StudentAidUnit 6: Vectors in Plane — Exercise 6 3
10th Class Mathematics · Unit 6: Vectors in Plane
6.3.1.Plot A(-5,4), B(-6,1) and C(-3,1) to form a triangle ABC. Also translate triangle ABC to triangle A'B'C' by translation vector 8i - 6j.
Given
Points
A(-5,4), B(-6,1), C(-3,1)
Formula
Translation rule
(x,y)to(x+8, y-6)
A'=(-5+8,4-6)=(3,-2); B'=(-6+8,1-6)=(2,-5); C'=(-3+8,1-6)=(5,-5)
Result
A'(3,-2), B'(2,-5), C'(5,-5)
6.3.2.Plot A(-6,-2), B(-6,-5), C(-3,-5) and D(-3,-2) to form a square ABCD. Also translate square ABCD to square A'B'C'D' by translation vector 9i + 7j.
Given
Points
A(-6,-2), B(-6,-5), C(-3,-5), D(-3,-2)
Formula
Translation rule
(x,y)to(x+9, y+7)
A'=(3,5); B'=(3,2); C'=(6,2); D'=(6,5)
Result
A'(3,5), B'(3,2), C'(6,2), D'(6,5)
6.3.3.Plot A(-6,3), B(-4,0), C(-2,3) and D(-4,4) to form a kite ABCD. Also translate kite ABCD to kite A'B'C'D' by translation vector 6i - 6j.
Given
Points
A(-6,3), B(-4,0), C(-2,3), D(-4,4)
Formula
Translation rule
(x,y)to(x+6, y-6)
A'=(0,-3); B'=(2,-6); C'=(4,-3); D'=(2,-2)
Result
A'(0,-3), B'(2,-6), C'(4,-3), D'(2,-2)
6.3.4.The coordinates of A, B and D are (1,2), (6,3) and (2,8) respectively. Find the coordinates of C by using vector method if ABCD is a parallelogram.
Formula
vec{AC}=vec{AB}+vec{AD}
vec{AB}=B-A=(5,1);quad vec{AD}=D-A=(1,6)
vec{AC}=(5,1)+(1,6)=(6,7)
Result
C=A+vec{AC}=(1+6, 2+7)=(7,9)
6.3.5.In parallelogram ABCD, the vectors representing two opposite sides are AB = 6i + 2j, DC = -6i - 2j. Show that the opposite sides are equal in magnitude and parallel.
Magnitude
|vec{AB}|=sqrt{36+4}=sqrt{40}=2sqrt{10};quad |vec{DC}|=sqrt{36+4}=2sqrt{10}
Parallel
vec{DC}=-6hat i-2hat j=-(6hat i+2hat j)=-vec{AB}
Result
|vec{AB}|=|vec{DC}|=2sqrt{10}; vec{AB}parallelvec{DC}
6.3.6.Points A(1,2), B(4,6) and C(7,2) form a triangle. Check whether triangle ABC is an isosceles by using vector magnitude.
vec{AB}=(3,4), |vec{AB}|=sqrt{9+16}=5
vec{AC}=(6,0), |vec{AC}|=sqrt{36}=6
vec{BC}=(3,-4), |vec{BC}|=sqrt{9+16}=5
Result
|vec{AB}|=|vec{BC}|=5ne|vec{AC}|=6 ⇒ Triangle ABC is isosceles.
6.3.7.Use vectors to show that PQRS is a parallelogram, where the points P, Q, R and S have coordinates (1,2), (5,2), (7,6) and (3,6) respectively.
vec{PQ}=(4,0);quad vec{RS}=(-4,0);quad vec{PQ}=-vec{RS} ⇒ vec{PQ}parallelvec{RS}
vec{QR}=(2,4);quad vec{PS}=(2,4);quad vec{QR}=vec{PS} ⇒ vec{QR}parallelvec{PS}
Result
Both pairs of opposite sides are parallel (and equal), so PQRS is a parallelogram.
6.3.8.Use vectors to show that triangle XYZ is an isosceles, where the points X, Y, Z have the coordinates (0,0), (2,0) and (1,3) respectively.
vec{XY}=(2,0), |vec{XY}|=2
vec{XZ}=(1,3), |vec{XZ}|=sqrt{10}
vec{YZ}=(-1,3), |vec{YZ}|=sqrt{10}
Result
|vec{XZ}|=|vec{YZ}|=sqrt{10} ⇒ Triangle XYZ is isosceles.
6.3.9.A ball is projected with an initial velocity vector v0 = 10i + 20j. The horizontal component is in the x-direction and gravity is g = 0i - 10j. Find the maximum height and horizontal range.
Given
ux=10 m/s, uy=20 m/s, g=10 m/s2 (downward)
Formula
Maximum height
vy2=uy2-2gh
Working
0=(20)2-2(10)h
20h=400 ⇒ h=20 m
Formula
Horizontal range
T=2uyg,quad R=ux × T
T=2(20)10=4 s;quad R=10 × 4=40 m
Result
Maximum height=20 m, Horizontal range=40 m
6.3.10.A car enters a loop with velocity vector v = 30j and exits with velocity v' = 30i. What is the change in velocity vector?
Given
vec v=0hat i+30hat j,quad vec v'=30hat i+0hat j
Δvec v=vec v'-vec v=(30hat i+0hat j)-(0hat i+30hat j)
Result
Δvec v=30hat i-30hat j
6.3.11.An aeroplane has airspeed vp = 20j and there is a crosswind vw = 50i. Find the resultant velocity and its magnitude.
Formula
vec v=vec vp+vec vw
Working
vec v=20hat j+50hat i=50hat i+20hat j
Magnitude
|vec v|=sqrt{(50)2+(20)2}=sqrt{2500+400}=sqrt{2900}=10sqrt{29}
Result
vec v=50hat i+20hat j,quad |vec v|=10sqrt{29} m/s