Vector Algebra
Projection and Components of Vectors
Grade None
Question:
<p>Let \(ABCD\) be a parallelogram such that \(\overrightarrow{AB} = \vec{q}\), \(\overrightarrow{AD} = \vec{p}\) 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</p>
<p>\(\vec{r} = 3\vec{q} - \dfrac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{q})}\vec{p}\)</p>
<p>\(\vec{r} = -\vec{q} + \left(\dfrac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right)\vec{p}\)</p>
<p>\(\vec{r} = \vec{q} - \left(\dfrac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right)\vec{p}\)</p>
<p>\(\vec{r} = -3\vec{q} + \dfrac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})}\vec{p}\)</p>
Step-by-Step Solution
Key Concept: The altitude from B to AD is perpendicular to AD, so we project AB onto the direction perpendicular to AD using the condition that $\vec{r} \cdot \vec{p} = 0$. The altitude vector can be expressed as $\vec{r} = \vec{q} - \frac{\vec{q} \cdot \vec{p}}{|\vec{p}|^2}\vec{p}$.
Step 1: The altitude from B to side AD is perpendicular to AD. If we decompose $\vec{AB} = \vec{q}$ into components parallel and perpendicular to $\vec{AD} = \vec{p}$: $\vec{q} = \text{(parallel component)} + \vec{r}$ Step 2: The parallel component of $\vec{q}$ along $\vec{p}$ is: $\frac{\vec{q} \cdot \vec{p}}{|\vec{p}|^2}\vec{p}$ Step 3: Therefore, the perpendicular component (altitude vector) is: $\vec{r} = \vec{q} - \frac{\vec{q} \cdot \vec{p}}{|\vec{p}|^2}\vec{p}$ Verification: Check that $\vec{r} \cdot \vec{p} = 0$: $\left(\vec{q} - \frac{\vec{q} \cdot \vec{p}}{|\vec{p}|^2}\vec{p}\right) \cdot \vec{p} = \vec{q} \cdot \vec{p} - \frac{\vec{q} \cdot \vec{p}}{|\vec{p}|^2}|\vec{p}|^2 = 0$ ✓ ∴ Answer: B
Correct Answer: B