A triangle is constructed on the vectors vec a=5vec i+2vec j…

A triangle is constructed on the vectors $\vec a=5\vec i+2\vec j$ and $\vec b=2\vec i-4\vec j$ ($\vec i,\vec j$ orthonormal). Find the altitude vector $\vec h_a$, drawn to the side $\vec a$, and the angle it makes with the side $\vec b$.

Explanation of Method

The altitude vector of a triangle and its angle with a side.

Step 1: Setup

Vertices $O(0,0)$, $A(5,2)$, $B(2,-4)$, with $\overrightarrow{OA}=\vec a$ and $\overrightarrow{OB}=\vec b$. The side $\vec a$ lies opposite $B$, so $\vec h_a$ is the altitude from $B$ to the line $OA$.

Step 2: Altitude vector (Problem 61)

\[ \vec a\cdot\vec b=10-8=2,\qquad |\vec a|^2=29, \]

\[ \vec h_a=\vec b-\frac{\vec a\cdot\vec b}{|\vec a|^2}\vec a=(2,-4)-\frac{2}{29}(5,2)=\left(\frac{48}{29},-\frac{120}{29}\right). \]

\[\vec h_a=\frac{24}{29}(2\vec i-5\vec j)=\overrightarrow{FB},\qquad F=\frac{2}{29}(5,2)\]

Check: $\vec h_a\cdot\vec a=\tfrac{24}{29}(10-10)=0$ $\checkmark$

$O$ $A(5,2)$ $B(2,-4)$ $F$ $\vec a$ $\vec b$ $\vec h_a$

The altitude $\vec h_a=\overrightarrow{FB}$ from $B$ to the side $\vec a=\overrightarrow{OA}$. Its foot $F=\tfrac{2}{29}(5,2)$ is very close to $O$, because $\angle(\vec a,\vec b)\approx85.2^\circ$ is almost a right angle.

Step 3: Length (checked with the area)

\[ |\vec h_a|^2=\frac{24^2(4+25)}{29^2}=\frac{576}{29},\qquad |\vec h_a|=\frac{24}{\sqrt{29}}\approx4.46. \]

\[ S=\tfrac12|5\cdot(-4)-2\cdot2|=12=\tfrac12\,|\vec a|\,|\vec h_a|=\tfrac12\sqrt{29}\cdot\tfrac{24}{\sqrt{29}}.\ \checkmark \]

Step 4: Angle with $\vec b$

\[ \vec h_a\cdot\vec b=\frac{24}{29}(4+20)=\frac{576}{29},\qquad |\vec b|=2\sqrt5, \]

\[ \cos\theta=\frac{576/29}{\frac{24}{\sqrt{29}}\cdot2\sqrt5}=\frac{12}{\sqrt{145}}\approx0.9965. \]

\[\theta=\arccos\frac{12}{\sqrt{145}}\approx4.76^\circ\]

Check: $\cos\angle(\vec a,\vec b)=\dfrac{2}{\sqrt{29}\cdot2\sqrt5}=\dfrac1{\sqrt{145}}$, which gives $\angle(\vec a,\vec b)\approx85.24^\circ$. Since $\vec h_a\perp\vec a$, $\theta=90^\circ-85.24^\circ=4.76^\circ$ $\checkmark$

$85.2^\circ$ $\theta\approx4.76^\circ$ $O$ $\vec a$ $\vec b$ $\vec h_a$

The same vectors from a common origin: $\vec h_a\perp\vec a$, so $\theta=90^\circ-\angle(\vec a,\vec b)\approx4.76^\circ$.

Result

\[\boxed{\,\vec h_a=\frac{24}{29}(2\vec i-5\vec j)=\overrightarrow{FB},\qquad F=\frac{2}{29}(5,2)\,}\]

\[\boxed{\,\theta=\arccos\frac{12}{\sqrt{145}}\approx4.76^\circ\,}\]

Remark

Direction. The formula gives the vector from the foot $F$ to the vertex $B$. The source's figure draws the arrow from $B$ to $F$, which is $-\vec h_a$; here the arrow is drawn $F\to B$.