Vector Algebra
Vector projection
Grade 12
Question:
<p>In the figure, <span>\(\overrightarrow{AE}\)</span> is the vector component of <span>\(\vec{q}\)</span> on <span>\(\vec{p}\)</span>. From <span>\(\triangle ABE\)</span>, we have <span>\(\overrightarrow{AB} + \overrightarrow{BE} = \overrightarrow{AE}\)</span>. If <span>\(\vec{q} + \vec{r} = \dfrac{(\vec{p}\cdot\vec{q})}{(\vec{p}\cdot\vec{q})}\vec{p}\)</span>, find <span>\(\vec{r}\)</span>:</p>
<p>\(\vec{r} = -\vec{q} + \dfrac{(\vec{p}\cdot\vec{q})}{(\vec{p}\cdot\vec{p})}\vec{p}\)</p>
<p>\(\vec{r} = \vec{q} - \dfrac{(\vec{p}\cdot\vec{q})}{(\vec{p}\cdot\vec{p})}\vec{p}\)</p>
<p>\(\vec{r} = \vec{q} + \dfrac{(\vec{p}\cdot\vec{q})}{(\vec{p}\cdot\vec{p})}\vec{p}\)</p>
<p>\(\vec{r} = -\vec{q} - \dfrac{(\vec{p}\cdot\vec{q})}{(\vec{p}\cdot\vec{p})}\vec{p}\)</p>
Step-by-Step Solution
Key Concept: The vector component of q along p is the projection formula: (q·p/p·p)p. Since q + r equals this projection, r is the difference between q and its projection on p, which represents the perpendicular component of q.
Step 1: Recognize that AE represents the vector component (projection) of q on p, given by: AE = (q·p)/(p·p) · p Step 2: From the given equation: q + r = (q·p)/(p·p) · p Step 3: Solve for r by rearranging: r = (q·p)/(p·p) · p - q Step 4: This can be rewritten as the component of q perpendicular to p: r = -[q - (q·p)/(p·p) · p] or equivalently: r = q - (q·p)/(p·p) · p Note: The vector r represents the perpendicular component of q with respect to p, satisfying r⊥p (i.e., r·p = 0). ∴ Answer: A
Correct Answer: A