Class 12 Vector Algebra · Case Study4 Markscross_product_applications
Class 12 Maths · Vector Algebra · Case Study #123
A surveyor models a roof supporting solar panels by its four corners, in metres: P₁=(6,8,4), P₂=(21,8,4), P₃=(21,16,10), P₄=(6,16,10), listed around its boundary. Let A=P₂−P₁ and B=P₄−P₁.
Part (1) 1 Marks
Find the components of both edge vectors A and B.
Answer: A=(15,0,0) m; B=(0,8,6) m.
- Use A=P₂−P₁ and B=P₄−P₁.
- A=(21−6,8−8,4−4)=(15,0,0); B=(6−6,16−8,10−4)=(0,8,6), in metres.
Part (2) 1 Marks
Find the magnitudes of A and B.
Answer: |A|=15 m; |B|=10 m.
- Use |v|=√(v₁²+v₂²+v₃²).
- |A|=√225=15 m and |B|=√(64+36)=10 m.
Part (3) 2 Marks
Find a nonzero normal vector N perpendicular to A,B and to the roof surface.
Answer: N=A×B=(0,−90,120); any nonzero scalar multiple is a normal.
- The roof is spanned by A=(15,0,0), B=(0,8,6); a normal is their cross product.
- Compute A×B=(0·6−0·8,0·0−15·6,15·8−0·0)=(0,−90,120). Its dot products with both edges are zero.
- Thus N=(0,−90,120) is a valid normal. Dividing by its norm 150 gives unit normals ±(0,−3/5,4/5).