AIEEE 2012. Previous Year Question.
Problem
Let $ABCD$ be a parallelogram such that $\vec{AB}=\vec{q}$, $\vec{AD}=\vec{p}$, and $\angle BAD$ be an acute angle. If $\vec{r}$ is the vector that coincides with the altitude directed from the vertex $B$ to the side $AD$, then $\vec{r}$ is given by:
Key insight. The foot of the altitude from $B$ onto line $AD$ is exactly the projection of $\vec{AB}$ onto $\vec{AD}$ — a standard vector formula. The altitude vector itself, directed from $B$ to that foot, is then just this projection minus $\vec{AB}$.
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Approach
Place $A$ at the origin, so $B$ and $D$ have position vectors $\vec{q}$ and $\vec{p}$ respectively. Find $F$, the foot of the perpendicular from $B$ to line $AD$, using the standard vector projection formula. The required vector $\vec{r}=\vec{BF}$ is then $F-B$, expressed in terms of $\vec{p}$ and $\vec{q}$.
Solution
Step 1 — Set up position vectors with A as the origin
$$\vec{AB} = \vec{q}, \qquad \vec{AD} = \vec{p}$$
Step 2 — Find the foot of the perpendicular F on line AD
$F$ is the projection of $B$ onto the line through $A$ in the direction $\vec{p}$:
$$\vec{AF} = \left(\frac{\vec{AB}\cdot\vec{AD}}{\vec{AD}\cdot\vec{AD}}\right)\vec{AD} = \left(\frac{\vec{q}\cdot\vec{p}}{\vec{p}\cdot\vec{p}}\right)\vec{p}$$
Step 3 — Find the altitude vector BF
The altitude directed from $B$ to $F$ is:
$$\vec{r} = \vec{BF} = \vec{AF}-\vec{AB} = \left(\frac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}\right)\vec{p} – \vec{q}$$
Step 4 — Rewrite in the given form
$$\vec{r} = -\vec{q} + \left(\frac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}\right)\vec{p}$$
Answer
$$\vec{r} = -\vec{q} + \left(\frac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}\right)\vec{p}$$
Common mistakes
- Computing $\vec{FB}$ instead of $\vec{BF}$. The problem specifically asks for the altitude “directed from the vertex $B$” — meaning the vector starts at $B$ and points toward $AD$, i.e. $\vec{BF}=F-B$, not the reverse $B-F$, which would flip every sign in the final answer.
- Forgetting why $\angle BAD$ being acute matters. This condition ensures the foot of the perpendicular $F$ actually lies on the segment $AD$ itself (not on its extension) — a detail confirming the geometric picture is as described, though it doesn’t change the algebra of the projection formula itself.
Practise next
- In a parallelogram $PQRS$ with $\vec{PQ}=\vec{u}$, $\vec{PS}=\vec{v}$, and acute $\angle QPS$, find the vector for the altitude directed from vertex $Q$ to side $PS$, using the same projection method.
Show answer
$\dfrac{\vec u\cdot\vec v}{|\vec v|^2}\,\vec v-\vec u$.
The foot of the perpendicular from $Q$ onto the line $PS$ is the projection of $\vec{PQ}=\vec u$ onto $\vec v$, namely $P+\dfrac{\vec u\cdot\vec v}{|\vec v|^2}\vec v$.
The altitude runs from $Q$ to that foot, so it is the foot minus $Q$:
$$\left(P+\frac{\vec u\cdot\vec v}{|\vec v|^2}\vec v\right)-(P+\vec u)=\frac{\vec u\cdot\vec v}{|\vec v|^2}\vec v-\vec u.$$
The angle being acute makes $\vec u\cdot\vec v>0$, so the foot lies on the side $PS$ itself rather than on its extension.
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