Components of Vectors — Question 10

Question 10

Slanted axes do not make a right triangle

Two unit-length directions in the plane make a $60^\circ$ angle: one is east and one is $60^\circ$ north of east. A vector is constructed by following 4 m along each direction. Here “4 and 4 are coefficients of the two slanted directions, not Cartesian projections.

1. Use the perpendicular east–north axes to find the constructed vector components.

2. Find its magnitude and direction. Why is $\sqrt{4^2+4^2}$ incorrect?

3. Find the signed projection of the resultant onto each slanted direction and compare those projections with the construction coefficients.

Question 10 - Solution

Slanted axes do not make a right triangle

(a) Project each geometric leg.

$$\vec A=(4,0)+4(\cos60^\circ,\sin60^\circ)=\boxed{(6,2\sqrt3)\,\mathrm m}$$ The east-directed leg contributes $(4,0)$ m. The slanted leg has horizontal projection $4\cos60^\circ=2$ m and vertical projection $4\sin60^\circ=2\sqrt3$ m. The endpoint is therefore $(6,2\sqrt3)$ m.

(b) Apply Pythagoras only to perpendicular projections.

$$A=\sqrt{36+12}=4\sqrt3\,\mathrm m,\quad\theta=30^\circ,\quad A^2=4^2+4^2+2(4)(4)\cos60^\circ$$ $A=\sqrt{36+12}=4\sqrt3\approx6.928$ m and $\theta=\tan^{-1}(2\sqrt3/6)=30^\circ$. The original two construction directions are not perpendicular. Their coefficients do not form the legs of a right triangle, so squaring and adding those coefficients is unjustified.

(c) Project the complete arrow.

$$A\cos30^\circ=4\sqrt3\frac{\sqrt3}{2}=6\,\mathrm m\ne4\,\mathrm m$$ The resultant is $30^\circ$ from either slanted direction. Each projection is $A\cos30^\circ=6$ m, not 4 m. In an orthonormal coordinate system coefficients and signed projections coincide; this oblique construction illustrates why the perpendicular-axis assumption matters.