between them is 90 deg and we can again conclude that the triangle is a right triangle. The triangle is isosceles if two of the sides are equal, i.e., if the magnitudes of two of the vectors are equal. It is possible for a triangle to 'pass' both of these tests, in which case the triangle would be both right and isosceles, with angles 45 deg, 45 deg and 90 deg.
The test for orthogonality of two vectors is that their dot product be zero. The dot product of two given vectors will be a single number or algebraic expression. Setting this expression equal to zero will give us an equation which expresses condition that the two vectors are orthogonal.
To find a vector orthogonal to two given vectors, it's easy to use the cross product. Dividing the cross product by its magnitude we get a unit vector orthogonal to the two given vectors. The negative of this vector is also orthogonal to the two given vectors. Given three nonzero vectors, none parallel to any of the others, we can use this reasoning to find a vector perpendicular to the first two. If the third is also perpendicular to this vector, then it lies in the same plane as the first two.
To find a vector orthogonal to two given vectors, we can also use the dot product. Let the two given vectors be u and v, and let desired vector be u = <a, b, c> , where a, b and c are parameters to be evaluated. The condition of orthogonality tells us that u dot w = 0, and u dot v = 0. This gives us two equations in the three unknowns a, b and c. For example if v = <1, 2, -2> and w = <1, 1, -2> , then we get the two equations
a + 2 b - 2 c = 0 and
a + b - 2 c = 0.
These equations can be solved for any two of the parameters a, b and c, in terms of the third.
All vectors in a plane are perpendicular to any vector perpendicular to the plane. To determine whether 3 vectors, none of which are parallel to any of the others, are all in the same plane, first find a vector, which we will call N, perpendicular to two of the three. Then figure out whether the third vector is also perpendicular to N. To find a vector perpendicular to the first two you can find their cross product. The third vector will be perpendicular to the cross product N if its dot product with N is zero.
Given vectors v and w , you can find the value of s such that s v - w is orthogonal to v using the criteria that the dot product of orthogonal vectors is zero. The condition becomes v dot (s v - w) = 0. Plugging in the components of the two vectors we get a linear equation in the single variable s, which unless our two original vectors are parallel we can easily solve (if v and w are parallel our equation will be inconsistent and no solution for s will be possible).
The work done by a force F in the direction of the displacement vector `ds is equal to the product of the component of F in the direction of `ds, multiplied by the magnitude of `ds. The component of F in the direction of `ds is || F || cos(theta), where theta is the angle between F and `ds. Thus the work is || F || cos(theta) * || `ds || = || F || || `ds || cos(theta). This is equal to the dot product of F and `ds : W = F dot `ds The text doesn't use the notation `ds , but rather uses the displacement vector PQ between two points, so there the work is expressed by the product F dot `PQ.