Vector Algebra
Perpendicular Projection
Grade 12
Question:
<p>Let ABCD be a parallelogram such that <strong>AB</strong> = <strong>q</strong>, <strong>AD</strong> = <strong>p</strong> and ∠BAD be an acute angle. If <strong>r</strong> is the vector that coincides with the altitude directed from the vertex B to the side AD, then <strong>r</strong> is given by</p>
<p>(a) <strong>r</strong> = <strong>q</strong> + \(\frac{3(\mathbf{p} \cdot \mathbf{q})}{\mathbf{p} \cdot \mathbf{p}}\)<strong>p</strong></p>
<p>(b) <strong>r</strong> = −<strong>q</strong> + \(\frac{\mathbf{p} \cdot \mathbf{q}}{\mathbf{p} \cdot \mathbf{p}}\)<strong>p</strong></p>
<p>(c) <strong>r</strong> = <strong>q</strong> − \(\frac{3(\mathbf{p} \cdot \mathbf{q})}{\mathbf{p} \cdot \mathbf{p}}\)<strong>p</strong></p>
<p>(d) Option D not clearly shown</p>
Step-by-Step Solution
Key Concept: The altitude vector from B to AD is the component of AB perpendicular to AD. We decompose AB into components parallel and perpendicular to AD, where the perpendicular component is the altitude vector r.
Step 1: Set up the problem using position vectors. Let A be the origin. Then:
• Position vector of B: q (since AB = q )
• Direction along AD: p (since AD = p )
• The altitude from B is perpendicular to AD Step 2: Decompose AB into parallel and perpendicular components. We can write:
AB = (component along AD) + (component perpendicular to AD)
q = (projection of q onto p ) + r Step 3: Calculate the projection of q onto p. The projection of q along p is:
projection = $\frac{\mathbf{p} \cdot \mathbf{q}}{\mathbf{p} \cdot \mathbf{p}}\mathbf{p}$ Step 4: Find the altitude vector r. Since r is the perpendicular component:
r = q − (projection of q onto p )
r = q − $\frac{\mathbf{p} \cdot \mathbf{q}}{\mathbf{p} \cdot \mathbf{p}}\mathbf{p}$
r = − q + $\frac{\mathbf{p} \cdot \mathbf{q}}{\mathbf{p} \cdot \mathbf{p}}\mathbf{p}$ Step 5: Verify the result. Check that r ⊥ p :
r · p = (− q + $\frac{\mathbf{p} \cdot \mathbf{q}}{\mathbf{p} \cdot \mathbf{p}}\mathbf{p}$) · p
= − q · p + $\frac{\mathbf{p} \cdot \mathbf{q}}{\mathbf{p} \cdot \mathbf{p}}$( p · p )
= − p · q + p · q = 0 ✓ ∴ Answer: B
Correct Answer: B