Find the projection of the vector a=2p-3q onto b=p+q , given…

Find the projection of the vector $\vec{a}=2\vec{p}-3\vec{q}$ onto $\vec{b}=\vec{p}+\vec{q}$, given $|\vec p|=2$, $|\vec q|=3$ and $\angle(\vec p,\vec q)=60^\circ$.

Explanation of Method

Projection of vector $\vec a$ onto vector $\vec b$.

Step 1: Formula

\[ \mathrm{Pr}_{\vec b}\vec a=\frac{\vec a\cdot\vec b}{|\vec b|}. \]

Step 2: Basic products

\[ \vec p\cdot\vec p=4,\qquad \vec q\cdot\vec q=9,\qquad \vec p\cdot\vec q=2\cdot3\cdot\cos60^\circ=3. \]

Step 3: The scalar product $\vec a\cdot\vec b$

\[ \vec a\cdot\vec b=(2\vec p-3\vec q)\cdot(\vec p+\vec q)=2\vec p^{\,2}-\vec p\cdot\vec q-3\vec q^{\,2}=8-3-27=-22. \]

Step 4: The length $|\vec b|$

\[ |\vec b|^2=\vec p^{\,2}+2\,\vec p\cdot\vec q+\vec q^{\,2}=4+6+9=19,\qquad |\vec b|=\sqrt{19}. \]

Step 5: Result

\[\mathrm{Pr}_{\vec b}\vec a=-\frac{22}{\sqrt{19}}=-\frac{22\sqrt{19}}{19}\approx-5.05\]

Check. $|\vec a|^2=4\cdot4-12\cdot3+9\cdot9=61$, so

\[ \cos\angle(\vec a,\vec b)=\frac{-22}{\sqrt{61}\sqrt{19}}\approx-0.646,\qquad \angle(\vec a,\vec b)\approx130.3^\circ, \] an obtuse angle — that is why the projection is negative.

$\approx130^\circ$ $O$ $\vec b$ $\vec a$ $\mathrm{Pr}_{\vec b}\vec a\approx-5.05$

$\vec a=2\vec p-3\vec q,\ |\vec a|=\sqrt{61}$
$\vec b=\vec p+\vec q,\ |\vec b|=\sqrt{19}$

Drawn to scale with $\vec b$ horizontal. The foot of the perpendicular from the tip of $\vec a$ lies on the opposite side of $O$: the projection is negative.

Result

\[\boxed{\,\mathrm{Pr}_{\vec b}\vec a=-\frac{22}{\sqrt{19}}=-\frac{22\sqrt{19}}{19}\approx-5.05\,}\]